Maharashtra Board Class 12 Maths Part 1 Chapter 2 Matrices 2.5 Solutions

Question 1. Apply the given elementary transformation on each of the following matrices:
(i)

, \(R_1 \leftrightarrow R_2\)
(ii)

, \(C_1 \leftrightarrow C_2\)
(iii)

\(3R_2\) and \(C_2 \to C_2 - 4C_1\)
Answer:
(i) Let \(A = \begin{pmatrix} 3 & -4 \\ 2 & 2 \end{pmatrix}\)
By \(R_1 \leftrightarrow R_2\), we get
\(A \sim \begin{pmatrix} 2 & 2 \\ 3 & -4 \end{pmatrix}\)
(ii) Let \(B = \begin{pmatrix} 2 & 4 \\ 1 & -5 \end{pmatrix}\)
By \(C_1 \leftrightarrow C_2\), we get
\(B \sim \begin{pmatrix} 4 & 2 \\ -5 & 1 \end{pmatrix}\)
(iii) Let \(C = \begin{pmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{pmatrix}\)
By \(3R_2\), we get
\(C \sim \begin{pmatrix} 3 & 1 & -1 \\ 3 & 9 & 3 \\ -1 & 1 & 3 \end{pmatrix}\)
By \(C_2 - 4C_1\) on C, we get
\(C \sim \begin{pmatrix} 3 & -11 & -1 \\ 1 & -1 & 1 \\ -1 & 5 & 3 \end{pmatrix}\)
In simple words: This question demonstrates how to perform elementary row and column transformations on matrices, which are fundamental operations used to simplify matrices or solve systems of linear equations. Each part applies a specific transformation like swapping rows, swapping columns, or scaling and subtracting columns.

🎯 Exam Tip: Mastering elementary transformations is crucial as they are the building blocks for finding inverses, solving systems of equations, and reducing matrices to simpler forms. Pay close attention to the row/column indices and the order of operations.

Question 2. Transform

into an upper triangular matrix by suitable row transformations.
Answer:
Let \(A = \begin{pmatrix} 1 & -1 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{pmatrix}\)
By \(R_2 - 2R_1\) and \(R_3 - 3R_1\), we get
\(A \sim \begin{pmatrix} 1 & -1 & 2 \\ 0 & 3 & -1 \\ 0 & 5 & -2 \end{pmatrix}\)
By \(\frac{1}{3}R_2\), we get
\(A \sim \begin{pmatrix} 1 & -1 & 2 \\ 0 & 1 & -1/3 \\ 0 & 5 & -2 \end{pmatrix}\)
By \(R_3 - 5R_2\), we get
\(A \sim \begin{pmatrix} 1 & -1 & 2 \\ 0 & 1 & -1/3 \\ 0 & 0 & -1/3 \end{pmatrix}\)
This is an upper triangular matrix.
In simple words: An upper triangular matrix has all elements below the main diagonal equal to zero. We achieve this by systematically applying row operations to eliminate elements in the lower triangle, transforming the original matrix step-by-step.

🎯 Exam Tip: When converting to an upper triangular matrix, always aim to make the elements below the main diagonal zero, working column by column from left to right. Gaussian elimination is the common method for this process.

Question 3. Find the cofactor matrix of the following matrices:
(i)

(ii)

