How can students understand matrices easily in Class 12 Maths?

NCERT Solutions for Class 12 Maths Chapter 3 – Matrices provides step-by-step solutions that help students understand matrix operations, properties, and problem solving techniques effectively.

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NCERT Solutions for Class 12 Maths Chapter 3 Matrices

NCERT – Exercise 3.1

In the matrix $A = \left[ {\begin{array}{*{20}{c}}2&5&{19}&{ – 7}\\{35}&{ – 2}&{5/2}&{12}\\{\sqrt 3 }&1&{ – 5}&{17}\end{array}} \right],$ write:
1. The order of the matrix,
2. The number of elements,
3. Write the elements ${a_{13}},{a_{21}},{a_{33}},{a_{24}},{a_{23}}.$
SOLUTION
1. The matrix A has 3 rows and 4 columns. Thus, order of the matrix A is $3 \times 4$.
2. There are $3 \times 4 = 12$ elements in the matrix A.
3. ${a_{13}} = 19, {a_{21}} = 35, {a_{33}} = – 5, {a_{24}} = 12, {a_{23}} = \cfrac{5}{2}.$
If a matrix has 24 elements, what are the possible orders it can have? What if, it has 13 elements?

SOLUTION: We know that a matrix of order $m \times n$, has $mn$ elements. Thus, all possible ordered pairs are $(1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), (6, 4)$. The matrix with 13 elements has possible order $1 \times 13$ and $13 \times 1$.
If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?

SOLUTION: We know that a matrix of order $m \times n$ has $mn$ elements. Then the possible orders are $1 \times 18, 18 \times 1, 2 \times 9, 9 \times 2, 3 \times 6, 6 \times 3$. If a matrix has 5 elements, then possible orders are $1 \times 5$ and $5 \times 1$.
Construct a $2 \times 2$ matrix, $A = [{a_{ij}}]$, whose elements are given by:
1. ${a_{ij}} = \cfrac{{{{(i + j)}^2}}}{2}$
2. ${a_{ij}} = \cfrac{i}{j}$
3. ${a_{ij}} = \cfrac{{{{(i + 2j)}^2}}}{2}$
SOLUTION
1. A $2 \times 2$ matrix has 2 rows and 2 columns. So, it is given by $A = {[{a_{ij}}]_{2 \times 2}} = \left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}\\{{a_{21}}}&{{a_{22}}}\end{array}} \right]$
  ${a_{11}} = \cfrac{{{{(1 + 1)}^2}}}{2} = 2, {a_{12}} = \cfrac{{{{(1 + 2)}^2}}}{2} = \cfrac{9}{2}$,
  
  ${a_{21}} = \cfrac{{{{(2 + 1)}^2}}}{2} = \cfrac{9}{2}, {a_{22}} = \cfrac{{{{(2 + 2)}^2}}}{2} = 8$
  
  $\therefore A = \left[ {\begin{array}{*{20}{c}}2&{\cfrac{9}{2}}\\{\cfrac{9}{2}}&8\end{array}} \right]$
2. For ${a_{ij}} = \cfrac{i}{j},$ we have
  ${a_{11}} = 1, {a_{12}} = \cfrac{1}{2}, {a_{21}} = 2, {a_{22}} = 1$
  
  $\therefore A = \left[ {\begin{array}{*{20}{c}}1&{\cfrac{1}{2}}\\2&1\end{array}} \right]$
3. For ${a_{ij}} = \cfrac{{{{(i + 2j)}^2}}}{2},$ we have
  ${a_{11}} = \cfrac{9}{2}, {a_{12}} = \cfrac{{25}}{2}, {a_{21}} = 8, {a_{22}} = 18$
  
  $\therefore A = \left[ {\begin{array}{*{20}{c}}{\cfrac{9}{2}}&{\cfrac{{25}}{2}}\\8&{18}\end{array}} \right]$
Construct a $3 \times 4$ matrix, whose elements are given by:
1. ${a_{ij}} = \cfrac{1}{2}| – 3i + j|$
2. ${a_{ij}} = 2i – j$
SOLUTION: A $3 \times 4$ matrix has 3 rows and 4 columns. In general, $3 \times 4$ matrix is given by $$A = {[{a_{ij}}]_{3 \times 4}} = \left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}&{{a_{14}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}&{{a_{24}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}&{{a_{34}}}\end{array}} \right]$$
1. For ${a_{ij}} = \cfrac{1}{2}| – 3i + j|,$ we have:
  ${a_{11}} = 1, {a_{12}} = \cfrac{1}{2}, {a_{13}} = 0, {a_{14}} = \cfrac{1}{2}$
  
  ${a_{21}} = \cfrac{5}{2}, {a_{22}} = 2, {a_{23}} = \cfrac{3}{2}, {a_{24}} = 1$
  
  ${a_{31}} = 4, {a_{32}} = \cfrac{7}{2}, {a_{33}} = 3, {a_{34}} = \cfrac{5}{2}.$
  
  $\therefore A = \left[ {\begin{array}{*{20}{c}}1&{\cfrac{1}{2}}&0&{\cfrac{1}{2}}\\{\cfrac{5}{2}}&2&{\cfrac{3}{2}}&1\\4&{\cfrac{7}{2}}&3&{\cfrac{5}{2}}\end{array}} \right]$
2. For ${a_{ij}} = 2i – j,$ we have:
  ${a_{11}} = 1, {a_{12}} = 0, {a_{13}} = – 1, {a_{14}} = – 2$
  
  ${a_{21}} = 3, {a_{22}} = 2, {a_{23}} = 1, {a_{24}} = 0$
  
  ${a_{31}} = 5, {a_{32}} = 4, {a_{33}} = 3, {a_{34}} = 2$
  
  Hence, $A = \left[ {\begin{array}{*{20}{c}}1&0&{ – 1}&{ – 2}\\3&2&1&0\\5&4&3&2\end{array}} \right]$
Find the values of $x, y$ and $z$ from the following equations:
1. $\left[ {\begin{array}{*{20}{c}}4&3\\x&5\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}y&z\\1&5\end{array}} \right]$
2. $\left[ {\begin{array}{*{20}{c}}{x + y}&2\\{5 + z}&{xy}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}6&2\\5&8\end{array}} \right]$
3. $\left[ {\begin{array}{*{20}{c}}{x + y + z}\\{x + z}\\{y + z}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}9\\5\\7\end{array}} \right]$
SOLUTION
1. Since the corresponding elements of equal matrices are equal, we have $x = 1, y = 4, z = 3.$
2. From the equality: $5 + z = 5 \Rightarrow z = 0$. Also, $x + y = 6 \Rightarrow y = 6 – x$ and $xy = 8$. Solving, we get $x(6 – x) = 8 \Rightarrow {x^2} – 6x + 8 = 0 \Rightarrow (x – 4)(x – 2) = 0$.
  Hence, $x = 2, y = 4, z = 0$ OR $x = 4, y = 2, z = 0$.
3. We have: $x + y + z = 9$ (i), $x + z = 5$ (ii), $y + z = 7$ (iii).
  From (i) and (ii): $y + 5 = 9 \Rightarrow y = 4$. From (i) and (iii): $x + 7 = 9 \Rightarrow x = 2$. From (ii): $2 + z = 5 \Rightarrow z = 3$.
  
  Hence, $x = 2, y = 4, z = 3$.
Find the values of $a, b, c$ and $d$ from the equation:

$$\left[ {\begin{array}{*{20}{c}}{a – b}&{2a + c}\\{2a – b}&{3c + d}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ – 1}&5\\0&{13}\end{array}} \right]$$

SOLUTION: From the given matrix, we have $a – b = – 1$ and $2a – b = 0$. Solving these, we get $a = 1$ and $b = 2$. Similarly, $2a + c = 5 \Rightarrow 2(1) + c = 5 \Rightarrow c = 3$. Finally, $3c + d = 13 \Rightarrow 3(3) + d = 13 \Rightarrow d = 4$.

Hence, $a = 1, b = 2, c = 3, d = 4$.
$A = {[{a_{ij}}]_{m \times n}}$ is a square matrix, if:
1. $m < n$
2. $m > n$
3. $m = n$
4. None of these
SOLUTION: (c) For a square matrix, we have $m = n$.
Which of the given values of $x$ and $y$ make the following pair of matrices equal?

$$\left[ {\begin{array}{*{20}{c}}{3x + 7}&5\\{y + 1}&{2 – 3x}\end{array}} \right], \left[ {\begin{array}{*{20}{c}}0&{y – 2}\\8&4\end{array}} \right]$$
1. $x = \cfrac{{ – 1}}{3}, y = 7$
2. Not possible to find
3. $y = 7, x = \cfrac{{ – 2}}{3}$
4. $x = \cfrac{{ – 1}}{3}, y = \cfrac{{ – 2}}{3}$
SOLUTION: (b) Equating elements: $3x + 7 = 0 \Rightarrow x = -7/3$. However, $2 – 3x = 4 \Rightarrow -3x = 2 \Rightarrow x = -2/3$. Since $x$ cannot have two different values simultaneously, it is not possible to find the values.
The number of all possible matrices of order $3 \times 3$ with each entry 0 or 1 is:
1. 27
2. 18
3. 81
4. 512
SOLUTION: (d) The matrix has $3 \times 3 = 9$ elements. Each element has 2 choices (0 or 1). Total possible matrices = $2^9 = 512$.

