OP Malhotra Class 12 Maths Solutions Chapter 6 Matrices Exercise 6 (I)

Question 1. Using elementary transformation, find the inverse of the following matrices, if it exists. A = \( \left[\begin{array}{rr} 1 & -1 \\ 2 & 3 \end{array}\right] \)
Answer:
We are given the matrix \( A = \left[\begin{array}{rr} 1 & -1 \\ 2 & 3 \end{array}\right] \). To find its inverse using elementary transformations, we start by writing \( A = IA \), where \( I \) is the identity matrix.
So, \( \left[\begin{array}{rr} 1 & -1 \\ 2 & 3 \end{array}\right]=\left[\begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}\right]A \)
Now, we perform row operations to transform the left matrix into the identity matrix. The same operations are applied to the right matrix, which will then become \( A^{-1} \).
First, we operate \( R_2 \rightarrow R_2 - 2R_1 \):
\( \left[\begin{array}{rr} 1 & -1 \\ 2 - 2(1) & 3 - 2(-1) \end{array}\right]=\left[\begin{array}{rr} 1 & 0 \\ 0 - 2(1) & 1 - 2(0) \end{array}\right]A \)
\( \left[\begin{array}{rr} 1 & -1 \\ 0 & 5 \end{array}\right]=\left[\begin{array}{rr} 1 & 0 \\ -2 & 1 \end{array}\right]A \)
Next, we operate \( R_2 \rightarrow \frac { 1 }{ 5 }R_2 \):
\( \left[\begin{array}{rr} 1 & -1 \\ 0/5 & 5/5 \end{array}\right]=\left[\begin{array}{rr} 1 & 0 \\ -2/5 & 1/5 \end{array}\right]A \)
\( \left[\begin{array}{rr} 1 & -1 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{rr} 1 & 0 \\ -2/5 & 1/5 \end{array}\right]A \)
Finally, we operate \( R_1 \rightarrow R_1 + R_2 \):
\( \left[\begin{array}{rr} 1 + 0 & -1 + 1 \end{array}\right]=\left[\begin{array}{rr} 1 + (-2/5) & 0 + 1/5 \end{array}\right]A \)
\( \left[\begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 3/5 & 1/5 \\ -2/5 & 1/5 \end{array}\right]A \)
Since the left side is now the identity matrix, the matrix on the right is the inverse of A.
Therefore, \( A^{-1} = \left[\begin{array}{cc} 3/5 & 1/5 \\ -2/5 & 1/5 \end{array}\right] \). This method is often used to solve systems of linear equations.
In simple words: We changed the given matrix step by step using simple row operations until it became the identity matrix. The same changes made to the identity matrix on the other side gave us the inverse matrix.

🎯 Exam Tip: Remember to apply each row operation to both sides of the equation \( A = IA \) correctly. A common mistake is to apply it only to one side or to forget to multiply/divide all elements in the row.

Question 2. Find the inverse of the matrix A = \( \left[\begin{array}{rr} 1 & 2 \\ 2 & -1 \end{array}\right] \) using elementary transformation, if it exists.
Answer:
We are given the matrix \( A = \left[\begin{array}{rr} 1 & 2 \\ 2 & -1 \end{array}\right] \). To find its inverse using elementary transformations, we write \( A = IA \), where \( I \) is the identity matrix.
So, \( \left[\begin{array}{rr} 1 & 2 \\ 2 & -1 \end{array}\right]=\left[\begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}\right]A \)
We will transform the left matrix into the identity matrix using row operations.
First, operate \( R_2 \rightarrow R_2 - 2R_1 \):
\( \left[\begin{array}{rr} 1 & 2 \\ 2 - 2(1) & -1 - 2(2) \end{array}\right]=\left[\begin{array}{rr} 1 & 0 \\ 0 - 2(1) & 1 - 2(0) \end{array}\right]A \)
\( \left[\begin{array}{rr} 1 & 2 \\ 0 & -5 \end{array}\right]=\left[\begin{array}{rr} 1 & 0 \\ -2 & 1 \end{array}\right]A \)
Next, operate \( R_2 \rightarrow \frac { -1 }{ 5 }R_2 \):
\( \left[\begin{array}{rr} 1 & 2 \\ 0/(-5) & -5/(-5) \end{array}\right]=\left[\begin{array}{rr} 1 & 0 \\ -2/(-5) & 1/(-5) \end{array}\right]A \)
\( \left[\begin{array}{rr} 1 & 2 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{rr} 1 & 0 \\ 2/5 & -1/5 \end{array}\right]A \)
Finally, operate \( R_1 \rightarrow R_1 - 2R_2 \):
\( \left[\begin{array}{rr} 1 - 2(0) & 2 - 2(1) \end{array}\right]=\left[\begin{array}{rr} 1 - 2(2/5) & 0 - 2(-1/5) \end{array}\right]A \)
\( \left[\begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{rr} 1 - 4/5 & 0 + 2/5 \\ 2/5 & -1/5 \end{array}\right]A \)
\( \left[\begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{rr} 1/5 & 2/5 \\ 2/5 & -1/5 \end{array}\right]A \)
Thus, the inverse of A is \( A^{-1} = \left[\begin{array}{rr} 1/5 & 2/5 \\ 2/5 & -1/5 \end{array}\right] \). The identity matrix \(I\) has ones on its main diagonal and zeros everywhere else.
In simple words: We used basic row changes to turn the starting matrix into a special matrix called the identity matrix. The same changes applied to the identity matrix on the other side gave us the inverse matrix we were looking for.