Answer:
(i) Let \(A = \begin{pmatrix} 1 & 2 \\ 5 & -8 \end{pmatrix}\)
Here, \(a_{11} = 1\), \(M_{11} = -8\)
\(\therefore A_{11} = (-1)^{1+1} M_{11} = -8\)
\(a_{12} = 2\), \(M_{12} = 5\)
\(\therefore A_{12} = (-1)^{1+2} M_{12} = -1(5) = -5\)
\(a_{21} = 5\), \(M_{21} = 2\)
\(\therefore A_{21} = (-1)^{2+1} M_{21} = -1(2) = -2\)
\(a_{22} = -8\), \(M_{22} = 1\)
\(\therefore A_{22} = (-1)^{2+2} M_{22} = 1\).
\(\therefore \text{cofactor matrix} = \begin{pmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{pmatrix}\)
\( = \begin{pmatrix} -8 & -5 \\ -2 & 1 \end{pmatrix}\)
(ii) Let \(A = \begin{pmatrix} 5 & 8 & 7 \\ -1 & -2 & 1 \\ -2 & 1 & 1 \end{pmatrix}\)
The cofactor of \(a_{ij}\) is given by \(A_{ij} = (-1)^{i+j} M_{ij}\)
Now, \(M_{11} = \begin{vmatrix} -2 & 1 \\ 1 & 1 \end{vmatrix} = -2-1 = -3\)
\(\therefore A_{11} = (-1)^{1+1} M_{11} = 1(-3) = -3\)
\(M_{12} = \begin{vmatrix} -1 & 1 \\ -2 & 1 \end{vmatrix} = -1-(-2) = 1\)
\(\therefore A_{12} = (-1)^{1+2} M_{12} = -1(1) = -1\)
\(M_{13} = \begin{vmatrix} -1 & -2 \\ -2 & 1 \end{vmatrix} = -1-4 = -5\)
\(\therefore A_{13} = (-1)^{1+3} M_{13} = 1(-5) = -5\)
\(M_{21} = \begin{vmatrix} 8 & 7 \\ 1 & 1 \end{vmatrix} = 8-7 = 1\)
\(\therefore A_{21} = (-1)^{2+1} M_{21} = -1(1) = -1\)
\(M_{22} = \begin{vmatrix} 5 & 7 \\ -2 & 1 \end{vmatrix} = 5+14 = 19\)
\(\therefore A_{22} = (-1)^{2+2} M_{22} = 1(19) = 19\)
\(M_{23} = \begin{vmatrix} 5 & 8 \\ -2 & 1 \end{vmatrix} = 5-(-16) = 21\)
\(\therefore A_{23} = (-1)^{2+3} M_{23} = -1(21) = -21\)
\(M_{31} = \begin{vmatrix} 8 & 7 \\ -2 & 1 \end{vmatrix} = 8-(-14) = 22\)
\(\therefore A_{31} = (-1)^{3+1} M_{31} = 1(22) = 22\)
\(M_{32} = \begin{vmatrix} 5 & 7 \\ -1 & 1 \end{vmatrix} = 5-(-7) = 12\)
\(\therefore A_{32} = (-1)^{3+2} M_{32} = -1(12) = -12\)
\(M_{33} = \begin{vmatrix} 5 & 8 \\ -1 & -2 \end{vmatrix} = -10-(-8) = -2\)
\(\therefore A_{33} = (-1)^{3+3} M_{33} = 1(-2) = -2\)
\(\therefore \text{cofactor matrix} = \begin{pmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{pmatrix}\)
\( = \begin{pmatrix} -3 & -1 & -5 \\ -1 & 19 & -21 \\ 22 & -12 & -2 \end{pmatrix}\)
In simple words: The cofactor of an element in a matrix is found by calculating the determinant of the submatrix remaining after removing the row and column of that element, and then multiplying it by \((-1)^{i+j}\), where \(i\) and \(j\) are the row and column indices. The cofactor matrix is simply a matrix formed by replacing each element with its corresponding cofactor.

🎯 Exam Tip: Accuracy in calculating minor determinants and applying the sign change \((-1)^{i+j}\) is paramount for correctly finding the cofactor matrix. A single error can cascade through the entire calculation.

Question 4. Find the adjoint of the following matrices:
(i)

(ii)

