NCERT Solutions Class-12-Chapter-3-Matrices
Miscellaneous Exercise on Chapter 3
Note: This exercise covers various properties of matrices including symmetric and skew-symmetric matrices, matrix multiplication, and finding unknown matrices. Recall that for a symmetric matrix \( A' = A \) and for a skew-symmetric matrix \( A' = -A \).
Q1
If A and B are symmetric matrices, prove that \( AB - BA \) is a skew symmetric matrix.
Solution:
Given A and B are symmetric, so \( A' = A \) and \( B' = B \).
We need to check the transpose of \( (AB - BA) \).
\[ \begin{aligned} (AB - BA)' &= (AB)' - (BA)' \\ &= B'A' - A'B' \quad (\text{Since } (XY)' = Y'X') \\ &= BA - AB \quad (\text{Using given conditions}) \\ &= -(AB - BA) \end{aligned} \]
Since \( (AB - BA)' = -(AB - BA) \), the matrix \( AB - BA \) is skew symmetric.
Q2
Show that the matrix \( B'AB \) is symmetric or skew symmetric according as A is symmetric or skew symmetric.
Solution:
Let \( P = B'AB \). Then \( P' = (B'AB)' = B'A'(B')' = B'A'B \).
Case 1: If A is symmetric (\( A' = A \)).
\( P' = B'AB = P \). So, \( B'AB \) is symmetric.
Case 2: If A is skew symmetric (\( A' = -A \)).
\( P' = B'(-A)B = -B'AB = -P \). So, \( B'AB \) is skew symmetric.
Q3
Find the values of \( x, y, z \) if the matrix \( A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix} \) satisfy the equation \( A'A = I \).
Solution:
\( A' = \begin{bmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{bmatrix} \).
\[ A'A = \begin{bmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{bmatrix} \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix} = \begin{bmatrix} 2x^2 & 0 & 0 \\ 0 & 6y^2 & 0 \\ 0 & 0 & 3z^2 \end{bmatrix} \]
Given \( A'A = I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \).
Equating diagonal elements:
- \( 2x^2 = 1 \Rightarrow x = \pm \frac{1}{\sqrt{2}} \)
- \( 6y^2 = 1 \Rightarrow y = \pm \frac{1}{\sqrt{6}} \)
- \( 3z^2 = 1 \Rightarrow z = \pm \frac{1}{\sqrt{3}} \)
Q4
For what values of \( x \): \( \begin{bmatrix} 1 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{bmatrix} \begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} = O \)?
Solution:
First multiply the first two matrices:
\[ \begin{bmatrix} 1+4+1 & 2+0+0 & 0+2+2 \end{bmatrix} = \begin{bmatrix} 6 & 2 & 4 \end{bmatrix} \]
Now multiply with the third matrix:
\[ \begin{bmatrix} 6 & 2 & 4 \end{bmatrix} \begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} = [0 + 4 + 4x] = [4 + 4x] \]
Given this equals the zero matrix \( O \).
\( 4 + 4x = 0 \Rightarrow 4x = -4 \Rightarrow x = -1 \).
Q5
If \( A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \), show that \( A^2 - 5A + 7I = 0 \).
Solution:
Calculate \( A^2 \):
\[ A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 9-1 & 3+2 \\ -3-2 & -1+4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} \]
Calculate \( A^2 - 5A + 7I \):
\[ \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} 8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \]
Thus, proved.
Q6
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:
First product:
\[ \begin{bmatrix} x-2 & -10 & 2x-5-3 \end{bmatrix} = \begin{bmatrix} x-2 & -10 & 2x-8 \end{bmatrix} \]
Second product:
\[ \begin{bmatrix} x-2 & -10 & 2x-8 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} = [ x(x-2) - 40 + (2x-8) ] = [ x^2 - 2x - 40 + 2x - 8 ] = [ x^2 - 48 ] \]
\( x^2 - 48 = 0 \Rightarrow x^2 = 48 \Rightarrow x = \pm 4\sqrt{3} \).
Q7
A manufacturer produces three products x, y, z which he sells in two markets. Annual sales are indicated below:
Market I: 10,000, 2,000, 18,000
Market II: 6,000, 20,000, 8,000
(a) If unit sale prices of x, y and z are ₹ 2.50, ₹ 1.50 and ₹ 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 ₹ 2.00, ₹ 1.00 and 50 paise respectively. Find the gross profit.
