CBSE Class 12 Mathematics HOTs Determinants Set 02

Question. Let \(A\) be a \(3 \times 3\) matrix such that \(|\text{adj } A| = 81\). Then, \(|A|\) is equal to
(a) 9 only
(b) -9 only
(c) 81
(d) 9 or -9
Answer: (d) 9 or -9

Question. If \(A = \begin{bmatrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{bmatrix}\), then \(|\text{adj } A|\) is equal to
(a) \(a^{27}\)
(b) \(a^9\)
(c) \(a^6\)
(d) \(a^2\)
Answer: (c) \(a^6\)

Question. If \(A\) is a square matrix of order 3 such that the value of \(|\text{adj } A| = 8\), then the value of \(|A^T|\) is
(a) \(\sqrt{2}\)
(b) \(-\sqrt{2}\)
(c) 8
(d) \(2\sqrt{2}\)
Answer: (d) \(2\sqrt{2}\)

Question. Let \(A\) be a square matrix of order 3. If \(|A| = 5\), then \(|\text{adj } A|\) is
(a) 5
(b) 125
(c) 25
(d) -5
Answer: (c) 25

Question. If \(A = \begin{bmatrix} 0 & 1 & -1 \\ 1 & 2 & 1 \\ 0 & 3 & -2 \end{bmatrix}\), then the value of \(|A \text{ adj } (A)|\) is
(a) -1
(b) 1
(c) 2
(d) 3
Answer: (a) -1

Question. If \(\begin{bmatrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 3 & a & 1 \end{bmatrix}\) is non-singular matrix and \(a \in \mathbb{R}\), then the set \(A\) is
(a) \(\mathbb{R}\)
(b) \(\{0\}\)
(c) \(\{4\}\)
(d) \(\mathbb{R} - \{4\}\)
Answer: (d) \(\mathbb{R} - \{4\}\)

Question. If \(A = \begin{bmatrix} 2 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{bmatrix}\), then \(A^{-1}\) exists, if
(a) \(\lambda = 2\)
(b) \(\lambda \neq 2\)
(c) \(\lambda \neq -2\)
(d) None of the options
Answer: (d) None of the options

Question. If \(|A| = 2\), where \(A\) is a \(2 \times 2\) matrix, then \(|4A^{-1}|\) equal to
(a) 4
(b) 2
(c) 8
(d) \(\frac{1}{32}\)
Answer: (c) 8

Question. For the matrix \(A = \begin{bmatrix} 2 & -1 & 1 \\ \lambda & 2 & 0 \\ 1 & -2 & 3 \end{bmatrix}\) to be invertible, the value of \(\lambda\) is
(a) 0
(b) 10
(c) \(\mathbb{R} - \{10\}\)
(d) \(\mathbb{R} - \{-10\}\)
Answer: (d) \(\mathbb{R} - \{-10\}\)

Question. If \(A\) and \(B\) are invertible square matrices of the same order, then which of the following is not correct?
(a) \(\text{adj } A = |A| \cdot A^{-1}\)
(b) \(\det(A^{-1}) = [\det(A)]^{-1}\)
(c) \((AB)^{-1} = B^{-1} A^{-1}\)
(d) \((A + B)^{-1} = B^{-1} + A^{-1}\)
Answer: (d) \((A + B)^{-1} = B^{-1} + A^{-1}\)

Question. If for a square matrix \(A\), \(A^2 - 3A + I = O\) and \(A^{-1} = xA + yI\), then the value of \(x + y\) is
(a) -2
(b) 2
(c) 3
(d) -3
Answer: (b) 2

Assertion-Reason Based Questions

Question. Assertion (A): If \(A\) is a non-singular matrix, then \(A^{-1}\) exist.
Reason (R): Determinant of a non-singular matrix is zero.
(a) Both A and R are correct; R is the correct explanation of A
(b) Both A and R are correct; R is not the correct explanation of A
(c) A is correct; R is incorrect
(d) R is correct; A is incorrect
Answer: (c) A is correct; R is incorrect