NCERT – Exercise 3.2

Let $ A = \left[ {\begin{array}{*{20}{c}}2&4\\3&2\end{array}} \right], B = \left[ {\begin{array}{*{20}{c}}1&3\\-2&5\end{array}} \right], C = \left[ {\begin{array}{*{20}{c}}-2&5\\3&4\end{array}} \right] $. Find each of the following:
1. $ A + B $
2. $ A – B $
3. $ 3A – C $
4. $ AB $
5. $ BA $
Solution:
1. \[ A + B = \left[ {\begin{array}{*{20}{c}}2&4\\3&2\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}1&3\\-2&5\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3&7\\1&7\end{array}} \right] \]
2. \[ A – B = \left[ {\begin{array}{*{20}{c}}2&4\\3&2\end{array}} \right] – \left[ {\begin{array}{*{20}{c}}1&3\\-2&5\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&1\\5&-3\end{array}} \right] \]
3. \[ 3A – C = 3\left[ {\begin{array}{*{20}{c}}2&4\\3&2\end{array}} \right] – \left[ {\begin{array}{*{20}{c}}-2&5\\3&4\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}8&7\\6&2\end{array}} \right] \]
4. \[ AB = \left[ {\begin{array}{*{20}{c}}2&4\\3&2\end{array}} \right] \left[ {\begin{array}{*{20}{c}}1&3\\-2&5\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}-6&26\\-1&19\end{array}} \right] \]
5. \[ BA = \left[ {\begin{array}{*{20}{c}}1&3\\-2&5\end{array}} \right] \left[ {\begin{array}{*{20}{c}}2&4\\3&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}11&10\\11&2\end{array}} \right] \]
Compute the following:
1. $ \left[ {\begin{array}{*{20}{c}}a&b\\-b&a\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}a&b\\b&a\end{array}} \right] $
2. $ \left[ {\begin{array}{*{20}{c}}a^2 + b^2&b^2 + c^2\\a^2 + c^2&a^2 + b^2\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}2ab&2bc\\-2ac&-2ab\end{array}} \right] $
3. $ \left[ {\begin{array}{*{20}{c}}-1&4&-6\\8&5&16\\2&8&5\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}12&7&6\\8&0&5\\3&2&4\end{array}} \right] $
4. $ \left[ {\begin{array}{*{20}{c}}\cos^2 x&\sin^2 x\\\sin^2 x&\cos^2 x\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}\sin^2 x&\cos^2 x\\\cos^2 x&\sin^2 x\end{array}} \right] $
Solution:
1. \[ \left[ {\begin{array}{*{20}{c}}a&b\\-b&a\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}a&b\\b&a\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2a&2b\\0&2a\end{array}} \right] \]
2. \[ \left[ {\begin{array}{*{20}{c}}a^2 + b^2&b^2 + c^2\\a^2 + c^2&a^2 + b^2\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}2ab&2bc\\-2ac&-2ab\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}(a + b)^2&(b + c)^2\\(a – c)^2&(a – b)^2\end{array}} \right] \]
3. \[ \left[ {\begin{array}{*{20}{c}}-1&4&-6\\8&5&16\\2&8&5\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}12&7&6\\8&0&5\\3&2&4\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}11&11&0\\16&5&21\\5&10&9\end{array}} \right] \]
4. \[ \left[ {\begin{array}{*{20}{c}}\cos^2 x&\sin^2 x\\\sin^2 x&\cos^2 x\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}\sin^2 x&\cos^2 x\\\cos^2 x&\sin^2 x\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&1\\1&1\end{array}} \right] \]
Compute the following products:
1. $ \left[ {\begin{array}{*{20}{c}}a&b\\-b&a\end{array}} \right] \left[ {\begin{array}{*{20}{c}}a&-b\\b&a\end{array}} \right] $
2. $ \left[ {\begin{array}{*{20}{c}}1\\2\\3\end{array}} \right] \left[ {\begin{array}{*{20}{c}}2&3&4\end{array}} \right] $
3. $ \left[ {\begin{array}{*{20}{c}}1&-2\\2&3\end{array}} \right] \left[ {\begin{array}{*{20}{c}}1&2&3\\2&3&1\end{array}} \right] $
4. $ \left[ {\begin{array}{*{20}{c}}2&3&4\\3&4&5\\4&5&6\end{array}} \right] \left[ {\begin{array}{*{20}{c}}1&-3&5\\0&2&4\\3&0&5\end{array}} \right] $
5. $ \left[ {\begin{array}{*{20}{c}}2&1\\3&2\\-1&1\end{array}} \right] \left[ {\begin{array}{*{20}{c}}1&0&1\\-1&2&1\end{array}} \right] $
6. $ \left[ {\begin{array}{*{20}{c}}3&-1&3\\-1&0&2\end{array}} \right] \left[ {\begin{array}{*{20}{c}}2&-3\\1&0\\3&1\end{array}} \right] $
Solution:
1. \[ \left[ {\begin{array}{*{20}{c}}a&b\\-b&a\end{array}} \right] \left[ {\begin{array}{*{20}{c}}a&-b\\b&a\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}a^2 + b^2&0\\0&a^2 + b^2\end{array}} \right] \]
2. \[ \left[ {\begin{array}{*{20}{c}}1\\2\\3\end{array}} \right] \left[ {\begin{array}{*{20}{c}}2&3&4\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2&3&4\\4&6&8\\6&9&12\end{array}} \right] \]
3. \[ \left[ {\begin{array}{*{20}{c}}1&-2\\2&3\end{array}} \right] \left[ {\begin{array}{*{20}{c}}1&2&3\\2&3&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}-3&-4&1\\8&13&9\end{array}} \right] \]
4. \[ \left[ {\begin{array}{*{20}{c}}2&3&4\\3&4&5\\4&5&6\end{array}} \right] \left[ {\begin{array}{*{20}{c}}1&-3&5\\0&2&4\\3&0&5\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}14&0&42\\18&-1&56\\22&-2&70\end{array}} \right] \]
5. \[ \left[ {\begin{array}{*{20}{c}}2&1\\3&2\\-1&1\end{array}} \right] \left[ {\begin{array}{*{20}{c}}1&0&1\\-1&2&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&2&3\\1&4&5\\-2&2&0\end{array}} \right] \]
6. \[ \left[ {\begin{array}{*{20}{c}}3&-1&3\\-1&0&2\end{array}} \right] \left[ {\begin{array}{*{20}{c}}2&-3\\1&0\\3&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}14&-6\\4&5\end{array}} \right] \]
If $ A = \left[ {\begin{array}{*{20}{c}}1&2&-3\\5&0&2\\1&-1&1\end{array}} \right], B = \left[ {\begin{array}{*{20}{c}}3&-1&2\\4&2&5\\2&0&3\end{array}} \right] $, and $ C = \left[ {\begin{array}{*{20}{c}}4&1&2\\0&3&2\\1&-2&3\end{array}} \right] $, then compute $ (A + B) $ and $ (B – C) $. Also, verify that $ A + (B – C) = (A + B) – C $.

Solution:

\[ A + B = \left[ {\begin{array}{*{20}{c}}4&1&-1\\9&2&7\\3&-1&4\end{array}} \right] \]

\[ B – C = \left[ {\begin{array}{*{20}{c}}-1&-2&0\\4&-1&3\\1&2&0\end{array}} \right] \]

\[ A + (B – C) = \left[ {\begin{array}{*{20}{c}}0&0&-3\\9&-1&5\\2&1&1\end{array}} \right] \]

\[ (A + B) – C = \left[ {\begin{array}{*{20}{c}}0&0&-3\\9&-1&5\\2&1&1\end{array}} \right] \]

Hence, $ A + (B – C) = (A + B) – C $.
If $ A + \left[ {\begin{array}{*{20}{c}}\frac{2}{3}&1&\frac{5}{3}\\\frac{1}{3}&\frac{2}{3}&\frac{4}{3}\\\frac{7}{3}&2&\frac{2}{3}\end{array}} \right] $ and $ B = \left[ {\begin{array}{*{20}{c}}\frac{2}{5}&\frac{3}{5}&1\\\frac{1}{5}&\frac{2}{5}&\frac{4}{5}\\\frac{7}{5}&\frac{6}{5}&\frac{2}{5}\end{array}} \right] $, then compute $ 3A – 5B $.

