NCERT Solution class 12 maths chapter 3 miscellaneous Matrices

NCERT Solution class 12 maths chapter 3 miscellaneous Matrices

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. \( 1 + \alpha^2 + \beta\gamma = 0 \).
2. \( 1 – \alpha^2 + \beta\gamma = 0 \).
3. \( 1 – \alpha^2 – \beta\gamma = 0 \).
4. \( 1 + \alpha^2 – \beta\gamma = 0 \).

Solution:

(C) Given \( A = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} \).

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

1. A is a diagonal matrix.
2. A is a zero matrix.
3. A is a square matrix.
4. 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

1. A.
2. \( I – A \).
3. I.
4. 3A.

Solution:

(C) We are given that \( A^2 = A \).

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

\( = I + A^2 + 3A(I + A) – 7A \).

\( = I + A + 3A + 3A^2 – 7A \).

\( = I + 4A^2 – 4A \).

\( = I + 4A – 4A = I \).