GSEB Class 12 Maths Solutions Chapter 3 Matrices Exercise 3.4

Question 1. \( \begin{pmatrix} 1 & -1 \\ 2 & 3 \end{pmatrix} \)
Answer: Let \( A = \begin{pmatrix} 1 & -1 \\ 2 & 3 \end{pmatrix} \). We know that \( A = IA \).
So, \( \begin{pmatrix} 1 & -1 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} A \).
Applying \( R_2 \rightarrow R_2 - 2R_1 \):
\( \begin{pmatrix} 1 & -1 \\ 0 & 5 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ -2 & 1 \end{pmatrix} A \)
Applying \( R_2 \rightarrow \frac{1}{5}R_2 \):
\( \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ -\frac{2}{5} & \frac{1}{5} \end{pmatrix} A \)
Applying \( R_1 \rightarrow R_1 + R_2 \):
\( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} \frac{3}{5} & \frac{1}{5} \\ -\frac{2}{5} & \frac{1}{5} \end{pmatrix} A \)
Therefore, the inverse matrix \( A^{-1} \) is \( \begin{pmatrix} \frac{3}{5} & \frac{1}{5} \\ -\frac{2}{5} & \frac{1}{5} \end{pmatrix} \).
In simple words: We used basic row operations to change the left matrix into an identity matrix. The same operations were applied to the identity matrix on the right, which then transformed it into the inverse of the original matrix.

Exam Tip: Remember to perform elementary row operations on both the matrix A and the identity matrix I simultaneously until A becomes I. The matrix I then becomes A^-1.

Question 2. \( \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} \)
Answer: Let \( A = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} \). We apply the formula \( A = IA \).
So, \( \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} A \).
Applying \( R_1 \leftrightarrow R_2 \):
\( \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} A \)
Applying \( R_2 \rightarrow R_2 - 2R_1 \):
\( \begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & -2 \end{pmatrix} A \)
Applying \( R_2 \rightarrow (-1)R_2 \):
\( \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 2 \end{pmatrix} A \)
Applying \( R_1 \rightarrow R_1 - R_2 \):
\( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix} A \)
Hence, the inverse matrix \( A^{-1} \) is \( \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix} \).
In simple words: By swapping rows and performing arithmetic operations on the rows, we changed the initial matrix into an identity matrix. The same changes applied to the right-hand identity matrix gave us the inverse.

Exam Tip: Swapping rows is a useful elementary operation to get a '1' in the top-left corner, making subsequent calculations simpler.

Question 3. \( \begin{pmatrix} 1 & 3 \\ 2 & 7 \end{pmatrix} \)
Answer: Let \( A = \begin{pmatrix} 1 & 3 \\ 2 & 7 \end{pmatrix} \). We apply the relation \( A = IA \).
So, \( \begin{pmatrix} 1 & 3 \\ 2 & 7 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} A \).
Applying \( R_2 \rightarrow R_2 - 2R_1 \):
\( \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ -2 & 1 \end{pmatrix} A \)
Applying \( R_1 \rightarrow R_1 - 3R_2 \):
\( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 7 & -3 \\ -2 & 1 \end{pmatrix} A \)
Therefore, the inverse matrix \( A^{-1} \) is \( \begin{pmatrix} 7 & -3 \\ -2 & 1 \end{pmatrix} \).
In simple words: We used basic row operations to change the given matrix into an identity matrix. The same operations, when applied to the initial identity matrix, resulted in the inverse matrix.

Exam Tip: Aim to create zeros below the leading '1' in each column, then work upwards to create zeros above the leading '1's.

Question 4. \( \begin{pmatrix} 2 & 3 \\ 5 & 7 \end{pmatrix} \)
Answer: Let \( A = \begin{pmatrix} 2 & 3 \\ 5 & 7 \end{pmatrix} \). We apply the principle \( A = IA \).
So, \( \begin{pmatrix} 2 & 3 \\ 5 & 7 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} A \).
Applying \( R_1 \leftrightarrow R_2 \):
\( \begin{pmatrix} 5 & 7 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} A \)
Applying \( R_1 \rightarrow R_1 - 2R_2 \):
\( \begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ 1 & 0 \end{pmatrix} A \)
Applying \( R_2 \rightarrow R_2 - 2R_1 \):
\( \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ 5 & -2 \end{pmatrix} A \)
Applying \( R_1 \rightarrow R_1 - R_2 \):
\( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} -7 & 3 \\ 5 & -2 \end{pmatrix} A \)
Hence, the inverse matrix \( A^{-1} \) is \( \begin{pmatrix} -7 & 3 \\ 5 & -2 \end{pmatrix} \).
In simple words: We perform row operations to transform the original matrix into an identity matrix. The corresponding operations applied to the identity matrix yield the inverse.