🎯 Exam Tip: Pay close attention to negative signs and fractions during calculations. Even a small error can lead to a completely wrong inverse matrix.

Question 3. Find the inverse of the matrix A = \( \left[\begin{array}{rr} 9 & 5 \\ 7 & 4 \end{array}\right] \) using elementary transformation, if it exists.
Answer:
Given the matrix \( A = \left[\begin{array}{rr} 9 & 5 \\ 7 & 4 \end{array}\right] \). We write \( A = IA \) to find its inverse using elementary row operations.
So, \( \left[\begin{array}{rr} 9 & 5 \\ 7 & 4 \end{array}\right]=\left[\begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}\right]A \)
We perform row operations to make the left matrix into the identity matrix.
First, operate \( R_1 \rightarrow R_1 - R_2 \):
\( \left[\begin{array}{rr} 9 - 7 & 5 - 4 \\ 7 & 4 \end{array}\right]=\left[\begin{array}{rr} 1 - 0 & 0 - 1 \\ 0 & 1 \end{array}\right]A \)
\( \left[\begin{array}{rr} 2 & 1 \\ 7 & 4 \end{array}\right]=\left[\begin{array}{rr} 1 & -1 \\ 0 & 1 \end{array}\right]A \)
Next, operate \( R_2 \rightarrow R_2 - 4R_1 \):
\( \left[\begin{array}{rr} 2 & 1 \\ 7 - 4(2) & 4 - 4(1) \end{array}\right]=\left[\begin{array}{rr} 1 & -1 \\ 0 - 4(1) & 1 - 4(-1) \end{array}\right]A \)
\( \left[\begin{array}{rr} 2 & 1 \\ -1 & 0 \end{array}\right]=\left[\begin{array}{rr} 1 & -1 \\ -4 & 5 \end{array}\right]A \)
Then, operate \( R_1 \rightarrow R_1 + 2R_2 \):
\( \left[\begin{array}{rr} 2 + 2(-1) & 1 + 2(0) \\ -1 & 0 \end{array}\right]=\left[\begin{array}{rr} 1 + 2(-4) & -1 + 2(5) \\ -4 & 5 \end{array}\right]A \)
\( \left[\begin{array}{rr} 0 & 1 \\ -1 & 0 \end{array}\right]=\left[\begin{array}{rr} -7 & 9 \\ -4 & 5 \end{array}\right]A \)
Now, operate \( R_1 \leftrightarrow R_2 \) (swap Row 1 and Row 2):
\( \left[\begin{array}{rr} -1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{rr} -4 & 5 \\ -7 & 9 \end{array}\right]A \)
Finally, operate \( R_1 \rightarrow -R_1 \):
\( \left[\begin{array}{rr} -1(-1) & -1(0) \\ 0 & 1 \end{array}\right]=\left[\begin{array}{rr} -1(-4) & -1(5) \\ -7 & 9 \end{array}\right]A \)
\( \left[\begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{rr} 4 & -5 \\ -7 & 9 \end{array}\right]A \)
Therefore, the inverse matrix is \( A^{-1} = \left[\begin{array}{rr} 4 & -5 \\ -7 & 9 \end{array}\right] \). This process of using row operations helps transform the matrix into an identity matrix while simultaneously building the inverse.
In simple words: We used different row operations to change the left matrix into the identity matrix. The new matrix formed on the right side from these operations is the inverse of the original matrix.