Answer:
(i) Let \(A = \begin{pmatrix} 2 & -3 \\ 3 & 5 \end{pmatrix}\)
Here, \(a_{11} = 2\), \(M_{11} = 5\)
\(\therefore A_{11} = (-1)^{1+1}(5) = 5\)
\(a_{12} = -3\), \(M_{12} = 3\)
\(\therefore A_{12} = (-1)^{1+2}(3) = -3\)
\(a_{21} = 3\), \(M_{21} = -3\)
\(\therefore A_{21} = (-1)^{2+1}(-3) = 3\)
\(a_{22} = 5\), \(M_{22} = 2\)
\(\therefore A_{22} = (-1)^{2+2}(2) = 2\)
\(\therefore \text{the cofactor matrix} = \begin{pmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{pmatrix}\)
\( = \begin{pmatrix} 5 & -3 \\ 3 & 2 \end{pmatrix}\)
\(\therefore \text{adj A} = \begin{pmatrix} 5 & 3 \\ -3 & 2 \end{pmatrix}\)
(ii) Let \(A = \begin{pmatrix} 1 & -1 & 2 \\ -2 & 3 & 5 \\ -2 & 0 & -1 \end{pmatrix}\)
The cofactor of \(a_{ij}\) is given by \(A_{ij} = (-1)^{i+j} M_{ij}\)
Now, \(M_{11} = \begin{vmatrix} 3 & 5 \\ 0 & -1 \end{vmatrix} = -3-0 = -3\)
\(\therefore A_{11} = (-1)^{1+1}(-3) = -3\)
\(M_{12} = \begin{vmatrix} -2 & 5 \\ -2 & -1 \end{vmatrix} = 2+10 = 12\)
\(\therefore A_{12} = (-1)^{1+2}(12) = -12\)
\(M_{13} = \begin{vmatrix} -2 & 3 \\ -2 & 0 \end{vmatrix} = 0+6 = 6\)
\(\therefore A_{13} = (-1)^{1+3}(6) = 6\)
\(M_{21} = \begin{vmatrix} -1 & 2 \\ 0 & -1 \end{vmatrix} = 1-0 = 1\)
\(\therefore A_{21} = (-1)^{2+1}(1) = -1\)
\(M_{22} = \begin{vmatrix} 1 & 2 \\ -2 & -1 \end{vmatrix} = -1+4 = 3\)
\(\therefore A_{22} = (-1)^{2+2}(3) = 3\)
\(M_{23} = \begin{vmatrix} 1 & -1 \\ -2 & 0 \end{vmatrix} = 0-2 = -2\)
\(\therefore A_{23} = (-1)^{2+3}(-2) = 2\)
\(M_{31} = \begin{vmatrix} -1 & 2 \\ 3 & 5 \end{vmatrix} = -5-6 = -11\)
\(\therefore A_{31} = (-1)^{3+1}(-11) = -11\)
\(M_{32} = \begin{vmatrix} 1 & 2 \\ -2 & 5 \end{vmatrix} = 5+4 = 9\)
\(\therefore A_{32} = (-1)^{3+2}(9) = -9\)
\(M_{33} = \begin{vmatrix} 1 & -1 \\ -2 & 3 \end{vmatrix} = 3-2 = 1\)
\(\therefore A_{33} = (-1)^{3+3}(1) = 1\).
\(\therefore \text{the cofactor matrix} = \begin{pmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{pmatrix}\)
\( = \begin{pmatrix} -3 & -1 & -5 \\ -1 & 19 & -21 \\ 22 & -12 & -2 \end{pmatrix}\)
\(\therefore \text{adj A} = \begin{pmatrix} -3 & -1 & -11 \\ -12 & 3 & -9 \\ 6 & 2 & 1 \end{pmatrix}\)
In simple words: The adjoint of a matrix is the transpose of its cofactor matrix. This means you first find the cofactor of each element, arrange them into a cofactor matrix, and then swap its rows and columns to get the adjoint matrix.

🎯 Exam Tip: Remember that the adjoint matrix (adj A) is explicitly the transpose of the cofactor matrix. A common mistake is forgetting this transposition step, which leads to an incorrect result.