Exam Tip: If your determinant is non-zero, an inverse will exist. Check your calculations carefully if you get a row of zeros when an inverse should exist.

Question 5. \( \begin{pmatrix} 2 & 1 \\ 7 & 4 \end{pmatrix} \)
Answer: Let \( A = \begin{pmatrix} 2 & 1 \\ 7 & 4 \end{pmatrix} \). We set up the equation \( A = IA \).
So, \( \begin{pmatrix} 2 & 1 \\ 7 & 4 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} A \).
Applying \( R_1 \leftrightarrow R_2 \):
\( \begin{pmatrix} 7 & 4 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} A \)
Applying \( R_1 \rightarrow R_1 - 3R_2 \):
\( \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} -3 & 1 \\ 1 & 0 \end{pmatrix} A \)
Applying \( R_2 \rightarrow R_2 - 2R_1 \):
\( \begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} -3 & 1 \\ 7 & -2 \end{pmatrix} A \)
Applying \( R_2 \rightarrow -R_2 \):
\( \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} -3 & 1 \\ -7 & 2 \end{pmatrix} A \)
Applying \( R_1 \rightarrow R_1 - R_2 \):
\( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 4 & -1 \\ -7 & 2 \end{pmatrix} A \)
Therefore, the inverse matrix \( A^{-1} \) is \( \begin{pmatrix} 4 & -1 \\ -7 & 2 \end{pmatrix} \).
In simple words: We changed the original matrix into an identity matrix using a sequence of row operations. These same operations applied to the right-hand identity matrix gave us the inverse.

Exam Tip: Be careful with signs, especially when multiplying rows by negative numbers. A small error can lead to a completely different result.

Question 6. \( \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix} \)
Answer: Let \( A = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix} \). We start with the equation \( A = IA \).
So, \( \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} A \).
Applying \( R_1 \leftrightarrow R_2 \):
\( \begin{pmatrix} 1 & 3 \\ 2 & 5 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} A \)
Applying \( R_2 \rightarrow R_2 - 2R_1 \):
\( \begin{pmatrix} 1 & 3 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & -2 \end{pmatrix} A \)
Applying \( R_2 \rightarrow -R_2 \):
\( \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 2 \end{pmatrix} A \)
Applying \( R_1 \rightarrow R_1 - 3R_2 \):
\( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix} A \)
Therefore, the inverse matrix \( A^{-1} \) is \( \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix} \).
In simple words: We perform row operations on the original matrix to get an identity matrix. The same operations on the identity matrix on the right produce the inverse matrix.

Exam Tip: Always verify your final answer by multiplying \( A \times A^{-1} \) to make sure you obtain the identity matrix.

Question 7. \( \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix} \)
Answer: Let \( A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix} \). We use the setup \( A = IA \).
So, \( \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} A \).
Applying \( R_1 \rightarrow 2R_1 - R_2 \):
\( \begin{pmatrix} 1 & 0 \\ 5 & 2 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ 0 & 1 \end{pmatrix} A \)
Applying \( R_2 \rightarrow R_2 - 5R_1 \):
\( \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -10 & 6 \end{pmatrix} A \)
Applying \( R_2 \rightarrow \frac{1}{2}R_2 \):
\( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} A \)
Therefore, the inverse matrix \( A^{-1} \) is \( \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} \).
In simple words: We transformed the matrix into an identity matrix using row operations. The same steps applied to the identity matrix on the right gave us the inverse.

Exam Tip: Sometimes, combining row operations (like \( 2R_1 - R_2 \)) can help you achieve a '1' or '0' more quickly. Practice helps you identify these shortcuts.

Question 8. \( \begin{pmatrix} 4 & 5 \\ 3 & 4 \end{pmatrix} \)
Answer: Let \( A = \begin{pmatrix} 4 & 5 \\ 3 & 4 \end{pmatrix} \). We start with the equation \( A = IA \).
So, \( \begin{pmatrix} 4 & 5 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} A \).
Applying \( R_1 \rightarrow R_1 - R_2 \):
\( \begin{pmatrix} 1 & 1 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix} A \)
Applying \( R_2 \rightarrow R_2 - 3R_1 \):
\( \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ -3 & 4 \end{pmatrix} A \)
Applying \( R_1 \rightarrow R_1 - R_2 \):
\( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 4 & -5 \\ -3 & 4 \end{pmatrix} A \)
Therefore, the inverse matrix \( A^{-1} \) is \( \begin{pmatrix} 4 & -5 \\ -3 & 4 \end{pmatrix} \).
In simple words: We used row operations to change the given matrix into an identity matrix. The corresponding operations applied to the initial identity matrix produced the inverse matrix.