🎯 Exam Tip: When performing elementary row operations, ensure you apply them consistently to all elements in the row. Swapping rows is a useful tool to get a leading 1 in the top-left position.

Question 4. Find the inverse of the matrix A = \( \left[\begin{array}{rr} 10 & -2 \\ -5 & 1 \end{array}\right] \) using elementary transformation, if it exists.
Answer:
Given the matrix \( A = \left[\begin{array}{rr} 10 & -2 \\ -5 & 1 \end{array}\right] \). We write \( A = IA \) to find its inverse using elementary row operations.
So, \( \left[\begin{array}{rr} 10 & -2 \\ -5 & 1 \end{array}\right]=\left[\begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}\right]A \)
Let's perform row operations. First, operate \( R_1 \rightarrow R_1 + 2R_2 \):
\( \left[\begin{array}{rr} 10 + 2(-5) & -2 + 2(1) \\ -5 & 1 \end{array}\right]=\left[\begin{array}{rr} 1 + 2(0) & 0 + 2(1) \\ 0 & 1 \end{array}\right]A \)
\( \left[\begin{array}{rr} 0 & 0 \\ -5 & 1 \end{array}\right]=\left[\begin{array}{rr} 1 & 2 \\ 0 & 1 \end{array}\right]A \)
Since all elements in the first row (\(R_1\)) of the left side matrix are zeros, the inverse of matrix A does not exist. A matrix has no inverse if its determinant is zero, and a row of zeros in the row-reduced form indicates this.
In simple words: When we tried to change the matrix using row operations, one whole row became all zeros. This means the matrix does not have an inverse.

🎯 Exam Tip: If any row (or column) of the matrix reduces to all zeros during elementary transformations, the inverse does not exist. This is a crucial condition to recognize early.