Question. Assertion (A): If \(|A| = 4\), then \(|A^{-1}| = -4\).
Reason (R): \(|A^{-1}| = \frac{1}{|A|}\).
(a) Both A and R are correct; R is the correct explanation of A
(b) Both A and R are correct; R is not the correct explanation of A
(c) A is correct; R is incorrect
(d) R is correct; A is incorrect
Answer: (d) R is correct; A is incorrect

Question. Assertion (A): If \(A\) is invertible \(3 \times 3\) matrix, then \(|A^{-1} \text{ adj } A| = |A|\).
Reason (R): If \(A\) and \(B\) are to invertible matrices such that \(B\) is the inverse of \(A\), then \(AB = BA = I\).
(a) Both A and R are correct; R is the correct explanation of A
(b) Both A and R are correct; R is not the correct explanation of A
(c) A is correct; R is incorrect
(d) R is correct; A is incorrect
Answer: (b) Both A and R are correct; R is not the correct explanation of A

Question. Assertion (A): If \(A = \begin{bmatrix} p & 0 & 0 \\ 0 & q & 0 \\ 0 & 0 & r \end{bmatrix}\), then \(A^{-1} = \begin{bmatrix} p^{-1} & 0 & 0 \\ 0 & q^{-1} & 0 \\ 0 & 0 & r^{-1} \end{bmatrix}\).
Reason (R): Inverse of a invertible diagonal matrix is always a diagonal matrix.
(a) Both A and R are correct; R is the correct explanation of A
(b) Both A and R are correct; R is not the correct explanation of A
(c) A is correct; R is incorrect
(d) R is correct; A is incorrect
Answer: (b) Both A and R are correct; R is not the correct explanation of A

Case Study Based Questions

Two organisations, P and Q, plan to reward their selected individuals for demonstrating the value of Integrity, Diligence and Timeliness. Organisation P plans to award ₹x, ₹y and ₹z for each of the three respective values to 3, 2 and 1 individuals, respectively with a total award amount ₹2200. Organisation Q plans to spend ₹3100 to reward 4, 1 and 3 individuals for the same values respectively with the same award amounts (₹x, ₹y and ₹z). Additionally, the total prize amount for one individual for all three values combined is ₹1200.

Question. Write the system of linear equation in variables x,y and z.
Answer: Based on the given information, we can translate the statements into the following system of linear equations:
\(3x + 2y + z = 2200\)
\(4x + y + 3z = 3100\)
\(x + y + z = 1200\)

Question. Write the matrix equation represented by the above situation.
Answer: The matrix equation of the system is \(AX = B\), where:
\(A = \begin{bmatrix} 3 & 2 & 1 \\ 4 & 1 & 3 \\ 1 & 1 & 1 \end{bmatrix}\), \(X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\), and \(B = \begin{bmatrix} 2200 \\ 3100 \\ 1200 \end{bmatrix}\).

Question. Find the values of x, y and z.
Answer: First, find the determinant \(|A|\):
\(|A| = 3(1 - 3) - 2(4 - 3) + 1(4 - 1) = 3(-2) - 2(1) + 1(3) = -6 - 2 + 3 = -5 \neq 0\).
Next, we calculate the cofactors of \(|A|\):
\(A_{11} = -2\), \(A_{12} = -1\), \(A_{13} = 3\)
\(A_{21} = -1\), \(A_{22} = 2\), \(A_{23} = -1\)
\(A_{31} = 5\), \(A_{32} = -5\), \(A_{33} = -5\)
The adjoint is:
\(\text{adj } A = \begin{bmatrix} -2 & -1 & 5 \\ -1 & 2 & -5 \\ 3 & -1 & -5 \end{bmatrix}\).
The inverse is:
\(A^{-1} = -\frac{1}{5} \begin{bmatrix} -2 & -1 & 5 \\ -1 & 2 & -5 \\ 3 & -1 & -5 \end{bmatrix}\).
Solving for \(X = A^{-1}B\):
\(X = -\frac{1}{5} \begin{bmatrix} -2 & -1 & 5 \\ -1 & 2 & -5 \\ 3 & -1 & -5 \end{bmatrix} \begin{bmatrix} 2200 \\ 3100 \\ 1200 \end{bmatrix} = \begin{bmatrix} 300 \\ 400 \\ 500 \end{bmatrix}\).
Hence, \(x = 300\), \(y = 400\), and \(z = 500\).