Exam Tip: For 2x2 matrices, finding the determinant \( (ad-bc) \) can confirm if the inverse exists before you start, as \( \frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \).

Question 9. \( \begin{pmatrix} 3 & 10 \\ 2 & 7 \end{pmatrix} \)
Answer: Let \( A = \begin{pmatrix} 3 & 10 \\ 2 & 7 \end{pmatrix} \). We begin with the equation \( A = IA \).
So, \( \begin{pmatrix} 3 & 10 \\ 2 & 7 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} A \).
Applying \( R_1 \rightarrow R_1 - R_2 \):
\( \begin{pmatrix} 1 & 3 \\ 2 & 7 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix} A \)
Applying \( R_2 \rightarrow R_2 - 2R_1 \):
\( \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ -2 & 3 \end{pmatrix} A \)
Applying \( R_1 \rightarrow R_1 - 3R_2 \):
\( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 7 & -10 \\ -2 & 3 \end{pmatrix} A \)
Therefore, the inverse matrix \( A^{-1} \) is \( \begin{pmatrix} 7 & -10 \\ -2 & 3 \end{pmatrix} \).
In simple words: We used elementary row operations to change the original matrix into an identity matrix. The same operations, applied to the identity matrix on the right, produced the inverse.

Exam Tip: Performing \( R_1 \rightarrow R_1 - R_2 \) when the difference between elements in column 1 is 1 often simplifies the matrix quickly.

Question 10. \( \begin{pmatrix} 3 & -1 \\ -4 & 2 \end{pmatrix} \)
Answer: Let \( A = \begin{pmatrix} 3 & -1 \\ -4 & 2 \end{pmatrix} \). We write \( A = IA \).
So, \( \begin{pmatrix} 3 & -1 \\ -4 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} A \).
Applying \( R_1 \leftrightarrow R_2 \):
\( \begin{pmatrix} -4 & 2 \\ 3 & -1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} A \)
Applying \( R_1 \rightarrow (-1)R_1 \):
\( \begin{pmatrix} 4 & -2 \\ 3 & -1 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} A \)
Applying \( R_1 \rightarrow R_1 - R_2 \):
\( \begin{pmatrix} 1 & -1 \\ 3 & -1 \end{pmatrix} = \begin{pmatrix} -1 & -1 \\ 1 & 0 \end{pmatrix} A \)
Applying \( R_2 \rightarrow R_2 - 3R_1 \):
\( \begin{pmatrix} 1 & -1 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} -1 & -1 \\ 4 & 3 \end{pmatrix} A \)
Applying \( R_2 \rightarrow \frac{1}{2}R_2 \):
\( \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} -1 & -1 \\ 2 & \frac{3}{2} \end{pmatrix} A \)
Applying \( R_1 \rightarrow R_1 + R_2 \):
\( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & \frac{1}{2} \\ 2 & \frac{3}{2} \end{pmatrix} A \)
Therefore, the inverse matrix \( A^{-1} \) is \( \begin{pmatrix} 1 & \frac{1}{2} \\ 2 & \frac{3}{2} \end{pmatrix} \).
In simple words: We used a series of row operations to turn the original matrix into an identity matrix. The same operations performed on the initial identity matrix yielded the inverse matrix.

Exam Tip: It is often easier to work towards a '1' in a pivot position (e.g., a₁₁, a₂₂) first, then use that '1' to create zeros in the rest of its column.

Question 11. \( \begin{pmatrix} 2 & -6 \\ 1 & -2 \end{pmatrix} \)
Answer: Let \( A = \begin{pmatrix} 2 & -6 \\ 1 & -2 \end{pmatrix} \). We use the property \( A = IA \).
So, \( \begin{pmatrix} 2 & -6 \\ 1 & -2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} A \).
Applying \( R_1 \leftrightarrow R_2 \):
\( \begin{pmatrix} 1 & -2 \\ 2 & -6 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} A \)
Applying \( R_2 \rightarrow R_2 - 2R_1 \):
\( \begin{pmatrix} 1 & -2 \\ 0 & -2 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & -2 \end{pmatrix} A \)
Applying \( R_2 \rightarrow -\frac{1}{2}R_2 \):
\( \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -\frac{1}{2} & 1 \end{pmatrix} A \)
Applying \( R_1 \rightarrow R_1 + 2R_2 \):
\( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} -1 & 3 \\ -\frac{1}{2} & 1 \end{pmatrix} A \)
Therefore, the inverse matrix \( A^{-1} \) is \( \begin{pmatrix} -1 & 3 \\ -\frac{1}{2} & 1 \end{pmatrix} \).
In simple words: We applied elementary row operations to change the matrix into an identity matrix. The same operations on the identity matrix produced the inverse.

Exam Tip: When fractions appear, handle them meticulously to avoid arithmetic errors. It's often better to introduce fractions later if possible.