Question 5. Find the inverse of the matrix A = \( \left[\begin{array}{rrr} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}\right] \) using elementary transformation, if it exists.
Answer:
We are given the matrix \( A = \left[\begin{array}{rrr} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}\right] \). To find its inverse using elementary transformations, we write \( A = IA \).
So, \( \left[\begin{array}{rrr} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}\right]=\left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]A \)
We will perform row operations to transform the left matrix into the identity matrix.
First, operate \( R_1 \leftrightarrow R_2 \) (swap Row 1 and Row 2) to get a leading 1:
\( \left[\begin{array}{rrr} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 3 & 1 & 1 \end{array}\right]=\left[\begin{array}{rrr} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right]A \)
Next, operate \( R_3 \rightarrow R_3 - 3R_1 \):
\( \left[\begin{array}{rrr} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 3 - 3(1) & 1 - 3(2) & 1 - 3(3) \end{array}\right]=\left[\begin{array}{rrr} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 - 3(0) & 0 - 3(1) & 1 - 3(0) \end{array}\right]A \)
\( \left[\begin{array}{rrr} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & -5 & -8 \end{array}\right]=\left[\begin{array}{rrr} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & -3 & 1 \end{array}\right]A \)
Now, operate \( R_3 \rightarrow R_3 + 5R_2 \):
\( \left[\begin{array}{rrr} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 + 5(0) & -5 + 5(1) & -8 + 5(2) \end{array}\right]=\left[\begin{array}{rrr} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 + 5(1) & -3 + 5(0) & 1 + 5(0) \end{array}\right]A \)
\( \left[\begin{array}{rrr} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{array}\right]=\left[\begin{array}{rrr} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 5 & -3 & 1 \end{array}\right]A \)
Next, operate \( R_3 \rightarrow \frac{1}{2}R_3 \):
\( \left[\begin{array}{rrr} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0/2 & 0/2 & 2/2 \end{array}\right]=\left[\begin{array}{rrr} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 5/2 & -3/2 & 1/2 \end{array}\right]A \)
\( \left[\begin{array}{rrr} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{rrr} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 5/2 & -3/2 & 1/2 \end{array}\right]A \)
Now, operate \( R_2 \rightarrow R_2 - 2R_3 \):
\( \left[\begin{array}{rrr} 1 & 2 & 3 \\ 0 - 2(0) & 1 - 2(0) & 2 - 2(1) \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{rrr} 0 & 1 & 0 \\ 1 - 2(5/2) & 0 - 2(-3/2) & 0 - 2(1/2) \\ 5/2 & -3/2 & 1/2 \end{array}\right]A \)
\( \left[\begin{array}{rrr} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{rrr} 0 & 1 & 0 \\ 1 - 5 & 0 + 3 & 0 - 1 \\ 5/2 & -3/2 & 1/2 \end{array}\right]A \)
\( \left[\begin{array}{rrr} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{rrr} 0 & 1 & 0 \\ -4 & 3 & -1 \\ 5/2 & -3/2 & 1/2 \end{array}\right]A \)
Finally, operate \( R_1 \rightarrow R_1 - 2R_2 \):
\( \left[\begin{array}{rrr} 1 - 2(0) & 2 - 2(1) & 3 - 2(0) \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{rrr} 0 - 2(-4) & 1 - 2(3) & 0 - 2(-1) \\ -4 & 3 & -1 \\ 5/2 & -3/2 & 1/2 \end{array}\right]A \)
\( \left[\begin{array}{rrr} 1 & 0 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{rrr} 0 + 8 & 1 - 6 & 0 + 2 \\ -4 & 3 & -1 \\ 5/2 & -3/2 & 1/2 \end{array}\right]A \)
\( \left[\begin{array}{rrr} 1 & 0 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{rrr} 8 & -5 & 2 \\ -4 & 3 & -1 \\ 5/2 & -3/2 & 1/2 \end{array}\right]A \)
Now, operate \( R_1 \rightarrow R_1 - 3R_3 \):
\( \left[\begin{array}{rrr} 1 - 3(0) & 0 - 3(0) & 3 - 3(1) \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{rrr} 8 - 3(5/2) & -5 - 3(-3/2) & 2 - 3(1/2) \\ -4 & 3 & -1 \\ 5/2 & -3/2 & 1/2 \end{array}\right]A \)
\( \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{rrr} 8 - 15/2 & -5 + 9/2 & 2 - 3/2 \\ -4 & 3 & -1 \\ 5/2 & -3/2 & 1/2 \end{array}\right]A \)
\( \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{rrr} 1/2 & -1/2 & 1/2 \\ -4 & 3 & -1 \\ 5/2 & -3/2 & 1/2 \end{array}\right]A \)
Therefore, the inverse is \( A^{-1} = \left[\begin{array}{rrr} 1/2 & -1/2 & 1/2 \\ -4 & 3 & -1 \\ 5/2 & -3/2 & 1/2 \end{array}\right] \). For 3x3 matrices, the process involves more steps but follows the same principles of row operations.
In simple words: By carefully using row operations, we changed the starting 3x3 matrix into the identity matrix. The new matrix that appeared on the right side from these steps is the inverse of the original matrix.

🎯 Exam Tip: When dealing with 3x3 matrices, it's helpful to first try to get zeros below the main diagonal, then turn the diagonal elements into ones, and finally get zeros above the diagonal. This systematic approach reduces errors.

