CBSE Class 12 Mathematics HOTs Matrices and Determinants Set 03

Question. If \( A \) is a square matrix such that \( A^2 = I \), then find the simplified value of \( (A - I)^3 + (A + I)^3 - 7A \).
Answer: Given, \( A^2 = I \)
Consider, \( (A - I)^3 + (A + I)^3 - 7A \)
\( = A^3 - I^3 - 3A^2 I + 3AI^2 + A^3 + I^3 + 3A^2 I + 3AI^2 - 7A \)
\( = 2A^3 + 6AI^2 - 7A = 2AA^2 + 6AI - 7A \)
\( = 2AI + 6A - 7A = 2A + 6A - 7A = A \)

Question. If \( \begin{bmatrix} 2 & 1 & 3 \end{bmatrix} \begin{bmatrix} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} = A \), then write the order of matrix \( A \).
Answer: Given, \( A = \begin{bmatrix} 2 & 1 & 3 \end{bmatrix} \begin{bmatrix} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} \)
\( = \begin{bmatrix} -2-1 & 0+1+3 & -2+0+3 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} \)
\( = \begin{bmatrix} -3 & 4 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} \)
\( = \begin{bmatrix} -3 + 0 - 1 \end{bmatrix} = \begin{bmatrix} -4 \end{bmatrix} \)
\(\therefore\) The order of matrix \( A = 1 \times 1 \).

Question. Solve the following matrix equation for \( x \):
\( \begin{bmatrix} x & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -2 & 0 \end{bmatrix} = O \)
Answer: Given, \( \begin{bmatrix} x & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -2 & 0 \end{bmatrix} = O \)
\( \Rightarrow \begin{bmatrix} x - 2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \end{bmatrix} \)
\( \Rightarrow x - 2 = 0 \Rightarrow x = 2 \)

Question. If \( 2 \begin{bmatrix} 3 & 4 \\ 5 & x \end{bmatrix} + \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix} \), find \( (x - y) \).
Answer: We have, \( 2 \begin{bmatrix} 3 & 4 \\ 5 & x \end{bmatrix} + \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 6 & 8 \\ 10 & 2x \end{bmatrix} + \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 7 & 8 + y \\ 10 & 2x + 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix} \)
On comparing corresponding elements of the matrices, we get
\( 8 + y = 0 \) and \( 2x + 1 = 5 \Rightarrow y = -8 \) and \( x = 2 \)
\(\therefore x - y = 2 - (-8) = 2 + 8 = 10 \)

Question. If \( A \) is a square matrix such that \( A^2 = A \), then write the value of \( 7A - (I + A)^3 \), where \( I \) is an identity matrix.
Answer: Here \( A^2 = A \)
Now, \( 7A - (I + A)^3 \)
\( = 7A - (I + A)(I + A)(I + A) \)
\( = 7A - (I + A)(I \cdot I + I \cdot A + A \cdot I + (A \cdot A)) \)
\( = 7A - (I + A)(I + A + A + A) \) (since \( I \cdot A = A \cdot I = A \) and \( A^2 = A \))
\( = 7A - (I + A)(I + 3A) \)
\( = 7A - (I \cdot I + I \cdot 3A + A \cdot I + A \cdot (3A)) \)
\( = 7A - (I + 3A + A + 3A) \) (since \( A \cdot (3A) = 3A^2 = 3A \))
\( = 7A - I - 7A = -I \)

Question. If \( \begin{bmatrix} 2x & 4 \end{bmatrix} \begin{bmatrix} x \\ -8 \end{bmatrix} = 0 \), find the positive value of \( x \).
Answer: Here, \( \begin{bmatrix} 2x & 4 \end{bmatrix} \begin{bmatrix} x \\ -8 \end{bmatrix} = 0 \)
\( \Rightarrow 2x \cdot x + 4 \cdot (-8) = 0 \Rightarrow 2x^2 - 32 = 0 \)
\( \Rightarrow x^2 = 16 = 4^2 \Rightarrow x = 4 \) which is the required positive value of \( x \).