Question. Find the value of \(\frac{x + y}{z}\).
Answer: Substituting the calculated values \(x = 300\), \(y = 400\), and \(z = 500\):
\(\frac{x + y}{z} = \frac{300 + 400}{500} = \frac{700}{500} = \frac{7}{5}\).

Very Short Answer Type Questions

Question. If \(A_{ij}\) is the cofactor of the element \(a_{ij}\) of the determinant \(\begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{vmatrix}\), then write the value of \(a_{32} \cdot A_{32}\).
Answer: Here, \(a_{32} = 5\).
Given, \(A_{ij}\) is the cofactor of the element \(a_{ij}\) of \(\Delta\).
\(\therefore A_{32} = (-1)^{3+2} \begin{vmatrix} 2 & 5 \\ 6 & 4 \end{vmatrix} = -1(8 - 30) = 22\)
\(\therefore a_{32} \cdot A_{32} = 5 \times 22 = 110\)

Question. For what value of \(x\), \(A = \begin{bmatrix} 2(x+1) & 2x \\ x & x-2 \end{bmatrix}\) is a singular matrix?
Answer: We know that a matrix \(A\) is said to be singular, if \(|A| = 0\).
\(\therefore \begin{vmatrix} 2x+2 & 2x \\ x & x-2 \end{vmatrix} = 0\)
\(\Rightarrow (2x+2)(x-2) - 2x^2 = 0\)
\(\Rightarrow 2x^2 - 4x + 2x - 4 - 2x^2 = 0\)
\(\Rightarrow -2x - 4 = 0\)
\(\Rightarrow -2x = 4\)
\(\Rightarrow x = -2\)

Question. If \(A\) is an invertible matrix of order 2 and \(\det A = 4\), then write the value of \(\det(A^{-1})\).
Answer: Given, \(|A| = 4\).
We know that \(AA^{-1} = I\)
\(\Rightarrow |AA^{-1}| = |I|\)
\(\Rightarrow |A||A^{-1}| = |I|\)
\(\Rightarrow |A||A^{-1}| = 1\)
\(\Rightarrow |A^{-1}| = \frac{1}{|A|}\)
\(\therefore |A^{-1}| = \frac{1}{4}\)

Short Answer Type Questions

Question. If the value of a third order determinant is 12, then find the value of the determinant formed by replacing each element by its cofactor.
Answer: Given, \(|A| = 12\) and the order \(n = 3\).
We know that the determinant formed by replacing each element by its cofactor is equal to \(|\text{adj } A|\).
Using the property \(|\text{adj } A| = |A|^{n-1}\):
\(|\text{adj } A| = |A|^{3-1} = |A|^2 = (12)^2 = 144\).