Question 6. Find the inverse of the matrix A = \( \left[\begin{array}{rrr} 1 & -1 & 0 \\ 2 & 5 & 3 \\ 0 & 2 & 1 \end{array}\right] \) using elementary transformation, if it exists.
Answer:
Given the matrix \( A = \left[\begin{array}{rrr} 1 & -1 & 0 \\ 2 & 5 & 3 \\ 0 & 2 & 1 \end{array}\right] \). We write \( A = IA \) to find its inverse using elementary row operations.
So, \( \left[\begin{array}{rrr} 1 & -1 & 0 \\ 2 & 5 & 3 \\ 0 & 2 & 1 \end{array}\right]=\left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]A \)
First, operate \( R_2 \rightarrow R_2 - 2R_1 \):
\( \left[\begin{array}{rrr} 1 & -1 & 0 \\ 2 - 2(1) & 5 - 2(-1) & 3 - 2(0) \\ 0 & 2 & 1 \end{array}\right]=\left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 - 2(1) & 1 - 2(0) & 0 - 2(0) \\ 0 & 0 & 1 \end{array}\right]A \)
\( \left[\begin{array}{rrr} 1 & -1 & 0 \\ 0 & 7 & 3 \\ 0 & 2 & 1 \end{array}\right]=\left[\begin{array}{rrr} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]A \)
Next, operate \( R_2 \rightarrow R_2 - 3R_3 \):
\( \left[\begin{array}{rrr} 1 & -1 & 0 \\ 0 - 3(0) & 7 - 3(2) & 3 - 3(1) \\ 0 & 2 & 1 \end{array}\right]=\left[\begin{array}{rrr} 1 & 0 & 0 \\ -2 - 3(0) & 1 - 3(0) & 0 - 3(1) \\ 0 & 0 & 1 \end{array}\right]A \)
\( \left[\begin{array}{rrr} 1 & -1 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{array}\right]=\left[\begin{array}{rrr} 1 & 0 & 0 \\ -2 & 1 & -3 \\ 0 & 0 & 1 \end{array}\right]A \)
Now, operate \( R_3 \rightarrow R_3 - 2R_2 \):
\( \left[\begin{array}{rrr} 1 & -1 & 0 \\ 0 & 1 & 0 \\ 0 - 2(0) & 2 - 2(1) & 1 - 2(0) \end{array}\right]=\left[\begin{array}{rrr} 1 & 0 & 0 \\ -2 & 1 & -3 \\ 0 - 2(-2) & 0 - 2(1) & 1 - 2(-3) \end{array}\right]A \)
\( \left[\begin{array}{rrr} 1 & -1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{rrr} 1 & 0 & 0 \\ -2 & 1 & -3 \\ 4 & -2 & 7 \end{array}\right]A \)
Finally, operate \( R_1 \rightarrow R_1 + R_2 \):
\( \left[\begin{array}{rrr} 1 + 0 & -1 + 1 & 0 + 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{rrr} 1 + (-2) & 0 + 1 & 0 + (-3) \\ -2 & 1 & -3 \\ 4 & -2 & 7 \end{array}\right]A \)
\( \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{rrr} -1 & 1 & -3 \\ -2 & 1 & -3 \\ 4 & -2 & 7 \end{array}\right]A \)
Therefore, the inverse is \( A^{-1} = \left[\begin{array}{rrr} -1 & 1 & -3 \\ -2 & 1 & -3 \\ 4 & -2 & 7 \end{array}\right] \). These transformations are powerful tools for solving systems of linear equations and other linear algebra problems.
In simple words: We used elementary row operations to change the left matrix into the identity matrix. The matrix that resulted on the right side from these operations is the inverse of the initial matrix.

🎯 Exam Tip: When performing multiple row operations, it's good practice to focus on one column at a time (e.g., getting zeros below the leading 1, then moving to the next column). This systematic approach helps prevent mistakes.