Market I: 10,000, 2,000, 18,000
Market II: 6,000, 20,000, 8,000
(a) If unit sale prices of x, y and z are ₹ 2.50, ₹ 1.50 and ₹ 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 ₹ 2.00, ₹ 1.00 and 50 paise respectively. Find the gross profit.
Solution:
(a) Total Revenue
Sales Matrix \( S = \begin{bmatrix} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix} \).
Price Matrix \( P = \begin{bmatrix} 2.50 \\ 1.50 \\ 1.00 \end{bmatrix} \).
Revenue \( R = SP = \begin{bmatrix} 10000(2.5) + 2000(1.5) + 18000(1) \\ 6000(2.5) + 20000(1.5) + 8000(1) \end{bmatrix} \).
\( R = \begin{bmatrix} 25000 + 3000 + 18000 \\ 15000 + 30000 + 8000 \end{bmatrix} = \begin{bmatrix} 46000 \\ 53000 \end{bmatrix} \).
Market I Revenue: ₹46,000. Market II Revenue: ₹53,000.
(b) Gross Profit
Cost Matrix \( C = \begin{bmatrix} 2.00 \\ 1.00 \\ 0.50 \end{bmatrix} \).
Total Cost \( TC = SC = \begin{bmatrix} 10000(2) + 2000(1) + 18000(0.5) \\ 6000(2) + 20000(1) + 8000(0.5) \end{bmatrix} = \begin{bmatrix} 31000 \\ 36000 \end{bmatrix} \).
Profit = Revenue - Cost.
Market I Profit = \( 46000 - 31000 = 15000 \).
Market II Profit = \( 53000 - 36000 = 17000 \).
Q8
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:
Let the order of X be \( m \times n \). The second matrix is \( 2 \times 3 \). The result is \( 2 \times 3 \).
Thus, \( n = 2 \) and \( m = 2 \). Let \( X = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \).
\[ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} a+4b & 2a+5b & 3a+6b \\ c+4d & 2c+5d & 3c+6d \end{bmatrix} \]
Equating to RHS:
- \( a+4b = -7 \)
- \( 2a+5b = -8 \)
Solving these: multiply (1) by 2: \( 2a+8b=-14 \). Subtract (2): \( 3b = -6 \Rightarrow b = -2 \).
\( a + 4(-2) = -7 \Rightarrow a = 1 \).
Similarly for c and d: \( c+4d = 2 \) and \( 2c+5d = 4 \).
Solving yields \( d = 0, c = 2 \).
So, \( X = \begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix} \).
Q9
If \( A = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} \) is such that \( A^2 = I \), then- \( 1 + \alpha^2 + \beta\gamma = 0 \)
- \( 1 - \alpha^2 + \beta\gamma = 0 \)
- \( 1 - \alpha^2 - \beta\gamma = 0 \)
- \( 1 + \alpha^2 - \beta\gamma = 0 \)
- \( 1 + \alpha^2 + \beta\gamma = 0 \)
- \( 1 - \alpha^2 + \beta\gamma = 0 \)
- \( 1 - \alpha^2 - \beta\gamma = 0 \)
- \( 1 + \alpha^2 - \beta\gamma = 0 \)
Answer: (C)
Solution:
\[ A^2 = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} = \begin{bmatrix} \alpha^2 + \beta\gamma & 0 \\ 0 & \beta\gamma + \alpha^2 \end{bmatrix} \]
Since \( A^2 = I \): \( \alpha^2 + \beta\gamma = 1 \).
Rearranging: \( 1 - \alpha^2 - \beta\gamma = 0 \).
Q10
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
- A is a diagonal matrix
- A is a zero matrix
- A is a square matrix
- None of these
Answer: (B)
Solution:
If A is symmetric, \( A' = A \).
If A is skew symmetric, \( A' = -A \).
Thus, \( A = -A \Rightarrow 2A = O \Rightarrow A = O \).
So, A is a zero matrix.
Q11
If A is square matrix such that \( A^2 = A \), then \( (I + A)^3 - 7A \) is equal to- A
- I - A
- I
- 3A
- A
- I - A
- I
- 3A
Answer: (C)
Solution:
Expand \( (I+A)^3 = I^3 + A^3 + 3I^2A + 3IA^2 \).
Since \( I^n = I \) and \( IA = A \), this becomes \( I + A^3 + 3A + 3A^2 \).
Given \( A^2 = A \). So \( A^3 = A^2 \cdot A = A \cdot A = A^2 = A \).
Expression becomes: \( I + A + 3A + 3A = I + 7A \).
So, \( (I+A)^3 - 7A = (I + 7A) - 7A = I \).