Question 5. Find the inverses of the following matrices by the adjoint method:
(i)

(ii)

(iii)

Answer:
(i) Let \(A = \begin{pmatrix} 3 & -1 \\ 2 & 1 \end{pmatrix}\)
Then \(|A| = \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix} = 3-(-2) = -1 \ne 0\)
\(\therefore A^{-1}\) exists.
First we have to find the cofactor matrix
\( = [A_{ij}]_{2 \times 2}\) where \(A_{ij} = (-1)^{i+j} M_{ij}\)
Now, \(A_{11} = (-1)^{1+1} M_{11} = -1\)
\(A_{12} = (-1)^{1+2} M_{12} = -2\)
\(A_{21} = (-1)^{2+1} M_{21} = -(-1) = 1\)
\(A_{22} = (-1)^{2+2} M_{22} = 3\)
\(\therefore \text{the cofactor matrix} = \begin{pmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{pmatrix}\)
\( = \begin{pmatrix} -1 & -2 \\ 1 & 3 \end{pmatrix}\)
\(\therefore \text{adj A} = \begin{pmatrix} -1 & 1 \\ -2 & 3 \end{pmatrix}\)
\(\therefore A^{-1} = \frac{1}{|A|} (\text{adj A}) = \frac{1}{-1} \begin{pmatrix} -1 & 1 \\ -2 & 3 \end{pmatrix}\)
\(\therefore A^{-1} = \begin{pmatrix} 1 & -1 \\ 2 & -3 \end{pmatrix}\)
(ii) Let \(A = \begin{pmatrix} 2 & -2 \\ 4 & 5 \end{pmatrix}\)
Then \(|A| = \begin{vmatrix} 2 & -2 \\ 4 & 5 \end{vmatrix} = 10+8 = 18 \ne 0\)
\(\therefore A^{-1}\) exists.
\(A_{11} = (-1)^{1+1} M_{11} = (1)(5) = 5\)
\(A_{12} = (-1)^{1+2} M_{12} = (-1)(4) = -4\)
\(A_{21} = (-1)^{2+1} M_{21} = (-1)(-2) = 2\)
\(A_{22} = (-1)^{2+2} M_{22} = (1)(2) = 2\)
\(\therefore \text{The matrix of the co-factors is}\)
\([A_{ij}]_{2 \times 2} = \begin{pmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{pmatrix}\)
\( = \begin{pmatrix} 5 & -4 \\ 2 & 2 \end{pmatrix}\)
Now \(\text{adj A} = [A_{ij}]^T_{2 \times 2} = \begin{pmatrix} 5 & 2 \\ -4 & 2 \end{pmatrix}\)
\(\therefore A^{-1} = \frac{1}{|A|} (\text{adj A})\)
\( = \frac{1}{18} \begin{pmatrix} 5 & 2 \\ -4 & 2 \end{pmatrix}\)
(iii) Let \(A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{pmatrix}\)
Then \(|A| = \begin{vmatrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{vmatrix}\)
\( = 1(10-0)-2(0-0)+3(0-0)\)
\( = 10 \ne 0\)
\(\therefore A^{-1}\) exist.
First we have to find the cofactor matrix
\( = [A_{ij}]_{3 \times 3}\), where \(A_{ij} = (-1)^{i+j} M_{ij}\)
Now, \(A_{11} = (-1)^{1+1} M_{11} = \begin{vmatrix} 2 & 4 \\ 0 & 5 \end{vmatrix} = 10-0 = 10\)
\(A_{12} = (-1)^{1+2} M_{12} = -\begin{vmatrix} 0 & 4 \\ 0 & 5 \end{vmatrix} = -(0-0) = 0\)
\(A_{13} = (-1)^{1+3} M_{13} = \begin{vmatrix} 0 & 2 \\ 0 & 0 \end{vmatrix} = 0-0 = 0\)
\(A_{21} = (-1)^{2+1} M_{21} = -\begin{vmatrix} 2 & 3 \\ 0 & 5 \end{vmatrix} = -(10-0) = -10\)
\(A_{22} = (-1)^{2+2} M_{22} = \begin{vmatrix} 1 & 3 \\ 0 & 5 \end{vmatrix} = 5-0 = 5\)
\(A_{23} = (-1)^{2+3} M_{23} = -\begin{vmatrix} 1 & 2 \\ 0 & 0 \end{vmatrix} = -(0-0) = 0\)
\(A_{31} = (-1)^{3+1} M_{31} = \begin{vmatrix} 2 & 3 \\ 2 & 4 \end{vmatrix} = 8-6 = 2\)
\(A_{32} = (-1)^{3+2} M_{32} = -\begin{vmatrix} 1 & 3 \\ 0 & 4 \end{vmatrix} = -(4-0) = -4\)
\(A_{33} = (-1)^{3+3} M_{33} = \begin{vmatrix} 1 & 2 \\ 0 & 2 \end{vmatrix} = 2-0 = 2\)
\(\therefore \text{the cofactor matrix}\)
\( = \begin{pmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{pmatrix}\)
\( = \begin{pmatrix} 10 & 0 & 0 \\ -10 & 5 & 0 \\ 2 & -4 & 2 \end{pmatrix}\)
\(\therefore \text{adj A} = \begin{pmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{pmatrix}\)
\(\therefore A^{-1} = \frac{1}{|A|} (\text{adj A})\)
\(A^{-1} = \frac{1}{10} \begin{pmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{pmatrix}\)
In simple words: The inverse of a matrix using the adjoint method involves three main steps: calculating the determinant of the matrix, finding its cofactor matrix, and then computing the transpose of the cofactor matrix (which is the adjoint). Finally, divide the adjoint matrix by the determinant. If the determinant is zero, the inverse does not exist.