Question 7. Find the inverse of the matrix A = \( \left[\begin{array}{rrr} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{array}\right] \) using elementary transformation, if it exists.
Answer:
We are given the matrix \( A = \left[\begin{array}{rrr} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{array}\right] \). To find its inverse using elementary transformations, we write \( A = IA \).
So, \( \left[\begin{array}{rrr} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{array}\right]=\left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]A \)
We perform row operations to transform the left matrix into the identity matrix.
First, operate \( R_2 \rightarrow R_2 + 3R_1 \) and \( R_3 \rightarrow R_3 - 2R_1 \):
\( \left[\begin{array}{rrr} 1 & 3 & -2 \\ -3+3(1) & 0+3(3) & -5+3(-2) \\ 2-2(1) & 5-2(3) & 0-2(-2) \end{array}\right]=\left[\begin{array}{rrr} 1 & 0 & 0 \\ 0+3(1) & 1+3(0) & 0+3(0) \\ 0-2(1) & 0-2(0) & 1-2(0) \end{array}\right]A \)
\( \left[\begin{array}{rrr} 1 & 3 & -2 \\ 0 & 9 & -11 \\ 0 & -1 & 4 \end{array}\right]=\left[\begin{array}{rrr} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 0 & 1 \end{array}\right]A \)
Next, operate \( R_2 \leftrightarrow R_3 \) (swap Row 2 and Row 3) to bring a smaller number to the pivot position:
\( \left[\begin{array}{rrr} 1 & 3 & -2 \\ 0 & -1 & 4 \\ 0 & 9 & -11 \end{array}\right]=\left[\begin{array}{rrr} 1 & 0 & 0 \\ -2 & 0 & 1 \\ 3 & 1 & 0 \end{array}\right]A \)
Then, operate \( R_2 \rightarrow -R_2 \) to make the leading element positive:
\( \left[\begin{array}{rrr} 1 & 3 & -2 \\ 0 & 1 & -4 \\ 0 & 9 & -11 \end{array}\right]=\left[\begin{array}{rrr} 1 & 0 & 0 \\ 2 & 0 & -1 \\ 3 & 1 & 0 \end{array}\right]A \)
Now, operate \( R_1 \rightarrow R_1 - 3R_2 \) and \( R_3 \rightarrow R_3 - 9R_2 \):
\( \left[\begin{array}{rrr} 1-3(0) & 3-3(1) & -2-3(-4) \\ 0 & 1 & -4 \\ 0-9(0) & 9-9(1) & -11-9(-4) \end{array}\right]=\left[\begin{array}{rrr} 1-3(2) & 0-3(0) & 0-3(-1) \\ 2 & 0 & -1 \\ 3-9(2) & 1-9(0) & 0-9(-1) \end{array}\right]A \)
\( \left[\begin{array}{rrr} 1 & 0 & 10 \\ 0 & 1 & -4 \\ 0 & 0 & 25 \end{array}\right]=\left[\begin{array}{rrr} -5 & 0 & 3 \\ 2 & 0 & -1 \\ -15 & 1 & 9 \end{array}\right]A \)
Next, operate \( R_3 \rightarrow \frac{1}{25}R_3 \):
\( \left[\begin{array}{rrr} 1 & 0 & 10 \\ 0 & 1 & -4 \\ 0/25 & 0/25 & 25/25 \end{array}\right]=\left[\begin{array}{rrr} -5 & 0 & 3 \\ 2 & 0 & -1 \\ -15/25 & 1/25 & 9/25 \end{array}\right]A \)
\( \left[\begin{array}{rrr} 1 & 0 & 10 \\ 0 & 1 & -4 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{rrr} -5 & 0 & 3 \\ 2 & 0 & -1 \\ -3/5 & 1/25 & 9/25 \end{array}\right]A \)
Finally, operate \( R_1 \rightarrow R_1 - 10R_3 \) and \( R_2 \rightarrow R_2 + 4R_3 \):
\( \left[\begin{array}{rrr} 1-10(0) & 0-10(0) & 10-10(1) \\ 0-4(0) & 1-4(0) & -4+4(1) \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{rrr} -5-10(-3/5) & 0-10(1/25) & 3-10(9/25) \\ 2+4(-3/5) & 0+4(1/25) & -1+4(9/25) \\ -3/5 & 1/25 & 9/25 \end{array}\right]A \)
\( \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{rrr} -5+6 & -10/25 & 3-18/5 \\ 2-12/5 & 4/25 & -1+36/25 \\ -3/5 & 1/25 & 9/25 \end{array}\right]A \)
\( \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{rrr} 1 & -2/5 & -3/5 \\ -2/5 & 4/25 & 11/25 \\ -3/5 & 1/25 & 9/25 \end{array}\right]A \)
Therefore, the inverse of A is \( A^{-1} = \left[\begin{array}{rrr} 1 & -2/5 & -3/5 \\ -2/5 & 4/25 & 11/25 \\ -3/5 & 1/25 & 9/25 \end{array}\right] \). These operations are chosen strategically to convert elements to 1s or 0s to reach the identity matrix.
In simple words: We used a series of row changes to turn the given matrix on the left into the identity matrix. The matrix that formed on the right side during these changes is the inverse of the original matrix.

🎯 Exam Tip: For 3x3 matrices, ensure you aim for a '1' on the main diagonal and '0's elsewhere in a systematic way (e.g., column by column) to avoid getting lost in the calculations. Double-check your arithmetic in each step.