Solution:

\[ 3A – 5B = \left[ {\begin{array}{*{20}{c}}0&0&0\\0&0&0\\0&0&0\end{array}} \right] \]
Simplify $ \cos \theta \left[ {\begin{array}{*{20}{c}}\cos \theta&\sin \theta\\-\sin \theta&\cos \theta\end{array}} \right] + \sin \theta \left[ {\begin{array}{*{20}{c}}\sin \theta&-\cos \theta\\\cos \theta&\sin \theta\end{array}} \right] $.

Solution:

\[ \cos \theta \left[ {\begin{array}{*{20}{c}}\cos \theta&\sin \theta\\-\sin \theta&\cos \theta\end{array}} \right] + \sin \theta \left[ {\begin{array}{*{20}{c}}\sin \theta&-\cos \theta\\\cos \theta&\sin \theta\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] \]
Find $ X $ and $ Y $, if:
1. $ X + Y = \left[ {\begin{array}{*{20}{c}}7&0\\2&5\end{array}} \right] $ and $ X – Y = \left[ {\begin{array}{*{20}{c}}3&0\\0&3\end{array}} \right] $
2. $ 2X + 3Y = \left[ {\begin{array}{*{20}{c}}2&3\\4&0\end{array}} \right] $ and $ 3X + 2Y = \left[ {\begin{array}{*{20}{c}}2&-2\\-1&5\end{array}} \right] $
Solution:
1. \[ X = \left[ {\begin{array}{*{20}{c}}5&0\\1&4\end{array}} \right], \quad Y = \left[ {\begin{array}{*{20}{c}}2&0\\1&1\end{array}} \right] \]
2. \[ X = \left[ {\begin{array}{*{20}{c}}\frac{2}{5}&\frac{-12}{5}\\\frac{-11}{5}&3\end{array}} \right], \quad Y = \left[ {\begin{array}{*{20}{c}}\frac{2}{5}&\frac{13}{5}\\\frac{14}{5}&-2\end{array}} \right] \]
Find $ X $, if $ Y = \left[ {\begin{array}{*{20}{c}}3&2\\1&4\end{array}} \right] $ and $ 2X + Y = \left[ {\begin{array}{*{20}{c}}1&0\\-3&2\end{array}} \right] $.

Solution:

\[ X = \left[ {\begin{array}{*{20}{c}}-1&-1\\-2&-1\end{array}} \right] \]
Find $ x $ and $ y $, if $ 2\left[ {\begin{array}{*{20}{c}}1&3\\0&x\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}y&0\\1&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}5&6\\1&8\end{array}} \right] $.

Solution:

$ x = 3 $ and $ y = 3 $.
Solve the equation for $ x, y, z $, and $ t $, if:

\[ 2\left[ {\begin{array}{*{20}{c}}x&z\\y&t\end{array}} \right] + 3\left[ {\begin{array}{*{20}{c}}1&-1\\0&2\end{array}} \right] = 3\left[ {\begin{array}{*{20}{c}}3&5\\4&6\end{array}} \right] \]

Solution:

$ x = 3, y = 6, z = 9 $, and $ t = 6 $.
If $ x\left[ {\begin{array}{*{20}{c}}2\\3\end{array}} \right] + y\left[ {\begin{array}{*{20}{c}}-1\\1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}10\\5\end{array}} \right] $, find the values of $ x $ and $ y $.

Solution:

$ x = 3 $ and $ y = -4 $.
Given $ 3\left[ {\begin{array}{*{20}{c}}x&y\\z&w\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}x&6\\-1&2w\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}4&x+y\\z+w&3\end{array}} \right] $, find the values of $ x, y, z $, and $ w $.

Solution:

$ x = 2, y = 4, z = 1 $, and $ w = 3 $.
If $ F(x) = \left[ {\begin{array}{*{20}{c}}\cos x&-\sin x&0\\\sin x&\cos x&0\\0&0&1\end{array}} \right] $, then show that $ F(x) \cdot F(y) = F(x + y) $.

Solution:

Verified to be true.
Show that:
1. $ \left[ {\begin{array}{*{20}{c}}5&-1\\6&7\end{array}} \right] \left[ {\begin{array}{*{20}{c}}2&1\\3&4\end{array}} \right] \neq \left[ {\begin{array}{*{20}{c}}2&1\\3&4\end{array}} \right] \left[ {\begin{array}{*{20}{c}}5&-1\\6&7\end{array}} \right] $
2. $ \left[ {\begin{array}{*{20}{c}}1&2&3\\0&1&0\\1&1&0\end{array}} \right] \left[ {\begin{array}{*{20}{c}}-1&1&0\\0&-1&1\\2&3&4\end{array}} \right] \neq \left[ {\begin{array}{*{20}{c}}-1&1&0\\0&-1&1\\2&3&4\end{array}} \right] \left[ {\begin{array}{*{20}{c}}1&2&3\\0&1&0\\1&1&0\end{array}} \right] $
Solution:

Both inequalities are verified to be true.
Find $ A^2 – 5A + 6I $, if $ A = \left[ {\begin{array}{*{20}{c}}2&0&1\\2&1&3\\1&-1&0\end{array}} \right] $.

Solution:

\[ A^2 – 5A + 6I = \left[ {\begin{array}{*{20}{c}}1&-1&-3\\-1&-1&-10\\-5&4&4\end{array}} \right] \]
If $ A = \left[ {\begin{array}{*{20}{c}}1&0&2\\0&2&1\\2&0&3\end{array}} \right] $, prove that $ A^3 – 6A^2 + 7A + 2I = O $.

Solution:

Verified to be true.
If $ A = \left[ {\begin{array}{*{20}{c}}3&-2\\4&-2\end{array}} \right] $ and $ I = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] $, find $ k $ so that $ A^2 = kA – 2I $.

Solution:

$ k = 1 $.
If $ A = \left[ {\begin{array}{*{20}{c}}0&-\tan \frac{\alpha}{2}\\\tan \frac{\alpha}{2}&0\end{array}} \right] $ and $ I $ is the identity matrix of order 2, then show that $ I + A = (I – A) \left[ {\begin{array}{*{20}{c}}\cos \alpha&-\sin \alpha\\\sin \alpha&\cos \alpha\end{array}} \right] $.

Solution:

Verified to be true.
A trust fund has Rs. 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs. 30,000 among the two types of bonds, if the trust fund obtains an annual total interest of:
1. Rs. 1800
2. Rs. 2000
Solution:
1. Invest Rs. 15,000 at 5% and Rs. 15,000 at 7%.
2. Invest Rs. 5,000 at 5% and Rs. 25,000 at 7%.
The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, and 10 dozen economics books. Their selling prices are Rs. 80, Rs. 60, and Rs. 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Solution:

Total amount received = Rs. 20,160.
Choose the correct answer:
1. The restriction on $ n, k $, and $ p $ so that $ PY + WY $ will be defined are:
2. If $ n = p $, then the order of the matrix $ 7X – 5Z $ is:

NCERT – Exercise 3.3

Find the transpose of each of the following matrices:
1. \[ \left[ {\begin{array}{*{20}{c}}5\\\frac{1}{2}\\-1\end{array}} \right] \]
2. \[ \left[ {\begin{array}{*{20}{c}}1&-1\\2&3\end{array}} \right] \]
3. \[ \left[ {\begin{array}{*{20}{c}}-1&5&6\\\sqrt{3}&5&6\\2&3&-1\end{array}} \right] \]
Solution:
1. \[ \left[ {\begin{array}{*{20}{c}}5&\frac{1}{2}&-1\end{array}} \right] \]
2. \[ \left[ {\begin{array}{*{20}{c}}1&2\\-1&3\end{array}} \right] \]
3. \[ \left[ {\begin{array}{*{20}{c}}-1&\sqrt{3}&2\\5&5&3\\6&6&-1\end{array}} \right] \]
If $ A = \left[ {\begin{array}{*{20}{c}}-1&2&3\\5&7&9\\-2&1&1\end{array}} \right] $ and $ B = \left[ {\begin{array}{*{20}{c}}-4&1&-5\\1&2&0\\1&3&1\end{array}} \right] $, then verify that:
1. $ (A + B)’ = A’ + B’ $
2. $ (A – B)’ = A’ – B’ $
Solution:
1. \[ A + B = \left[ {\begin{array}{*{20}{c}}-5&3&-2\\6&9&9\\-1&4&2\end{array}} \right] \] \[ (A + B)’ = \left[ {\begin{array}{*{20}{c}}-5&6&-1\\3&9&4\\-2&9&2\end{array}} \right] \] \[ A’ + B’ = \left[ {\begin{array}{*{20}{c}}-5&6&-1\\3&9&4\\-2&9&2\end{array}} \right] \] Hence, $ (A + B)’ = A’ + B’ $.
2. \[ A – B = \left[ {\begin{array}{*{20}{c}}3&1&8\\4&5&9\\-3&-2&0\end{array}} \right] \] \[ (A – B)’ = \left[ {\begin{array}{*{20}{c}}3&4&-3\\1&5&-2\\8&9&0\end{array}} \right] \] \[ A’ – B’ = \left[ {\begin{array}{*{20}{c}}3&4&-3\\1&5&-2\\8&9&0\end{array}} \right] \] Hence, $ (A – B)’ = A’ – B’ $.
If $ A’ = \left[ {\begin{array}{*{20}{c}}3&4\\-1&2\\0&1\end{array}} \right] $ and $ B = \left[ {\begin{array}{*{20}{c}}-1&2&1\\1&2&3\end{array}} \right] $, then verify that:
1. $ (A + B)’ = A’ + B’ $
2. $ (A – B)’ = A’ – B’ $
Solution:
1. \[ A + B = \left[ {\begin{array}{*{20}{c}}2&1&1\\5&4&4\end{array}} \right] \] \[ (A + B)’ = \left[ {\begin{array}{*{20}{c}}2&5\\1&4\\1&4\end{array}} \right] \] \[ A’ + B’ = \left[ {\begin{array}{*{20}{c}}2&5\\1&4\\1&4\end{array}} \right] \] Hence, $ (A + B)’ = A’ + B’ $.
2. \[ A – B = \left[ {\begin{array}{*{20}{c}}4&-3&-1\\3&0&-2\end{array}} \right] \] \[ (A – B)’ = \left[ {\begin{array}{*{20}{c}}4&3\\-3&0\\-1&-2\end{array}} \right] \] \[ A’ – B’ = \left[ {\begin{array}{*{20}{c}}4&3\\-3&0\\-1&-2\end{array}} \right] \] Hence, $ (A – B)’ = A’ – B’ $.
If $ A’ = \left[ {\begin{array}{*{20}{c}}-2&3\\1&2\end{array}} \right] $ and $ B = \left[ {\begin{array}{*{20}{c}}-1&0\\1&2\end{array}} \right] $, then find $ (A + 2B)’ $.

Solution:
\[ (A + 2B)’ = \left[ {\begin{array}{*{20}{c}}-4&5\\1&6\end{array}} \right] \]
For the matrices $ A $ and $ B $, verify that $ (AB)’ = B’A’ $, where:
1. $ A = \left[ {\begin{array}{*{20}{c}}1\\-4\\3\end{array}} \right], B = \left[ {\begin{array}{*{20}{c}}-1&2&1\end{array}} \right] $
2. $ A = \left[ {\begin{array}{*{20}{c}}0\\1\\2\end{array}} \right], B = \left[ {\begin{array}{*{20}{c}}1&5&7\end{array}} \right] $
Solution:
1. \[ AB = \left[ {\begin{array}{*{20}{c}}-1&2&1\\4&-8&-4\\-3&6&3\end{array}} \right] \] \[ (AB)’ = \left[ {\begin{array}{*{20}{c}}-1&4&-3\\2&-8&6\\1&-4&3\end{array}} \right] \] \[ B’A’ = \left[ {\begin{array}{*{20}{c}}-1&4&-3\\2&-8&6\\1&-4&3\end{array}} \right] \] Hence, $ (AB)’ = B’A’ $.
2. \[ AB = \left[ {\begin{array}{*{20}{c}}0&0&0\\1&5&7\\2&10&14\end{array}} \right] \] \[ (AB)’ = \left[ {\begin{array}{*{20}{c}}0&1&2\\0&5&10\\0&7&14\end{array}} \right] \] \[ B’A’ = \left[ {\begin{array}{*{20}{c}}0&1&2\\0&5&10\\0&7&14\end{array}} \right] \] Hence, $ (AB)’ = B’A’ $.
If:
1. $ A = \left[ {\begin{array}{*{20}{c}}\cos \alpha&\sin \alpha\\-\sin \alpha&\cos \alpha\end{array}} \right] $, then verify that $ A’A = I $.
2. $ A = \left[ {\begin{array}{*{20}{c}}\sin \alpha&\cos \alpha\\-\cos \alpha&\sin \alpha\end{array}} \right] $, then verify that $ A’A = I $.
Solution:
1. \[ A’A = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = I \] Hence, $ A’A = I $.
2. \[ A’A = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = I \] Hence, $ A’A = I $.
Show that the matrix $ A = \left[ {\begin{array}{*{20}{c}}1&-1&5\\-1&2&1\\5&1&3\end{array}} \right] $ is a symmetric matrix.

Solution:
\[ A’ = \left[ {\begin{array}{*{20}{c}}1&-1&5\\-1&2&1\\5&1&3\end{array}} \right] = A \] Hence, $ A $ is a symmetric matrix.
Show that the matrix $ A = \left[ {\begin{array}{*{20}{c}}0&1&-1\\-1&0&1\\1&-1&0\end{array}} \right] $ is a skew-symmetric matrix.

Solution:
\[ A’ = \left[ {\begin{array}{*{20}{c}}0&-1&1\\1&0&-1\\-1&1&0\end{array}} \right] = -A \] Hence, $ A $ is a skew-symmetric matrix.
For the matrix $ A = \left[ {\begin{array}{*{20}{c}}1&5\\6&7\end{array}} \right] $, verify that:
1. $ (A + A’) $ is a symmetric matrix.
2. $ (A – A’) $ is a skew-symmetric matrix.
Solution:
1. \[ A + A’ = \left[ {\begin{array}{*{20}{c}}2&11\\11&14\end{array}} \right] \] Hence, $ (A + A’) $ is a symmetric matrix.
2. \[ A – A’ = \left[ {\begin{array}{*{20}{c}}0&-1\\1&0\end{array}} \right] \] Hence, $ (A – A’) $ is a skew-symmetric matrix.
Find $ (A + A’) $ and $ \frac{1}{2}(A – A’) $, when $ A = \left[ {\begin{array}{*{20}{c}}0&a&b\\-a&0&c\\-b&-c&0\end{array}} \right] $.

Solution:
\[ A + A’ = \left[ {\begin{array}{*{20}{c}}0&0&0\\0&0&0\\0&0&0\end{array}} \right] \] \[ \frac{1}{2}(A – A’) = \left[ {\begin{array}{*{20}{c}}0&a&b\\-a&0&c\\-b&-c&0\end{array}} \right] \]
Express the following matrices as the sum of a symmetric and a skew-symmetric matrix:
1. $ \left[ {\begin{array}{*{20}{c}}3&5\\1&-1\end{array}} \right] $
2. $ \left[ {\begin{array}{*{20}{c}}6&-2&2\\-2&3&-1\\2&-1&3\end{array}} \right] $
3. $ \left[ {\begin{array}{*{20}{c}}3&3&-1\\-2&-2&1\\-4&-5&2\end{array}} \right] $
4. $ \left[ {\begin{array}{*{20}{c}}1&5\\-1&2\end{array}} \right] $
Solution:
1. \[ P = \left[ {\begin{array}{*{20}{c}}3&3\\3&-1\end{array}} \right], \quad Q = \left[ {\begin{array}{*{20}{c}}0&2\\-2&0\end{array}} \right] \] Hence, $ P + Q = A $.
2. \[ P = \left[ {\begin{array}{*{20}{c}}6&-2&2\\-2&3&-1\\2&-1&3\end{array}} \right], \quad Q = \left[ {\begin{array}{*{20}{c}}0&0&0\\0&0&0\\0&0&0\end{array}} \right] \] Hence, $ P + Q = A $.
3. \[ P = \left[ {\begin{array}{*{20}{c}}3&\frac{1}{2}&-\frac{5}{2}\\\frac{1}{2}&-2&-2\\-\frac{5}{2}&-2&2\end{array}} \right], \quad Q = \left[ {\begin{array}{*{20}{c}}0&\frac{5}{2}&\frac{3}{2}\\-\frac{5}{2}&0&3\\-\frac{3}{2}&-3&0\end{array}} \right] \] Hence, $ P + Q = A $.
4. \[ P = \left[ {\begin{array}{*{20}{c}}1&2\\2&2\end{array}} \right], \quad Q = \left[ {\begin{array}{*{20}{c}}0&3\\-3&0\end{array}} \right] \] Hence, $ P + Q = A $.
Choose the correct answer:
1. If $ A, B $ are symmetric matrices of the same order, then $ AB – BA $ is a:
2. If $ A = \left[ {\begin{array}{*{20}{c}}\cos \alpha&-\sin \alpha\\\sin \alpha&\cos \alpha\end{array}} \right] $, then $ A + A’ = I $, if the value of $ \alpha $ is:

NCERT – Exercise 3.4

Using elementary transformations, find the inverse of each of the matrices, if it exists in questions 1 to 17.

$$ \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\2&3\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\2&3\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\2&3\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – 2R_1$

$$ \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&5\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\{ – 2}&1\end{array}} \right]A $$

Applying $R_2 \to \frac{1}{5}R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\{\frac{{ – 2}}{5}}&{\frac{1}{5}}\end{array}} \right]A $$

Applying $R_1 \to R_1 + R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\frac{3}{5}}&{\frac{1}{5}}\\{\frac{{ – 2}}{5}}&{\frac{1}{5}}\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{c}}{\frac{3}{5}}&{\frac{1}{5}}\\{\frac{{ – 2}}{5}}&{\frac{1}{5}}\end{array}} \right]$
$$ \left[ {\begin{array}{*{20}{c}}2&1\\1&1\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{c}}2&1\\1&1\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{c}}2&1\\1&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]A $$

Applying $R_1 \to R_1 – R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\1&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – R_1$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\{ – 1}&2\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\{ – 1}&2\end{array}} \right]$
$$ \left[ {\begin{array}{*{20}{c}}1&3\\2&7\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{c}}1&3\\2&7\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{c}}1&3\\2&7\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – 2R_1$

$$ \left[ {\begin{array}{*{20}{c}}1&3\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\{ – 2}&1\end{array}} \right]A $$

Applying $R_1 \to R_1 – 3R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}7&{ – 3}\\{ – 2}&1\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{c}}7&{ – 3}\\{ – 2}&1\end{array}} \right]$
$$ \left[ {\begin{array}{*{20}{c}}2&3\\5&7\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{c}}2&3\\5&7\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{c}}2&3\\5&7\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – 2R_1$

$$ \left[ {\begin{array}{*{20}{c}}2&3\\1&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\{ – 2}&1\end{array}} \right]A $$

Applying $R_1 \to R_1 – R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&2\\1&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3&{ – 1}\\{ – 2}&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – R_1$

$$ \left[ {\begin{array}{*{20}{c}}1&2\\0&{ – 1}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3&{ – 1}\\{ – 5}&2\end{array}} \right]A $$

Applying $R_2 \to -R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&2\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3&{ – 1}\\5&{ – 2}\end{array}} \right]A $$

Applying $R_1 \to R_1 – 2R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ – 7}&3\\5&{ – 2}\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{c}}{ – 7}&3\\5&{ – 2}\end{array}} \right]$
$$ \left[ {\begin{array}{*{20}{c}}2&1\\7&4\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{c}}2&1\\7&4\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{c}}2&1\\7&4\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – 3R_1$

$$ \left[ {\begin{array}{*{20}{c}}2&1\\1&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\{ – 3}&1\end{array}} \right]A $$

Applying $R_1 \to R_1 – R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\1&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}4&{ – 1}\\{ – 3}&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – R_1$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}4&{ – 1}\\{ – 7}&2\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{c}}4&{ – 1}\\{ – 7}&2\end{array}} \right]$
$$ \left[ {\begin{array}{*{20}{c}}2&5\\1&3\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{c}}2&5\\1&3\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{c}}2&5\\1&3\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]A $$

Applying $R_1 \to R_1 – R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&2\\1&3\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – R_1$

$$ \left[ {\begin{array}{*{20}{c}}1&2\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\{ – 1}&2\end{array}} \right]A $$

Applying $R_1 \to R_1 – 2R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3&{ – 5}\\{ – 1}&2\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{c}}3&{ – 5}\\{ – 1}&2\end{array}} \right]$
$$ \left[ {\begin{array}{*{20}{c}}3&1\\5&2\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{c}}3&1\\5&2\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{c}}3&1\\5&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – R_1$

$$ \left[ {\begin{array}{*{20}{c}}3&1\\2&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\{ – 1}&1\end{array}} \right]A $$

Applying $R_1 \to R_1 – R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\2&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2&{ – 1}\\{ – 1}&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – 2R_1$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2&{ – 1}\\{ – 5}&3\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{c}}2&{ – 1}\\{ – 5}&3\end{array}} \right]$
$$ \left[ {\begin{array}{*{20}{c}}4&5\\3&4\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{c}}4&5\\3&4\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{c}}4&5\\3&4\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]A $$

Applying $R_1 \to R_1 – R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&1\\3&4\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – 3R_1$

$$ \left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\{ – 3}&4\end{array}} \right]A $$

Applying $R_1 \to R_1 – R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}4&{ – 5}\\{ – 3}&4\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{c}}4&{ – 5}\\{ – 3}&4\end{array}} \right]$
$$ \left[ {\begin{array}{*{20}{c}}3&{10}\\2&7\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{c}}3&{10}\\2&7\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{c}}3&{10}\\2&7\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]A $$

Applying $R_1 \to R_1 – R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&3\\2&7\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – 2R_1$

$$ \left[ {\begin{array}{*{20}{c}}1&3\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\{ – 2}&3\end{array}} \right]A $$

Applying $R_1 \to R_1 – 3R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}7&{ – 10}\\{ – 2}&3\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{c}}7&{ – 10}\\{ – 2}&3\end{array}} \right]$
$$ \left[ {\begin{array}{*{20}{c}}3&{ – 1}\\{ – 4}&2\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{c}}3&{ – 1}\\{ – 4}&2\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{c}}3&{ – 1}\\{ – 4}&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]A $$

Applying $R_1 \to R_1 + R_2$

$$ \left[ {\begin{array}{*{20}{c}}{ – 1}&1\\{ – 4}&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right]A $$

Applying $R_1 \to -R_1$

$$ \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\{ – 4}&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ – 1}&{ – 1}\\0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 + 4R_1$

$$ \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&{ – 2}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ – 1}&{ – 1}\\{ – 4}&{ – 3}\end{array}} \right]A $$

Applying $R_2 \to -R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ – 1}&{ – 1}\\4&3\end{array}} \right]A $$

Applying $R_2 \to \frac{1}{2}R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ – 1}&{ – 1}\\2&{\frac{3}{2}}\end{array}} \right]A $$

Applying $R_1 \to R_1 + R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{\frac{1}{2}}\\2&{\frac{3}{2}}\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{c}}1&{\frac{1}{2}}\\2&{\frac{3}{2}}\end{array}} \right]$
$$ \left[ {\begin{array}{*{20}{c}}2&{ – 6}\\1&{ – 2}\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{c}}2&{ – 6}\\1&{ – 2}\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{c}}2&{ – 6}\\1&{ – 2}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]A $$

Applying $R_1 \to R_1 – R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&{ – 4}\\1&{ – 2}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – R_1$

$$ \left[ {\begin{array}{*{20}{c}}1&{ – 4}\\0&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\{ – 1}&2\end{array}} \right]A $$

Applying $R_1 \to R_1 + 2R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\0&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ – 1}&3\\{ – 1}&2\end{array}} \right]A $$

Applying $R_2 \to \frac{1}{2}R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ – 1}&3\\{\frac{{ – 1}}{2}}&1\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{c}}{ – 1}&3\\{\frac{{ – 1}}{2}}&1\end{array}} \right]$
$$ \left[ {\begin{array}{*{20}{c}}6&{ – 3}\\{ – 2}&1\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{c}}6&{ – 3}\\{ – 2}&1\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{c}}6&{ – 3}\\{ – 2}&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]A $$

Applying $R_1 \to R_1 + 3R_2$

$$ \left[ {\begin{array}{*{20}{c}}0&0\\{ – 2}&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&3\\0&1\end{array}} \right]A $$

We have all zeroes in the first row of the left hand side matrix of the above equation.