Question. If \( \begin{bmatrix} 9 & -1 & 4 \\ -2 & 1 & 3 \end{bmatrix} = A + \begin{bmatrix} 1 & 2 & -1 \\ 0 & 4 & 9 \end{bmatrix} \), then find the matrix \( A \).
Answer: \( A = \begin{bmatrix} 9 & -1 & 4 \\ -2 & 1 & 3 \end{bmatrix} - \begin{bmatrix} 1 & 2 & -1 \\ 0 & 4 & 9 \end{bmatrix} \)
\( \Rightarrow A = \begin{bmatrix} 9-1 & -1-2 & 4-(-1) \\ -2-0 & 1-4 & 3-9 \end{bmatrix} = \begin{bmatrix} 8 & -3 & 5 \\ -2 & -3 & -6 \end{bmatrix} \)

Question. If matrix \( A = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \) and \( A^2 = kA \), then write the value of \( k \).
Answer: Here, \( A = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \) and \( A^2 = kA \)
\( \Rightarrow \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} = k \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} = k \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \)
\( \Rightarrow 2 \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} = k \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \Rightarrow k = 2 \)

Question. If matrix \( A = \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} \) and \( A^2 = pA \), then write the value of \( p \).
Answer: Here, \( A = \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} \) and \( A^2 = pA \)
\( \Rightarrow \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} = p \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 8 & -8 \\ -8 & 8 \end{bmatrix} = p \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} \)
\( \Rightarrow 4 \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} = p \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} \Rightarrow p = 4 \)

Question. If matrix \( A = \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} \) and \( A^2 = \lambda A \), then write the value of \( \lambda \).
Answer: Here, \( A = \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} \) and \( A^2 = \lambda A \)
\( \Rightarrow \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} = \lambda \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 18 & -18 \\ -18 & 18 \end{bmatrix} = \lambda \begin{bmatrix} 3 & -3 & 3 \end{bmatrix} \)
\( \Rightarrow 6 \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} = \lambda \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} \Rightarrow \lambda = 6 \)

Question. If \( 2 \begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} \), then write the value of \( (x + y) \).
Answer: Given, \( 2 \begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 2 + y & 6 \\ 1 & 2x + 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} \)
\( \Rightarrow 2 + y = 5, 2x + 2 = 8 \Rightarrow y = 3, x = 3 \)
\(\dots x + y = 6 \)

Question. Simplify :
\( \cos\theta \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} + \sin\theta \begin{bmatrix} \sin\theta & -\cos\theta \\ \cos\theta & \sin\theta \end{bmatrix} \)
Answer: \( = \begin{bmatrix} \cos^2\theta & \cos\theta\sin\theta \\ -\sin\theta\cos\theta & \cos^2\theta \end{bmatrix} + \begin{bmatrix} \sin^2\theta & -\sin\theta\cos\theta \\ \sin\theta\cos\theta & \sin^2\theta \end{bmatrix} \)
\( = \begin{bmatrix} \cos^2\theta + \sin^2\theta & \cos\theta\sin\theta - \sin\theta\cos\theta \\ -\sin\theta\cos\theta + \sin\theta\cos\theta & \cos^2\theta + \sin^2\theta \end{bmatrix} \)
\( = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)

Question. If \( \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix} \begin{bmatrix} 1 & -3 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} -4 & 6 \\ -9 & x \end{bmatrix} \), write the value of \( x \).
Answer: Given, \( \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix} \begin{bmatrix} 1 & -3 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} -4 & 6 \\ -9 & x \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 2-6 & -6+12 \\ 5-14 & -15+28 \end{bmatrix} = \begin{bmatrix} -4 & 6 \\ -9 & x \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} -4 & 6 \\ -9 & 13 \end{bmatrix} = \begin{bmatrix} -4 & 6 \\ -9 & x \end{bmatrix} \Rightarrow x = 13 \)

Question. Find the value of \( x + y \) from the following equation: \( 2 \begin{bmatrix} x & 5 \\ 7 & y - 3 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix} \)
Answer: Given, \( 2 \begin{bmatrix} x & 5 \\ 7 & y - 3 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 2x & 10 \\ 14 & 2y - 6 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 2x + 3 & 6 \\ 15 & 2y - 4 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix} \)
\( \Rightarrow 2x + 3 = 7, 2y - 4 = 14 \)
\( \Rightarrow 2x = 4, 2y = 18 \Rightarrow x = 2, y = 9 \)
\(\therefore x + y = 2 + 9 = 11 \)

Question. If \( 3A - B = \begin{bmatrix} 5 & 0 \\ 1 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix} \), then find the matrix \( A \).
Answer: Given, \( 3A - B = \begin{bmatrix} 5 & 0 \\ 1 & 1 \end{bmatrix} \), \( B = \begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix} \)
\(\therefore (3A - B) + B = \begin{bmatrix} 5 & 0 \\ 1 & 1 \end{bmatrix} + \begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix} \)
\( \Rightarrow 3A = \begin{bmatrix} 5+4 & 0+3 \\ 1+2 & 1+5 \end{bmatrix} = \begin{bmatrix} 9 & 3 \\ 3 & 6 \end{bmatrix} \)
\( \Rightarrow 3A = 3 \begin{bmatrix} 3 & 1 \\ 1 & 2 \end{bmatrix} \)
\( \Rightarrow A = \begin{bmatrix} 3 & 1 \\ 1 & 2 \end{bmatrix} \)