Question. Find the adjoint of the matrix \(A = \begin{bmatrix} 1 & 4 & 5 \\ 3 & 2 & 6 \\ 0 & 1 & 0 \end{bmatrix}\) and hence show that \(A(\text{adj } A) = |A|I_3\).
Answer: We have, \(A = \begin{bmatrix} 1 & 4 & 5 \\ 3 & 2 & 6 \\ 0 & 1 & 0 \end{bmatrix}\).
First, find the determinant of \(A\):
\(|A| = 1(0 - 6) - 4(0 - 0) + 5(3 - 0) = -6 + 15 = 9\).
The cofactors of the elements of \(|A|\) are:
\(C_{11} = \begin{vmatrix} 2 & 6 \\ 1 & 0 \end{vmatrix} = -6\), \(C_{12} = -\begin{vmatrix} 3 & 6 \\ 0 & 0 \end{vmatrix} = 0\), \(C_{13} = \begin{vmatrix} 3 & 2 \\ 0 & 1 \end{vmatrix} = 3\)
\(C_{21} = -\begin{vmatrix} 4 & 5 \\ 1 & 0 \end{vmatrix} = 5\), \(C_{22} = \begin{vmatrix} 1 & 5 \\ 0 & 0 \end{vmatrix} = 0\), \(C_{23} = -\begin{vmatrix} 1 & 4 \\ 0 & 1 \end{vmatrix} = -1\)
\(C_{31} = \begin{vmatrix} 4 & 5 \\ 2 & 6 \end{vmatrix} = 24 - 10 = 14\), \(C_{32} = -\begin{vmatrix} 1 & 5 \\ 3 & 6 \end{vmatrix} = -(6 - 15) = 9\), \(C_{33} = \begin{vmatrix} 1 & 4 \\ 3 & 2 \end{vmatrix} = 2 - 12 = -10\).
\(\therefore \text{adj } A = \begin{bmatrix} C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32} \\ C_{13} & C_{23} & C_{33} \end{bmatrix} = \begin{bmatrix} -6 & 5 & 14 \\ 0 & 0 & 9 \\ 3 & -1 & -10 \end{bmatrix}\).
Now, we check \(A(\text{adj } A)\):
\(A(\text{adj } A) = \begin{bmatrix} 1 & 4 & 5 \\ 3 & 2 & 6 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} -6 & 5 & 14 \\ 0 & 0 & 9 \\ 3 & -1 & -10 \end{bmatrix} = \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix} = 9 I_3 = |A|I_3\).
Hence proved.

Question. If \(A\) is a non-singular symmetric matrix, then write whether \(A^{-1}\) is symmetric or skew-symmetric.
Answer: Let \(A\) be an invertible symmetric matrix. Then, \(|A| \neq 0\) and \(A^T = A\).
Now, \((A^{-1})^T = (A^T)^{-1} = A^{-1}\).
Thus, \(A^{-1}\) is a symmetric matrix.

Question. If \(A = \begin{bmatrix} 1 & -2 & 3 \\ 0 & -1 & 4 \\ -2 & 2 & 1 \end{bmatrix}\), then find \((A')^{-1}\).
Answer: We have, \(A = \begin{bmatrix} 1 & -2 & 3 \\ 0 & -1 & 4 \\ -2 & 2 & 1 \end{bmatrix}\).
Now, \(|A| = 1(-1 - 8) + 2(0 + 8) + 3(0 - 2) = -9 + 16 - 6 = 1 \neq 0\).
So, \(A\) is non-singular and its inverse exists.
Cofactors of elements of \(|A|\) are:
\(A_{11} = +(-1-8) = -9\), \(A_{12} = -(0+8) = -8\), \(A_{13} = +(0-2) = -2\)
\(A_{21} = -(-2-6) = 8\), \(A_{22} = +(1+6) = 7\), \(A_{23} = -(2-4) = 2\)
\(A_{31} = +(-8+3) = -5\), \(A_{32} = -(4-0) = -4\), \(A_{33} = +(-1-0) = -1\).
Thus, \(\text{adj } A = \begin{bmatrix} -9 & 8 & -5 \\ -8 & 7 & -4 \\ -2 & 2 & -1 \end{bmatrix}\).
Hence, \(A^{-1} = \frac{1}{|A|} \text{adj } A = \begin{bmatrix} -9 & 8 & -5 \\ -8 & 7 & -4 \\ -2 & 2 & -1 \end{bmatrix}\).
Now, \((A')^{-1} = (A^{-1})' = \begin{bmatrix} -9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1 \end{bmatrix}\).

Question. If for the non-singular matrix \(A\), \(A^2 = I\), then find \(A^{-1}\).
Answer: Given, \(A^2 = I\).
Since \(A\) is a non-singular matrix, \(|A| \neq 0\), so \(A^{-1}\) exists.
Pre-multiplying both sides by \(A^{-1}\):
\(A^{-1}(A^2) = A^{-1}I\)
\(\Rightarrow (A^{-1}A)A = A^{-1}\)
\(\Rightarrow IA = A^{-1}\)
\(\Rightarrow A^{-1} = A\).