🎯 Exam Tip: The adjoint method for finding the inverse is calculation-intensive. Double-check your determinant, cofactor matrix, and adjoint matrix before the final division. A small error in any step will lead to an incorrect inverse.

Question 6. Find the inverses of the following matrices by the transformation method:
(i)

(ii)

Answer:
(i) Let \(A = \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix}\)
Then \(|A| = \begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} = -1-4 = -5 \ne 0\)
\(\therefore A^{-1}\) exists.
\(A^{-1}\) by Row transformations:
We write \(AA^{-1} = I\)
\(\therefore \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix} A^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)
By \(R_2 - 2R_1\), we get
\(\begin{pmatrix} 1 & 2 \\ 0 & -5 \end{pmatrix} A^{-1} = \begin{pmatrix} 1 & 0 \\ -2 & 1 \end{pmatrix}\)
By \((-\frac{1}{5})R_2\), we get
\(\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} A^{-1} = \begin{pmatrix} 1 & 0 \\ 2/5 & -1/5 \end{pmatrix}\)
By \(R_1 - 2R_2\), we get
\(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} A^{-1} = \begin{pmatrix} 1 & 2 \\ 2/5 & -1/5 \end{pmatrix}\)
\(\therefore A^{-1} = \frac{1}{5} \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix}\) ...(1)
\(A^{-1}\) by Column transformations:
We write \(AA^{-1} = I\)
\(\therefore A^{-1} \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)
By \(C_2 - 2C_1\), we get
\(A^{-1} \begin{pmatrix} 1 & 0 \\ 2 & -5 \end{pmatrix} = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}\)
By \((-\frac{1}{5})C_2\), we get
\(A^{-1} \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 2/5 \\ 0 & -1/5 \end{pmatrix}\)
By \(C_1 - 2C_2\), we get
\(A^{-1} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 2/5 & -1/5 \end{pmatrix}\)
\(\therefore A^{-1} = \frac{1}{5} \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix}\) ...(2)
From (1) and (2), which is unique.
(ii) Let \(A = \begin{pmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{pmatrix}\)
Then \(|A| = \begin{vmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{vmatrix}\)
\( = 2(3-0)-0(15-0)-1(5-0)\)
\( = 6-0-5 = 1 \ne 0\)
\(\therefore A^{-1}\) exists.
We write \(AA^{-1} = I\)
\(\begin{pmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{pmatrix} A^{-1} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\)
By \(3R_1\), we get
\(\begin{pmatrix} 6 & 0 & -3 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{pmatrix} A^{-1} = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\)
By \(R_1 - R_2\), we get
\(\begin{pmatrix} 1 & -1 & -3 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{pmatrix} A^{-1} = \begin{pmatrix} 3 & -1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\)
By \(R_2 - 5R_1\), we get
\(\begin{pmatrix} 1 & -1 & -3 \\ 0 & 6 & 15 \\ 0 & 1 & 3 \end{pmatrix} A^{-1} = \begin{pmatrix} 3 & -1 & 0 \\ -15 & 6 & 0 \\ 0 & 0 & 1 \end{pmatrix}\)
By \(R_2 - 5R_3\), we get
\(\begin{pmatrix} 1 & -1 & -3 \\ 0 & 1 & 0 \\ 0 & 1 & 3 \end{pmatrix} A^{-1} = \begin{pmatrix} 3 & -1 & 0 \\ -15 & 6 & -5 \\ 0 & 0 & 1 \end{pmatrix}\)
By \(R_1 + R_2\) and \(R_3 - R_2\), we get
\(\begin{pmatrix} 1 & 0 & -3 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{pmatrix} A^{-1} = \begin{pmatrix} -12 & 5 & -5 \\ -15 & 6 & -5 \\ 15 & -6 & 6 \end{pmatrix}\)
By \((\frac{1}{3})R_3\), we get
\(\begin{pmatrix} 1 & 0 & -3 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} A^{-1} = \begin{pmatrix} -12 & 5 & -5 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{pmatrix}\)
By \(R_1 + 3R_3\), we get
\(\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} A^{-1} = \begin{pmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{pmatrix}\)
\(\therefore A^{-1} = \begin{pmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{pmatrix}\)
In simple words: The transformation method for finding the inverse of a matrix involves augmented matrices where the given matrix A is placed next to an identity matrix I, forming \([A|I]\). Elementary row (or column) operations are then applied to transform A into I. The same operations simultaneously transform I into \(A^{-1}\), resulting in \([I|A^{-1}]\).