Therefore, $A^{-1}$ does not exist.
$$ \left[ {\begin{array}{*{20}{c}}2&{ – 3}\\{ – 1}&2\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{c}}2&{ – 3}\\{ – 1}&2\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{c}}2&{ – 3}\\{ – 1}&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]A $$

Applying $R_1 \to R_1 + R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\{ – 1}&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 + R_1$

$$ \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&1\\1&2\end{array}} \right]A $$

Applying $R_1 \to R_1 + R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2&3\\1&2\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{c}}2&3\\1&2\end{array}} \right]$
$$ \left[ {\begin{array}{*{20}{c}}2&1\\4&2\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{c}}2&1\\4&2\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{c}}2&1\\4&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – 2R_1$

$$ \left[ {\begin{array}{*{20}{c}}2&1\\0&0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\{ – 2}&1\end{array}} \right]A $$

We have all zeroes in the second row of the left hand side matrix of the above equation. Therefore, $A^{-1}$ does not exist.
$$ \left[ {\begin{array}{*{20}{r}}2&{ – 3}&3\\2&2&3\\3&{ – 2}&2\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{r}}2&{ – 3}&3\\2&2&3\\3&{ – 2}&2\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{r}}2&{ – 3}&3\\2&2&3\\3&{ – 2}&2\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]A $$

Applying $R_1 \to R_1 – R_2$

$$ \left[ {\begin{array}{*{20}{r}}0&{ – 5}&0\\2&2&3\\3&{ – 2}&2\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1&{ – 1}&0\\0&1&0\\0&0&1\end{array}} \right]A $$

Applying $R_3 \to R_3 – R_2$

$$ \left[ {\begin{array}{*{20}{r}}0&{ – 5}&0\\2&2&3\\1&{ – 4}&{ – 1}\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1&{ – 1}&0\\0&1&0\\0&{ – 1}&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – 2R_3$

$$ \left[ {\begin{array}{*{20}{r}}0&{ – 5}&0\\0&{10}&5\\1&{ – 4}&{ – 1}\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1&{ – 1}&0\\0&3&{ – 2}\\0&{ – 1}&1\end{array}} \right]A $$

Applying $R_1 \leftrightarrow R_3$

$$ \left[ {\begin{array}{*{20}{r}}1&{ – 4}&{ – 1}\\0&{10}&5\\0&{ – 5}&0\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}0&{ – 1}&1\\0&3&{ – 2}\\1&{ – 1}&0\end{array}} \right]A $$

Applying $R_3 \leftrightarrow R_2$

$$ \left[ {\begin{array}{*{20}{r}}1&{ – 4}&{ – 1}\\0&{ – 5}&0\\0&{10}&5\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}0&{ – 1}&1\\1&{ – 1}&0\\0&3&{ – 2}\end{array}} \right]A $$

Applying $R_3 \to R_3 + 2R_2$

$$ \left[ {\begin{array}{*{20}{r}}1&{ – 4}&{ – 1}\\0&{ – 5}&0\\0&0&5\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}0&{ – 1}&1\\1&{ – 1}&0\\2&1&{ – 2}\end{array}} \right]A $$

Applying $R_2 \to -R_2$

$$ \left[ {\begin{array}{*{20}{r}}1&{ – 4}&{ – 1}\\0&5&0\\0&0&5\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}0&{ – 1}&1\\{ – 1}&1&0\\2&1&{ – 2}\end{array}} \right]A $$

Applying $R_2 \to \frac{1}{5}R_2$ and $R_3 \to \frac{1}{5}R_3$

$$ \left[ {\begin{array}{*{20}{r}}1&{ – 4}&{ – 1}\\0&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}0&{ – 1}&1\\{\frac{{ – 1}}{5}}&{\frac{1}{5}}&0\\{\frac{2}{5}}&{\frac{1}{5}}&{\frac{{ – 2}}{5}}\end{array}} \right]A $$

Applying $R_1 \to R_1 + 4R_2$

$$ \left[ {\begin{array}{*{20}{r}}1&0&{ – 1}\\0&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}{ – \frac{4}{5}}&{ – \frac{1}{5}}&1\\{\frac{{ – 1}}{5}}&{\frac{1}{5}}&0\\{\frac{2}{5}}&{\frac{1}{5}}&{\frac{{ – 2}}{5}}\end{array}} \right]A $$

Applying $R_1 \to R_1 + R_3$

$$ \left[ {\begin{array}{*{20}{r}}1&0&0\\0&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}{\frac{{ – 2}}{5}}&0&{\frac{3}{5}}\\{\frac{{ – 1}}{5}}&{\frac{1}{5}}&0\\{\frac{2}{5}}&{\frac{1}{5}}&{\frac{{ – 2}}{5}}\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{r}}{\frac{{ – 2}}{5}}&0&{\frac{3}{5}}\\{\frac{{ – 1}}{5}}&{\frac{1}{5}}&0\\{\frac{2}{5}}&{\frac{1}{5}}&{\frac{{ – 2}}{5}}\end{array}} \right]$
$$ \left[ {\begin{array}{*{20}{r}}1&3&{ – 2}\\{ – 3}&0&{ – 5}\\2&5&0\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{r}}1&3&{ – 2}\\{ – 3}&0&{ – 5}\\2&5&0\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{r}}1&3&{ – 2}\\{ – 3}&0&{ – 5}\\2&5&0\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 + 3R_1$

$$ \left[ {\begin{array}{*{20}{r}}1&3&{ – 2}\\0&9&{ – 11}\\2&5&0\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1&0&0\\3&1&0\\0&0&1\end{array}} \right]A $$

Applying $R_3 \to R_3 – 2R_1$

$$ \left[ {\begin{array}{*{20}{r}}1&3&{ – 2}\\0&9&{ – 11}\\0&{ – 1}&4\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1&0&0\\3&1&0\\{ – 2}&0&1\end{array}} \right]A $$

Applying $R_1 \to R_1 + 3R_3$

$$ \left[ {\begin{array}{*{20}{r}}1&0&{10}\\0&9&{ – 11}\\0&{ – 1}&4\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}{ – 5}&0&3\\3&1&0\\{ – 2}&0&1\end{array}} \right]A $$

Interchanging $R_2$ and $R_3$

$$ \left[ {\begin{array}{*{20}{r}}1&0&{10}\\0&{ – 1}&4\\0&9&{ – 11}\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}{ – 5}&0&3\\{ – 2}&0&1\\3&1&0\end{array}} \right]A $$

Applying $R_3 \to R_3 + 9R_2$

$$ \left[ {\begin{array}{*{20}{r}}1&0&{10}\\0&{ – 1}&4\\0&0&{25}\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}{ – 5}&0&3\\{ – 2}&0&1\\{ – 15}&1&9\end{array}} \right]A $$

Applying $R_2 \to -R_2$

$$ \left[ {\begin{array}{*{20}{r}}1&0&{10}\\0&1&{ – 4}\\0&0&{25}\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}{ – 5}&0&3\\2&0&{ – 1}\\{ – 15}&1&9\end{array}} \right]A $$

Applying $R_3 \to \frac{1}{25}R_3$

$$ \left[ {\begin{array}{*{20}{r}}1&0&{10}\\0&1&{ – 4}\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}{ – 5}&0&3\\2&0&{ – 1}\\{\frac{{ – 15}}{25}}&{\frac{1}{25}}&{\frac{9}{25}}\end{array}} \right]A $$

Applying $R_1 \to R_1 – 10R_3$

$$ \left[ {\begin{array}{*{20}{r}}1&0&0\\0&1&{ – 4}\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1&{\frac{{ – 2}}{5}}&{\frac{{ – 3}}{5}}\\2&0&{ – 1}\\{\frac{{ – 3}}{5}}&{\frac{1}{25}}&{\frac{9}{25}}\end{array}} \right]A $$

Applying $R_2 \to R_2 + 4R_3$

$$ \left[ {\begin{array}{*{20}{r}}1&0&0\\0&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1&{\frac{{ – 2}}{5}}&{\frac{{ – 3}}{5}}\\{\frac{{ – 2}}{5}}&{\frac{4}{25}}&{\frac{11}{25}}\\{\frac{{ – 3}}{5}}&{\frac{1}{25}}&{\frac{9}{25}}\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{r}}1&{\frac{{ – 2}}{5}}&{\frac{{ – 3}}{5}}\\{\frac{{ – 2}}{5}}&{\frac{4}{25}}&{\frac{11}{25}}\\{\frac{{ – 3}}{5}}&{\frac{1}{25}}&{\frac{9}{25}}\end{array}} \right]$
$$ \left[ {\begin{array}{*{20}{r}}2&0&{ – 1}\\5&1&0\\0&1&3\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{r}}2&0&{ – 1}\\5&1&0\\0&1&3\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{r}}2&0&{ – 1}\\5&1&0\\0&1&3\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]A $$

Interchanging $R_1$ and $R_2$

$$ \left[ {\begin{array}{*{20}{r}}5&1&0\\2&0&{ – 1}\\0&1&3\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}0&1&0\\1&0&0\\0&0&1\end{array}} \right]A $$

Applying $R_1 \to R_1 – 2R_2$

$$ \left[ {\begin{array}{*{20}{r}}1&1&2\\2&0&{ – 1}\\0&1&3\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}{ – 2}&1&0\\1&0&0\\0&0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – 2R_1$

$$ \left[ {\begin{array}{*{20}{r}}1&1&2\\0&{ – 2}&{ – 5}\\0&1&3\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}{ – 2}&1&0\\5&{ – 2}&0\\0&0&1\end{array}} \right]A $$