Question. If \(A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}\), then find the value of \(\lambda\) so that \(A^2 = \lambda A - 2I\). Hence, find \(A^{-1}\).
Answer: We are given \(A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}\) and \(A^2 = \lambda A - 2I\).
First, find \(A^2\):
\(A^2 = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} = \begin{bmatrix} 9 - 8 & -6 + 4 \\ 12 - 8 & -8 + 4 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix}\).
Substitute into the equation \(\lambda A = A^2 + 2I\):
\(\lambda \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} + \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}\).
Comparing both sides, we get \(\lambda = 1\).
Thus, the matrix equation is \(A^2 = A - 2I\).
Pre-multiplying by \(A^{-1}\):
\(A^{-1}A^2 = A^{-1}A - 2A^{-1}I\)
\(\Rightarrow A = I - 2A^{-1}\)
\(\Rightarrow 2A^{-1} = I - A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} = \begin{bmatrix} -2 & 2 \\ -4 & 3 \end{bmatrix}\)
\(\dots A^{-1} = \frac{1}{2}\begin{bmatrix} -2 & 2 \\ -4 & 3 \end{bmatrix}\).

Question. If \(A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}\), then find \(A^{-1}\) and hence prove that \(A^2 - 4A - 5I = O\).
Answer: We have, \(|A| = 1(1-4) - 2(2-4) + 2(4-2) = -3 + 4 + 4 = 5 \neq 0\).
So \(A^{-1}\) exists.
The cofactors of the elements of \(|A|\) are:
\(A_{11} = -3\), \(A_{12} = 2\), \(A_{13} = 2\)
\(A_{21} = 2\), \(A_{22} = -3\), \(A_{23} = 2\)
\(A_{31} = 2\), \(A_{32} = 2\), \(A_{33} = -3\).
\(\therefore \text{adj } A = \begin{bmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{bmatrix}\).
\(\therefore A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{5} \begin{bmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{bmatrix}\).
To prove \(A^2 - 4A - 5I = O\):
Pre-multiplying both sides of the equation by \(A^{-1}\), we get:
\(A^{-1}(A^2 - 4A - 5I) = O \Rightarrow A - 4I - 5A^{-1} = O \Rightarrow A - 4I = 5A^{-1}\).
Now, \(A - 4I = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} - \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix} = \begin{bmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{bmatrix}\).
And \(5A^{-1} = 5 \cdot \frac{1}{5} \begin{bmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{bmatrix} = \begin{bmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{bmatrix}\).
Since \(A - 4I = 5A^{-1}\), it follows that \(A^2 - 4A - 5I = O\). Hence proved.

Long Answer Type Questions

Question. Find the matrix \(A\) satisfying the matrix equation \(\begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} A \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\).
Answer: Let \(B = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}\) and \(C = \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix}\).
The given matrix equation is \(BAC = I\).
We have, \(|B| = 4 - 3 = 1 \neq 0\) and \(|C| = 9 - 10 = -1 \neq 0\). Thus, both \(B^{-1}\) and \(C^{-1}\) exist.
Pre-multiplying both sides by \(B^{-1}\) in \(BAC = I\), we get:
\(B^{-1}(BAC) = B^{-1} I \Rightarrow (B^{-1}B)(AC) = B^{-1} \Rightarrow I(AC) = B^{-1} \Rightarrow AC = B^{-1}\).
Now post-multiplying both sides by \(C^{-1}\), we get:
\(AC C^{-1} = B^{-1} C^{-1} \Rightarrow A = B^{-1} C^{-1}\).
Let us find \(B^{-1}\) and \(C^{-1}\):
\(B^{-1} = \frac{1}{|B|} \text{adj } B = \frac{1}{1} \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}\).
\(C^{-1} = \frac{1}{|C|} \text{adj } C = \frac{1}{-1} \begin{bmatrix} -3 & -2 \\ -5 & -3 \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 5 & 3 \end{bmatrix}\).
From \(A = B^{-1} C^{-1}\), we get:
\(A = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} 3 & 2 \\ 5 & 3 \end{bmatrix} = \begin{bmatrix} 6-5 & 4-3 \\ -9+10 & -6+6 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}\).