🎯 Exam Tip: When using the transformation method, keep your goal in mind: convert A to an identity matrix while simultaneously performing the same operations on the identity matrix. Maintain consistency (either pure row or pure column operations) to avoid errors.

Question 7. Find the inverse of \(A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{pmatrix}\) by elementary column transformations.
Answer:
\(A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{pmatrix}\)
\(\therefore |A| = \begin{vmatrix} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{vmatrix}\)
\( = 1(2-6)-0+1(0-2)\)
\( = -4-2 = -6 \ne 0\)
\(\therefore A^{-1}\) exists.
We write \(A A^{-1} = I\)
\(\therefore \begin{pmatrix} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{pmatrix} A^{-1} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\)
By \(C_3 - C_1\), we get
\(\begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 3 \\ 1 & 2 & 0 \end{pmatrix} A^{-1} = \begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\)
By \((\frac{1}{2})C_2\), we get
\(\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 3 \\ 1 & 1 & 0 \end{pmatrix} A^{-1} = \begin{pmatrix} 1 & 0 & -1 \\ 0 & 1/2 & 0 \\ 0 & 0 & 1 \end{pmatrix}\)
By \(C_3 - 3C_2\), we get
\(\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & -3 \end{pmatrix} A^{-1} = \begin{pmatrix} 1 & 0 & -1 \\ 0 & 1/2 & -3/2 \\ 0 & 0 & 1 \end{pmatrix}\)
By \((-\frac{1}{3})C_3\), we get
\(\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & 1 \end{pmatrix} A^{-1} = \begin{pmatrix} 1 & 0 & 1/3 \\ 0 & 1/2 & 1/2 \\ 0 & 0 & -1/3 \end{pmatrix}\)
By \(C_1 - C_3\) and \(C_2 - C_3\), we get
\(\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} A^{-1} = \begin{pmatrix} 2/3 & -1/3 & 1/3 \\ -1/2 & 0 & 1/2 \\ 1/3 & 1/3 & -1/3 \end{pmatrix}\)
\(\therefore A^{-1} = \frac{1}{6} \begin{pmatrix} 4 & -2 & 2 \\ -3 & 0 & 3 \\ 2 & 2 & -2 \end{pmatrix}\)
In simple words: To find the inverse using elementary column transformations, you start with the augmented matrix \(\begin{pmatrix} A \\ I \end{pmatrix}\) and perform column operations to convert A into the identity matrix. The same operations applied to the original identity matrix will yield \(A^{-1}\).