Question. If \( \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ 2 & 5 \end{bmatrix} = \begin{bmatrix} 7 & 11 \\ k & 23 \end{bmatrix} \), then write the value of \( k \).
Answer: Given, \( \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ 2 & 5 \end{bmatrix} = \begin{bmatrix} 7 & 11 \\ k & 23 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 7 & 11 \\ 17 & 23 \end{bmatrix} = \begin{bmatrix} 7 & 11 \\ k & 23 \end{bmatrix} \)
On comparing the corresponding elements of matrices, we get \( k = 17 \)

Question. If \( A \) is a matrix of order \( 3 \times 4 \) and \( B \) is a matrix of order \( 4 \times 3 \), find the order of matrix \( (AB) \).
Answer: The order of matrix \( A \) is \( 3 \times 4 \).
The order of matrix \( B \) is \( 4 \times 3 \).
\(\therefore\) The order of matrix \( AB \) is \( 3 \times 3 \).

Question. Find the values of \( x \) and \( y \), if \( 2 \begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} \).
Answer: Given, \( 2 \begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 2 + y & 6 \\ 1 & 2x + 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} \)
On comparing corresponding elements, we get:
\( 2 + y = 5 \Rightarrow y = 3 \)
\( 2x + 2 = 8 \Rightarrow 2x = 6 \Rightarrow x = 3 \).

Question. If \( A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} \), then find \( A - B \).
Answer: \( A - B = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} = \begin{bmatrix} 2 - 1 & 4 - 3 \\ 3 - (-2) & 2 - 5 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 5 & -3 \end{bmatrix} \)

Question. If \( A = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} \), find \( A^2 - 5A + 4I \) and hence find a matrix \( X \) such that \( A^2 - 5A + 4I + X = O \).
Answer: \( A^2 = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} \)
\( = \begin{bmatrix} 2(2)+0(2)+1(1) & 2(0)+0(1)+1(-1) & 2(1)+0(3)+1(0) \\ 2(2)+1(2)+3(1) & 2(0)+1(1)+3(-1) & 2(1)+1(3)+3(0) \\ 1(2)+(-1)(2)+0(1) & 1(0)+(-1)(1)+0(-1) & 1(1)+(-1)(3)+0(0) \end{bmatrix} \)
\( = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix} \)
Now, \( A^2 - 5A + 4I = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix} - 5 \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} + 4 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix} - \begin{bmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{bmatrix} + \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix} \)
\( = \begin{bmatrix} 9 & -1 & 2 \\ 9 & 2 & 5 \\ 0 & -1 & 2 \end{bmatrix} - \begin{bmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{bmatrix} = \begin{bmatrix} -1 & -1 & -3 \\ -1 & -3 & -10 \\ -5 & 4 & 2 \end{bmatrix} \)
Since, \( A^2 - 5A + 4I + X = O \)
\( \Rightarrow X = -(A^2 - 5A + 4I) \)
\(\therefore X = \begin{bmatrix} 1 & 1 & 3 \\ 1 & 3 & 10 \\ 5 & -4 & -2 \end{bmatrix} \)

Question. Three schools \( A \), \( B \) and \( C \) organized a mela for collecting funds for helping the rehabilitation of flood victims. They sold hand made fans, mats and plates from recycled material at a cost of ₹ 25, ₹ 100 and ₹ 50 each. The number of articles sold are given below.

Find the funds collected by each school separately by selling the above articles. Also, find the total funds collected for the purpose. Write one value generated by the above situation.
Answer: The number of articles sold by each school can be written in the matrix form as:
\( X = \begin{bmatrix} 40 & 25 & 35 \\ 50 & 40 & 50 \\ 20 & 30 & 40 \end{bmatrix} \)
The cost of each article can be written in the matrix form as:
\( Y = \begin{bmatrix} 25 & 100 & 50 \end{bmatrix} \)
The fund collected by each school is given by:
\( YX = \begin{bmatrix} 25 & 100 & 50 \end{bmatrix} \begin{bmatrix} 40 & 25 & 35 \\ 50 & 40 & 50 \\ 20 & 30 & 40 \end{bmatrix} \)
\( = \begin{bmatrix} 7000 & 6125 & 7875 \end{bmatrix} \)
Therefore, the funds collected by schools \( A \), \( B \) and \( C \) are ₹ 7000, ₹ 6125 and ₹ 7875 respectively.
Thus, the total fund collected \( = ₹ (7000 + 6125 + 7875) = ₹ 21000 \).
The situation highlights the helping nature of the students.