NCERT 3.1 class 12 maths solutions

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 = $2^9 = 512$.