Question. Let \(A = \begin{bmatrix} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix}\), verify that \((\text{adj } A)^{-1} = \text{adj}(A^{-1})\).
Answer: We have, \(A = \begin{bmatrix} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix}\).
First, find the determinant of \(A\):
\(|A| = 1(15 - 1) + 2(-10 - 1) + 1(-2 - 3) = 14 - 22 - 5 = -13 \neq 0\).
Since \(|A| \neq 0\), \(A\) is invertible and so are its adjoint and inverse.
Using the matrix property \(\text{adj } A = |A| A^{-1}\), taking the inverse of both sides gives:
\((\text{adj } A)^{-1} = (|A| A^{-1})^{-1} = \frac{1}{|A|} (A^{-1})^{-1} = \frac{1}{|A|} A\)     ...(i)
Using the same property for the matrix \(A^{-1}\), we have:
\(\text{adj}(A^{-1}) = |A^{-1}| (A^{-1})^{-1} = \frac{1}{|A|} A\)     ...(ii)
From (i) and (ii), we clearly see:
\((\text{adj } A)^{-1} = \text{adj}(A^{-1}) = -\frac{1}{13} \begin{bmatrix} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix}\). Hence verified.

Question. Find matrix \(A\), if \(\begin{bmatrix} 2 & 4 \\ 1 & 3 \end{bmatrix} A \begin{bmatrix} 0 & 2 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 6 \\ 3 & -2 \end{bmatrix}\).
Answer: Let \(B = \begin{bmatrix} 2 & 4 \\ 1 & 3 \end{bmatrix}\), \(C = \begin{bmatrix} 0 & 2 \\ 1 & 3 \end{bmatrix}\), and \(D = \begin{bmatrix} 1 & 6 \\ 3 & -2 \end{bmatrix}\).
The given equation is \(BAC = D \Rightarrow A = B^{-1} D C^{-1}\).
First, find \(B^{-1}\) and \(C^{-1}\):
\(|B| = 2\), hence \(B^{-1} = \frac{1}{2} \begin{bmatrix} 3 & -4 \\ -1 & 2 \end{bmatrix}\).
\(|C| = -2\), hence \(C^{-1} = -\frac{1}{2} \begin{bmatrix} 3 & -2 \\ -1 & 0 \end{bmatrix}\).
Now, calculate the product \(B^{-1} D\):
\(B^{-1} D = \frac{1}{2} \begin{bmatrix} 3 & -4 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 6 \\ 3 & -2 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 3 - 12 & 18 + 8 \\ -1 + 6 & -6 - 4 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} -9 & 26 \\ 5 & -10 \end{bmatrix}\).
Now, compute \(A = (B^{-1} D) C^{-1}\):
\(A = \frac{1}{2} \begin{bmatrix} -9 & 26 \\ 5 & -10 \end{bmatrix} \left( -\frac{1}{2} \begin{bmatrix} 3 & -2 \\ -1 & 0 \end{bmatrix} \right) = -\frac{1}{4} \begin{bmatrix} -9(3) + 26(-1) & -9(-2) + 26(0) \\ 5(3) + (-10)(-1) & 5(-2) + (-10)(0) \end{bmatrix}\)
\(= -\frac{1}{4} \begin{bmatrix} -53 & 18 \\ 25 & -10 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} 53 & -18 \\ -25 & 10 \end{bmatrix}\).