Applying $R_2 \to -R_2$

$$ \left[ {\begin{array}{*{20}{r}}1&1&2\\0&2&5\\0&1&3\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}{ – 2}&1&0\\{ – 5}&2&0\\0&0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – R_3$

$$ \left[ {\begin{array}{*{20}{r}}1&1&2\\0&1&2\\0&1&3\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}{ – 2}&1&0\\{ – 5}&2&{ – 1}\\0&0&1\end{array}} \right]A $$

Applying $R_1 \to R_1 – R_2$

$$ \left[ {\begin{array}{*{20}{r}}1&0&0\\0&1&2\\0&1&3\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}3&{ – 1}&1\\{ – 5}&2&{ – 1}\\0&0&1\end{array}} \right]A $$

Applying $R_3 \to R_3 – R_2$

$$ \left[ {\begin{array}{*{20}{r}}1&0&0\\0&1&2\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}3&{ – 1}&1\\{ – 5}&2&{ – 1}\\5&{ – 2}&2\end{array}} \right]A $$

Applying $R_2 \to R_2 – 2R_3$

$$ \left[ {\begin{array}{*{20}{r}}1&0&0\\0&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}3&{ – 1}&1\\{ – 15}&6&{ – 5}\\5&{ – 2}&2\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{r}}3&{ – 1}&1\\{ – 15}&6&{ – 5}\\5&{ – 2}&2\end{array}} \right]$
Matrices A and B will be inverse of each other only if
1. AB = BA
2. AB = BA = 0
3. AB = 0, BA = I
4. AB = BA = I
Solution:

(D) Matrices A and B will be inverse of each other only if, AB = BA = I.

NCERT – Miscellaneous Exercise

1. Let $ A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} $, show that $ (aI + bA)^n = a^nI + na^{n-1}bA $, where I is the identity matrix of order 2 and $ n \in \mathbb{N} $.

Solution:

We have, $ A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} $ and $ (aI + bA)^n = a^nI + na^{n-1}bA $.

For $ n = 1 $,

$ (aI + bA)^1 = a^1I + 1 \cdot a^{1-1}bA $

$ \Rightarrow aI + bA = aI + bA $.

So, it is true for $ n = 1 $.

Assume that it is true for $ n = k $, i.e.,

$ (aI + bA)^k = a^kI + ka^{k-1}bA $.

Then, $ (aI + bA)^{k+1} = (aI + bA)^k \cdot (aI + bA) $

$ = (a^kI + ka^{k-1}bA)(aI + bA) $

$ = a^{k+1}I \cdot I + ka^k bAI + a^k bAI + ka^{k-1}b^2 A \cdot A $

$ = a^{k+1}I + ka^k bA + a^k bA + ka^{k-1}b^2 \cdot O $

$ = a^{k+1}I + (k+1)a^k bA $

$ = a^{k+1}I + (k+1)a^{(k+1)-1}bA $.

Thus, it is true for $ n = k+1 $.

Hence, by mathematical induction, it is true for all $ n \in \mathbb{N} $.

2. If $ A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix} $, prove that $ A^n = \begin{bmatrix} 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \end{bmatrix} $, $ n \in \mathbb{N} $.

Solution:

We shall prove it by mathematical induction.

To prove that $ n = 1 $ is true.

$ A^1 = \begin{bmatrix} 3^{1-1} & 3^{1-1} & 3^{1-1} \\ 3^{1-1} & 3^{1-1} & 3^{1-1} \\ 3^{1-1} & 3^{1-1} & 3^{1-1} \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix} = A $.

Thus, it is true for $ n = 1 $.

Assume that it is true for $ n = k $, i.e.,

$ A^k = \begin{bmatrix} 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \end{bmatrix} $.

Then, $ A^{k+1} = A^k \cdot A $

$ = \begin{bmatrix} 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix} $

$ = \begin{bmatrix} 3^{k-1} + 3^{k-1} + 3^{k-1} & 3^{k-1} + 3^{k-1} + 3^{k-1} & 3^{k-1} + 3^{k-1} + 3^{k-1} \\ 3^{k-1} + 3^{k-1} + 3^{k-1} & 3^{k-1} + 3^{k-1} + 3^{k-1} & 3^{k-1} + 3^{k-1} + 3^{k-1} \\ 3^{k-1} + 3^{k-1} + 3^{k-1} & 3^{k-1} + 3^{k-1} + 3^{k-1} & 3^{k-1} + 3^{k-1} + 3^{k-1} \end{bmatrix} $

$ = \begin{bmatrix} 3^k & 3^k & 3^k \\ 3^k & 3^k & 3^k \\ 3^k & 3^k & 3^k \end{bmatrix} $.

Thus, it is true for $ n = k+1 $.

So, by mathematical induction, $ A^n = \begin{bmatrix} 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \end{bmatrix} $, $ n \in \mathbb{N} $ is true.

3. If $ A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} $, then prove that $ A^n = \begin{bmatrix} 1 + 2n & -4n \\ n & 1 – 2n \end{bmatrix} $, where $ n $ is any positive integer.

Solution:

We have, $ A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} $ and $ A^n = \begin{bmatrix} 1 + 2n & -4n \\ n & 1 – 2n \end{bmatrix} $.

For $ n = 1 $,

$ A^1 = \begin{bmatrix} 1 + 2 \cdot 1 & -4 \cdot 1 \\ 1 & 1 – 2 \cdot 1 \end{bmatrix} = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} = A $.

So, it is true for $ n = 1 $.

Assume that it is true for $ n = k $, i.e.,

$ A^k = \begin{bmatrix} 1 + 2k & -4k \\ k & 1 – 2k \end{bmatrix} $.

Also, $ A^{k+1} = \begin{bmatrix} 1 + 2(k+1) & -4(k+1) \\ k+1 & 1 – 2(k+1) \end{bmatrix} $ for $ n = k+1 $.

$ \Rightarrow A^{k+1} = \begin{bmatrix} 2k + 3 & -4k – 4 \\ k + 1 & -2k – 1 \end{bmatrix} $.

Now, $ A^{k+1} = A \cdot A^k = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 1 + 2k & -4k \\ k & 1 – 2k \end{bmatrix} $

$ = \begin{bmatrix} 3 + 6k – 4k & -12k – 4 + 8k \\ 1 + k – k & -4k – 1 + 2k \end{bmatrix} $

$ = \begin{bmatrix} 2k + 3 & -4k – 4 \\ k + 1 & -2k – 1 \end{bmatrix} = A^{k+1} $.

So, it is true for $ n = k+1 $.

Hence, by mathematical induction, $ A^n = \begin{bmatrix} 1 + 2n & -4n \\ n & 1 – 2n \end{bmatrix} $ is true.

4. If A and B are symmetric matrices, prove that $ AB – BA $ is a skew symmetric matrix.

Solution:

Given: A and B are symmetric matrices, therefore $ A’ = A $, $ B’ = B $.

To prove: $ (AB – BA)’ = -(AB – BA) $.

Proof: $ (AB – BA)’ = (AB)’ – (BA)’ $

$ = B’A’ – A’B’ = BA – AB = -(AB – BA) $.

So, $ AB – BA $ is a skew-symmetric matrix.

5. Show that the matrix $ B’AB $ is symmetric or skew symmetric according as A is symmetric or skew symmetric.

Solution:

Case I: Given that A is symmetric. We will prove $ B’AB $ is symmetric.

As A is symmetric, so $ A’ = A $.

Now, $ (B’AB)’ = B’A'(B’)’ = B’A’B = B’AB $.

Thus, $ B’AB $ is a symmetric matrix.

Case II: Given A is skew symmetric, i.e., $ A’ = -A $. We will prove that $ B’AB $ is skew symmetric.

Now, $ (B’AB)’ = B’A'(B’)’ = B’A’B $

$ = B'(-A)B = -B’AB $.

Hence, $ B’AB $ is a skew-symmetric matrix.

6. Find the values of $ x, y, z $ if the matrix $ A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix} $ satisfies the equation $ A’A = I $.

Solution:

Given that, matrix $ A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix} $ and $ A’A = I $.

$ \Rightarrow \begin{bmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{bmatrix} \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $.

$ \Rightarrow \begin{bmatrix} 0 + x^2 + x^2 & 0 + xy – xy & 0 – zx + zx \\ 0 + xy – xy & 4y^2 + y^2 + y^2 & 2yz – yz – yz \\ 0 – zx + xz & 2yz – zy – zy & z^2 + z^2 + z^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $.

$ \Rightarrow \begin{bmatrix} 2x^2 & 0 & 0 \\ 0 & 6y^2 & 0 \\ 0 & 0 & 3z^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $.