🎯 Exam Tip: When using column transformations, ensure you apply them consistently across all columns of the augmented matrix. Be careful not to mix row and column operations within the same problem, as this will lead to an incorrect inverse.

Question 7. Find the inverse of A = \( \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{bmatrix} \) by elementary column transformations.
Answer: Solution: Let A = \( \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{bmatrix} \)
Therefore, \( |A| = \begin{vmatrix} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{vmatrix} \)
= \( 1(2-6) - 0 + 1(0-2) \)
= \( -4 - 2 = -6 \neq 0 \)
Therefore, \( A^{-1} \) exists. We write \( A^{-1}A = I \)
\( A^{-1} \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
By \( C_3-C_1 \), we get
\( A^{-1} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 3 \\ 1 & 2 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
By \( \left(\frac{1}{2}\right)C_2 \), we get
\( A^{-1} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 3/2 \\ 1 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1/2 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
By \( C_3-3C_2 \), we get
\( A^{-1} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1/2 & -3/2 \\ 0 & 0 & 1 \end{bmatrix} \)
By \( \left(-\frac{1}{3}\right)C_3 \), we get
\( A^{-1} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 1/3 \\ 0 & 1/2 & 1/2 \\ 0 & 0 & -1/3 \end{bmatrix} \)
By \( C_1-C_3 \) and \( C_2-C_3 \), we get
\( A^{-1} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 1 & 1/3 \\ -3 & -3 & 1 \\ 4 & -2 & -1/3 \end{bmatrix} \)
Therefore, \( A^{-1} = \frac{1}{6} \begin{bmatrix} 4 & 2 & 2 \\ -6 & -6 & 6 \\ 8 & -4 & -2 \end{bmatrix} \)
In simple words: To find the inverse of matrix A using elementary column transformations, we apply column operations to transform A into the identity matrix I, while simultaneously applying the same operations to an identity matrix I, which then becomes the inverse matrix \( A^{-1} \).

🎯 Exam Tip: Mastering elementary transformations is crucial for accurately finding matrix inverses; remember to apply operations consistently to both matrices to avoid errors.

Question 8. Find the inverse of \( \begin{bmatrix} 1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7 \end{bmatrix} \) by the elementary row transformations.
Answer: Solution: Let A = \( \begin{bmatrix} 1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7 \end{bmatrix} \)
Then \( |A|= \begin{vmatrix} 1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7 \end{vmatrix} \)
= \( 1(7-20) - 2(7-10) + 3(4-2) \)
= \( -13 + 6 + 6 = -1 \neq 0 \)
Therefore, \( A^{-1} \) exists. We write \( AA^{-1} = I \)
\( \begin{bmatrix} 1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7 \end{bmatrix} A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
By \( R_2-R_1 \) and \( R_3 -2R_1 \), we get
\( \begin{bmatrix} 1 & 2 & 3 \\ 0 & -1 & 2 \\ 0 & 0 & 1 \end{bmatrix} A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ -2 & 0 & 1 \end{bmatrix} \)
By \( (-1)R_2 \), we get
\( \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{bmatrix} A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 1 & -1 & 0 \\ -2 & 0 & 1 \end{bmatrix} \)
By \( R_1-2R_2 \), we get
\( \begin{bmatrix} 1 & 0 & 7 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{bmatrix} A^{-1} = \begin{bmatrix} -1 & 2 & 0 \\ 1 & -1 & 0 \\ -2 & 0 & 1 \end{bmatrix} \)
By \( R_1-7R_3 \) and \( R_2 + 2R_3 \), we get,
\( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A^{-1} = \begin{bmatrix} 15 & 2 & -7 \\ -3 & -1 & 2 \\ -2 & 0 & 1 \end{bmatrix} \)
Therefore, \( A^{-1} = \begin{bmatrix} 15 & 2 & -7 \\ -3 & -1 & 2 \\ -2 & 0 & 1 \end{bmatrix} \)
In simple words: To find the inverse of a matrix using elementary row transformations, we apply row operations to transform the original matrix into an identity matrix, while simultaneously performing the same operations on an identity matrix to obtain the inverse.