Question. If \(A = \begin{bmatrix} 1 & 2 & -3 \\ 3 & 2 & -2 \\ 2 & -1 & -1 \end{bmatrix}\), then find \(A^{-1}\) and use it to solve the system \(x + 3y + 2z = 6, 2x + 2y - z = 3 \text{ and } -3x - 2y - z = 5\).
Answer: We are given \(A = \begin{bmatrix} 1 & 2 & -3 \\ 3 & 2 & -2 \\ 2 & -1 & -1 \end{bmatrix}\).
First, find the determinant of \(A\):
\(|A| = 1(-2-2) - 2(-3+4) - 3(-3-4) = -4 - 2 + 21 = 15 \neq 0\).
The cofactors of the elements of \(|A|\) are:
\(A_{11} = -4\), \(A_{12} = -1\), \(A_{13} = -7\)
\(A_{21} = 5\), \(A_{22} = 5\), \(A_{23} = 5\)
\(A_{31} = 2\), \(A_{32} = -7\), \(A_{33} = -4\).
\(\therefore \text{adj } A = \begin{bmatrix} -4 & 5 & 2 \\ -1 & 5 & -7 \\ -7 & 5 & -4 \end{bmatrix}\).
Hence, \(A^{-1} = \frac{1}{15} \begin{bmatrix} -4 & 5 & 2 \\ -1 & 5 & -7 \\ -7 & 5 & -4 \end{bmatrix}\).
The given system of equations is:
\(x + 3y + 2z = 6\)
\(2x + 2y - z = 3\)
\(-3x - 2y - z = 5\)
In matrix form, the system is represented as \(A^T X = B\), where:
\(A^T = \begin{bmatrix} 1 & 3 & 2 \\ 2 & 2 & -1 \\ -3 & -2 & -1 \end{bmatrix}\), \(X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\), and \(B = \begin{bmatrix} 6 \\ 3 \\ 5 \end{bmatrix}\).
Therefore, \(X = (A^T)^{-1} B = (A^{-1})^T B\).
Using \((A^{-1})^T = \frac{1}{15} \begin{bmatrix} -4 & -1 & -7 \\ 5 & 5 & 5 \\ 2 & -7 & -4 \end{bmatrix}\), we compute:
\(X = \frac{1}{15} \begin{bmatrix} -4 & -1 & -7 \\ 5 & 5 & 5 \\ 2 & -7 & -4 \end{bmatrix} \begin{bmatrix} 6 \\ 3 \\ 5 \end{bmatrix} = \frac{1}{15} \begin{bmatrix} -24 - 3 - 35 \\ 30 + 15 + 25 \\ 12 - 21 - 20 \end{bmatrix} = \frac{1}{15} \begin{bmatrix} -62 \\ 70 \\ -29 \end{bmatrix}\).
\(\therefore x = -\frac{62}{15}\), \(y = \frac{14}{3}\), and \(z = -\frac{29}{15}\).

Question. Evaluate the product \(AB\), where \(A = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix}\) and \(B = \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix}\). Hence, solve the system of linear equations \(x - y = 3\), \(2x + 3y + 4z = 17\) and \(y + 2z = 7\).
Answer: We have, \(A = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix}\) and \(B = \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix}\).
Evaluating the product \(AB\):
\(AB = \begin{bmatrix} 1(2) - 1(-4) + 0 & 1(2) - 1(2) + 0 & 1(-4) - 1(-4) + 0 \\ 2(2) + 3(-4) + 4(2) & 2(2) + 3(2) + 4(-1) & 2(-4) + 3(-4) + 4(5) \\ 0 + 1(-4) + 2(2) & 0 + 1(2) + 2(-1) & 0 + 1(-4) + 2(5) \end{bmatrix}\)
\(= \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} = 6I\).
Since \(AB = 6I\), we have \(A \left(\frac{1}{6}B\right) = I\), which implies \(A^{-1} = \frac{1}{6} B\).
The system of equations can be written in matrix form as \(AX = D\), where:
\(A = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix}\), \(X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\), and \(D = \begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix}\).
Thus, the unique solution is given by \(X = A^{-1} D\):
\(X = \frac{1}{6} B D = \frac{1}{6} \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix} \begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix}\)
\(= \frac{1}{6} \begin{bmatrix} 6 + 34 - 28 \\ -12 + 34 - 28 \\ 6 - 17 + 35 \end{bmatrix} = \frac{1}{6} \begin{bmatrix} 12 \\ -6 \\ 24 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 4 \end{bmatrix}\).
\(\therefore x = 2\), \(y = -1\), and \(z = 4\).