$ \Rightarrow 2x^2 = 1, 6y^2 = 1, 3z^2 = 1 \Rightarrow x^2 = \frac{1}{2}, y^2 = \frac{1}{6}, z^2 = \frac{1}{3} $.

Hence, $ x = \pm \frac{1}{\sqrt{2}}, y = \pm \frac{1}{\sqrt{6}}, z = \pm \frac{1}{\sqrt{3}} $.

7. For what values of $ x $: $ \begin{bmatrix} 1 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} = O $?

Solution:

$ \begin{bmatrix} 1 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} = O $.

$ \Rightarrow \begin{bmatrix} 1 + 4 + 1 & 2 + 0 + 0 & 0 + 2 + 0 \end{bmatrix} \begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} = O $.

$ \Rightarrow \begin{bmatrix} 6 & 2 & 4 \end{bmatrix} \begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} = O $.

$ 0 + 4 + 4x = 0 \Rightarrow 4(x + 1) = 0 \Rightarrow x + 1 = 0 \Rightarrow x = -1 $.

8. If $ A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} $, then show that $ A^2 – 5A + 7I = O $.

Solution:

Given that, $ A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} $.

$ \Rightarrow A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} $

$ = \begin{bmatrix} 9 – 1 & 3 + 2 \\ -3 – 2 & -1 + 4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} $.

And $ 5A = 5 \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} $.

Now, substituting the values, we have

$ A^2 – 5A + 7I = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} – \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} $

$ = \begin{bmatrix} -7 & 0 \\ 0 & -7 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O $.

Hence, proved.

9. Find $ x $, if $ \begin{bmatrix} x & -5 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} = O $.

Solution:

$ \begin{bmatrix} x & -5 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} = O $.

$ \Rightarrow \begin{bmatrix} x & -5 & -1 \end{bmatrix} \begin{bmatrix} x + 2 \\ 8 + 1 \\ 2x + 3 \end{bmatrix} = O $.

$ \Rightarrow \begin{bmatrix} x & -5 & -1 \end{bmatrix} \begin{bmatrix} x + 2 \\ 9 \\ 2x + 3 \end{bmatrix} = O $.

$ \Rightarrow x(x + 2) – 45 – 2x – 3 = 0 \Rightarrow x^2 – 48 = 0 $.

$ \Rightarrow x = \pm 4\sqrt{3} $.

10. A manufacturer produces three products x, y, z which he sells in two markets. Annual sales are indicated as:

(a) If unit sale prices of x, y, and z are Rs 2.50, Rs 1.50, and Rs 1.00, respectively, find the total revenue in each market with the help of matrix algebra.

(b) If the unit costs of the above three commodities are Rs 2.00, Rs 1.00, and 50 paise respectively, find the gross profit.

Solution:

Let quantity matrix be $ A = \begin{bmatrix} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix} $.

(a) Selling Price $ B = \begin{bmatrix} 2.50 \\ 1.50 \\ 1.00 \end{bmatrix} $.

Now, Total Selling Price,

$ AB = \begin{bmatrix} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix} \begin{bmatrix} 2.50 \\ 1.50 \\ 1.00 \end{bmatrix} $.

$ = \begin{bmatrix} 10000 \times 2.50 + 2000 \times 1.50 + 18000 \times 1 \\ 6000 \times 2.50 + 20000 \times 1.50 + 8000 \times 1 \end{bmatrix} $.

$ = \begin{bmatrix} 25000 + 3000 + 18000 \\ 15000 + 30000 + 8000 \end{bmatrix} $.

Total revenue in market I = Rs. 46,000.

Total revenue in market II = Rs. 53,000.

(b) Now, cost price $ = \begin{bmatrix} 2.00 \\ 1.00 \\ 0.50 \end{bmatrix} $.

Total cost price $ = \begin{bmatrix} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \\ 0.5 \end{bmatrix} $.

$ = \begin{bmatrix} 10000 \times 2 + 2000 \times 1 + 18000 \times 0.5 \\ 6000 \times 2 + 20000 \times 1 + 8000 \times 0.5 \end{bmatrix} $.

Total cost price = 31000 + 36000 = Rs. 67,000.

Total selling price = 46000 + 53000 = Rs. 99,000.

Profit = S.P. – C.P. = 99,000 – 67,000 = Rs. 32,000.

11. Find the matrix X so that $ X \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix} $.

Solution:

$ X \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix} $.

We can say that X is a 2 × 2 matrix.

Let $ X = \begin{bmatrix} a & b \\ c & d \end{bmatrix} $.

$ \Rightarrow \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix} $.

$ \Rightarrow \begin{bmatrix} a + 4b & 2a + 5b & 3a + 6b \\ c + 4d & 2c + 5d & 3c + 6d \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix} $.

$ \Rightarrow a + 4b = -7 $ and $ c + 4d = 2 $.

$ 2a + 5b = -8 $ and $ 2c + 5d = 4 $.

Solving (i) and (iii), we get $ a = 1, b = -2 $.

Solving (ii) and (iv), we get $ c = 2, d = 0 $.

Hence, $ X = \begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix} $.

12. If A and B are square matrices of the same order such that $ AB = BA $, then prove by induction that $ AB^n = B^nA $. Further, prove that $ (AB)^n = A^nB^n $ for all $ n \in \mathbb{N} $.

Solution:

Given $ AB = BA $,

To prove:

(1) $ AB^n = B^nA $ and (2) $ (AB)^n = A^nB^n $ $ \forall n \in \mathbb{N} $.

We will prove it by mathematical induction.

(1) Given that $ AB = BA $.

We have to prove $ AB^n = B^nA $.

For $ n = 1 $, $ AB^1 = B^1A $.

$ \Rightarrow AB = BA $, which is true.

Let it be true for $ n = m $,

$ AB^m = B^mA $.

Then, for $ n = m + 1 $,

$ AB^{m+1} = A(B^mB) = (AB^m)B = (B^mA)B $.

$ = B^m(AB) = B^m(BA) $.

$ = (B^mB)A = B^{m+1}A $.

So, it is true for $ n = m + 1 $.

$ \therefore AB^n = B^nA $.

(2) For $ n = 1 $, $ (AB)^1 = A^1B^1 $.

$ \Rightarrow AB = BA $, which is true for $ n = 1 $.

Let (i) be true for a positive integer $ n = m $,

i.e., $ (AB)^m = A^mB^m $.

Then for $ n = m + 1 $, $ (AB)^{m+1} = (AB)^m(AB) $.

$ = (A^mB^m)(AB) $.

$ = A^m(B^mA)B $.

$ = A^m(AB^m)B $.

$ = (A^mA)(B^mB) = A^{m+1}B^{m+1} $.

So, it holds for $ n = m + 1 $.

Hence, $ (AB)^n = A^nB^n $ $ \forall n \in \mathbb{N} $.

Choose the correct answer in the following questions:

13. If $ A = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} $ is such that $ A^2 = I $, then

$ 1 + \alpha^2 + \beta\gamma = 0 $.
$ 1 – \alpha^2 + \beta\gamma = 0 $.
$ 1 – \alpha^2 – \beta\gamma = 0 $.
$ 1 + \alpha^2 – \beta\gamma = 0 $.

Solution:

Now, $ A^2 = I $.

$ \Rightarrow \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $.

$ \Rightarrow \begin{bmatrix} \alpha^2 + \beta\gamma & \alpha\beta – \alpha\beta \\ \gamma\alpha – \alpha\gamma & \gamma\beta + \alpha^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $.

$ \Rightarrow \begin{bmatrix} \alpha^2 + \beta\gamma & 0 \\ 0 & \gamma\beta + \alpha^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $.

$ \Rightarrow \alpha^2 + \beta\gamma = 1 \Rightarrow 1 – \alpha^2 – \gamma\beta = 0 $.

14. If the matrix A is both symmetric and skew symmetric, then

A is a diagonal matrix.
A is a zero matrix.
A is a square matrix.
None of these.

Solution:

(B) Consider the matrix A.

Clearly, $ A’ = A $ and $ A’ = -A $.

$ \therefore A = -A \Rightarrow 2A = 0 \Rightarrow A = 0 $.

$ \therefore A $ is a zero matrix.

15. If A is a square matrix such that $ A^2 = A $, then $ (I + A)^3 – 7A $ is equal to

A.
$ I – A $.
I.
3A.

Solution:

Now, $ (I + A)^3 – 7A = I^3 + A^3 + 3IA(I + A) – 7A $.

$ = I + A^2 + 3A(I + A) – 7A $.

$ = I + A + 3A + 3A^2 – 7A $.

$ = I + 4A^2 – 4A $.

$ = I + 4A – 4A = I $.

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