🎯 Exam Tip: Pay close attention to the order of operations and arithmetic to avoid calculation errors, as one mistake can propagate through the entire transformation process.

Question 9. If A = \( \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{bmatrix} \) and B = \( \begin{bmatrix} 1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7 \end{bmatrix} \), then find matrix X such that XA = B.
Answer: Solution: XA = B
Therefore, \( X \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7 \end{bmatrix} \)
By \( C_3-C_1 \), we get
\( X \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 3 \\ 1 & 2 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 2 \\ 1 & 1 & 4 \\ 2 & 4 & 5 \end{bmatrix} \)
By \( \left(\frac{1}{2}\right)C_2 \), we get
\( X \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 3/2 \\ 1 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 2 \\ 1 & 1/2 & 4 \\ 2 & 2 & 5 \end{bmatrix} \)
By \( C_3-3C_2 \), we get
\( X \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 1 & -1 \\ 1 & 1/2 & 5/2 \\ 2 & 2 & -1 \end{bmatrix} \)
By \( \left(-\frac{1}{3}\right)C_3 \), we get
\( X \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1/3 \\ 1 & 1/2 & -5/6 \\ 2 & 2 & 1/3 \end{bmatrix} \)
By \( C_1-C_3 \) and \( C_2-C_3 \), we get
\( X \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2/3 & 2/3 & 1/3 \\ 11/6 & 4/3 & -5/6 \\ 5/3 & 5/3 & 1/3 \end{bmatrix} \)
Therefore, \( X = \frac{1}{6} \begin{bmatrix} 4 & 4 & 2 \\ 11 & 8 & -5 \\ 10 & 10 & 2 \end{bmatrix} \)
In simple words: To find matrix X such that XA = B, we perform column transformations on the augmented matrix [A|B] to transform A into the identity matrix I, thereby transforming B into X.

🎯 Exam Tip: When solving matrix equations like XA=B, ensure transformations are applied consistently to both matrices; column operations are appropriate when the unknown matrix X is pre-multiplying A.

Question 10. Find matrix X, if AX = B, where A = \( \begin{bmatrix} 1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4 \end{bmatrix} \) and B = \( \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \).
Answer: Solution: AX = B
Therefore, \( \begin{bmatrix} 1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4 \end{bmatrix} X = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \)
By \( R_2 + R_1 \) and \( R_3 - R_1 \), we get
\( \begin{bmatrix} 1 & 2 & 3 \\ 0 & 3 & 5 \\ 0 & 0 & 1 \end{bmatrix} X = \begin{bmatrix} 1 \\ 3 \\ 2 \end{bmatrix} \)
By \( \left(\frac{1}{3}\right)R_2 \), we get
\( \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 5/3 \\ 0 & 0 & 1 \end{bmatrix} X = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} \)
By \( R_1-2R_2 \), we get
\( \begin{bmatrix} 1 & 0 & -1/3 \\ 0 & 1 & 5/3 \\ 0 & 0 & 1 \end{bmatrix} X = \begin{bmatrix} -1 \\ 1 \\ 2 \end{bmatrix} \)
By \( R_1 + \frac{1}{3}R_3 \) and \( R_2-\frac{5}{3}R_3 \), we get
\( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} X = \begin{bmatrix} -1/3 \\ -7/3 \\ 2 \end{bmatrix} \)
Therefore, \( X = \begin{bmatrix} -1/3 \\ -7/3 \\ 2 \end{bmatrix} \)
In simple words: To find matrix X in the equation AX = B, row transformations are applied to the augmented matrix [A|B] to transform A into the identity matrix, which simultaneously transforms B into X.

🎯 Exam Tip: For matrix equations of the form AX=B, use elementary row transformations on the augmented matrix [A|B] to convert A to identity; column transformations would be used if the equation was XA=B.