Question. If \(A = \begin{bmatrix} 1 & 3 & 4 \\ 2 & 1 & 2 \\ 5 & 1 & 1 \end{bmatrix}\), then find \(A^{-1}\). Hence, solve the system of equations \(x + 3y + 4z = 8\), \(2x + y + 2z = 5\) and \(5x + y + z = 7\).
Answer: We have, \(A = \begin{bmatrix} 1 & 3 & 4 \\ 2 & 1 & 2 \\ 5 & 1 & 1 \end{bmatrix}\).
First, find the determinant of \(A\):
\(|A| = 1(1 - 2) - 3(2 - 10) + 4(2 - 5) = -1 + 24 - 12 = 11 \neq 0\).
Since \(|A| \neq 0\), \(A\) is invertible. Now find the cofactors of \(|A|\):
\(A_{11} = -1\), \(A_{12} = 8\), \(A_{13} = -3\)
\(A_{21} = 1\), \(A_{22} = -19\), \(A_{23} = 14\)
\(A_{31} = 2\), \(A_{32} = 6\), \(A_{33} = -5\).
\(\therefore \text{adj } A = \begin{bmatrix} -1 & 1 & 2 \\ 8 & -19 & 6 \\ -3 & 14 & -5 \end{bmatrix}\).
Hence, \(A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{11} \begin{bmatrix} -1 & 1 & 2 \\ 8 & -19 & 6 \\ -3 & 14 & -5 \end{bmatrix}\).
The system can be written in matrix form as \(AX = B\), where:
\(X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\) and \(B = \begin{bmatrix} 8 \\ 5 \\ 7 \end{bmatrix}\).
The solution is \(X = A^{-1} B\):
\(X = \frac{1}{11} \begin{bmatrix} -1 & 1 & 2 \\ 8 & -19 & 6 \\ -3 & 14 & -5 \end{bmatrix} \begin{bmatrix} 8 \\ 5 \\ 7 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} -8 + 5 + 14 \\ 64 - 95 + 42 \\ -24 + 70 - 35 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} 11 \\ 11 \\ 11 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\).
\(\therefore x = 1\), \(y = 1\), and \(z = 1\).

Question. Using matrices, solve the system of equations: \(4x + 3y + 2z = 60\), \(x + 2y + 3z = 45\) and \(6x + 2y + 3z = 70\).
Answer: The given system of equations can be written in matrix form as \(AX = B\), where:
\(A = \begin{bmatrix} 4 & 3 & 2 \\ 1 & 2 & 3 \\ 6 & 2 & 3 \end{bmatrix}\), \(X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\), and \(B = \begin{bmatrix} 60 \\ 45 \\ 70 \end{bmatrix}\).
First, find the determinant of \(A\):
\(|A| = 4(6-6) - 3(3-18) + 2(2-12) = 0 + 45 - 20 = 25 \neq 0\).
So, \(A\) is a non-singular matrix and \(A^{-1}\) exists.
The cofactors of the elements of \(|A|\) are:
\(C_{11} = 0\), \(C_{12} = 15\), \(C_{13} = -10\)
\(C_{21} = -5\), \(C_{22} = 0\), \(C_{23} = 10\)
\(C_{31} = 5\), \(C_{32} = -10\), \(C_{33} = 5\).
\(\therefore \text{adj } A = \begin{bmatrix} 0 & -5 & 5 \\ 15 & 0 & -10 \\ -10 & 10 & 5 \end{bmatrix}\).
Thus, \(A^{-1} = \frac{1}{25} \begin{bmatrix} 0 & -5 & 5 \\ 15 & 0 & -10 \\ -10 & 10 & 5 \end{bmatrix}\).
The solution of the system is given by \(X = A^{-1} B\):
\(X = \frac{1}{25} \begin{bmatrix} 0 & -5 & 5 \\ 15 & 0 & -10 \\ -10 & 10 & 5 \end{bmatrix} \begin{bmatrix} 60 \\ 45 \\ 70 \end{bmatrix} = \frac{1}{25} \begin{bmatrix} 0 - 225 + 350 \\ 900 + 0 - 700 \\ -600 + 450 + 350 \end{bmatrix} = \frac{1}{25} \begin{bmatrix} 125 \\ 200 \\ 200 \end{bmatrix} = \begin{bmatrix} 5 \\ 8 \\ 8 \end{bmatrix}\).
Comparing the corresponding elements, we get:
\(x = 5\), \(y = 8\), and \(z = 8\).