Question 12. \( \begin{pmatrix} 6 & -3 \\ -2 & 1 \end{pmatrix} \)
Answer: Let \( A = \begin{pmatrix} 6 & -3 \\ -2 & 1 \end{pmatrix} \). We use the equation \( A = IA \).
So, \( \begin{pmatrix} 6 & -3 \\ -2 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} A \).
Applying \( R_1 \leftrightarrow R_2 \):
\( \begin{pmatrix} -2 & 1 \\ 6 & -3 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} A \)
Applying \( R_2 \rightarrow R_2 + 3R_1 \):
\( \begin{pmatrix} -2 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 3 \end{pmatrix} A \)
In the second row of the left-hand side matrix, all elements are zero. This indicates that the determinant of the original matrix is zero.
Therefore, the inverse \( A^{-1} \) does not exist.
In simple words: When doing row operations, if you get a whole row of zeros on the left side, it means the matrix cannot be inverted. This happens because the determinant is zero.

Exam Tip: If you encounter a row of zeros in the matrix on the left-hand side during elementary operations, you can immediately conclude that the inverse does not exist. This saves time and prevents further unnecessary calculations.

Question 13. \( \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix} \)
Answer: Let \( A = \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix} \). We use the relation \( A = IA \).
So, \( \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} A \).
Applying \( R_1 \rightarrow R_1 + R_2 \):
\( \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} A \)
Applying \( R_2 \rightarrow R_2 + R_1 \):
\( \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} A \)
Applying \( R_1 \rightarrow R_1 + R_2 \):
\( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix} A \)
Therefore, the inverse matrix \( A^{-1} \) is \( \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix} \).
In simple words: We applied a series of row operations to transform the initial matrix into an identity matrix. These same operations, when performed on the identity matrix, gave us the inverse.

Exam Tip: When the elements of the matrix are small, a combination of \( R_i \rightarrow R_i + R_j \) operations often quickly leads to the identity matrix.

Question 14. \( \begin{pmatrix} 2 & 1 \\ 4 & 2 \end{pmatrix} \)
Answer: Let \( A = \begin{pmatrix} 2 & 1 \\ 4 & 2 \end{pmatrix} \). We begin with the equation \( A = IA \).
So, \( \begin{pmatrix} 2 & 1 \\ 4 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} A \).
Applying \( R_2 \rightarrow R_2 - 2R_1 \):
\( \begin{pmatrix} 2 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ -2 & 1 \end{pmatrix} A \)
In the second row of the left-hand side matrix, all elements are zero. This means the determinant of the original matrix is zero.
Therefore, the inverse \( A^{-1} \) does not exist.
In simple words: When we tried to find the inverse using row operations, we ended up with a row full of zeros on the left side. This means the matrix cannot be inverted.

Exam Tip: Always calculate the determinant of a 2x2 matrix as \( ad-bc \). If it's zero, the inverse won't exist, and you can state this directly without further operations.

Question 15. \( \begin{pmatrix} 2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{pmatrix} \)
Answer: Let \( A = \begin{pmatrix} 2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{pmatrix} \). We use the setup \( A = IA \).
So, \( \begin{pmatrix} 2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} A \).
Applying \( R_1 \leftrightarrow R_3 \):
\( \begin{pmatrix} 3 & -2 & 2 \\ 2 & 2 & 3 \\ 2 & -3 & 3 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix} A \)
Applying \( R_1 \rightarrow R_1 - R_2 \):
\( \begin{pmatrix} 1 & -4 & -1 \\ 2 & 2 & 3 \\ 2 & -3 & 3 \end{pmatrix} = \begin{pmatrix} 0 & -1 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix} A \)
Applying \( R_2 \rightarrow R_2 - 2R_1 \) and \( R_3 \rightarrow R_3 - 2R_1 \):
\( \begin{pmatrix} 1 & -4 & -1 \\ 0 & 10 & 5 \\ 0 & 5 & 5 \end{pmatrix} = \begin{pmatrix} 0 & -1 & 1 \\ 0 & 3 & -2 \\ 1 & 2 & -2 \end{pmatrix} A \)
Applying \( R_2 \leftrightarrow R_3 \):
\( \begin{pmatrix} 1 & -4 & -1 \\ 0 & 5 & 5 \\ 0 & 10 & 5 \end{pmatrix} = \begin{pmatrix} 0 & -1 & 1 \\ 1 & 2 & -2 \\ 0 & 3 & -2 \end{pmatrix} A \)
Applying \( R_2 \rightarrow \frac{1}{5}R_2 \):
\( \begin{pmatrix} 1 & -4 & -1 \\ 0 & 1 & 1 \\ 0 & 10 & 5 \end{pmatrix} = \begin{pmatrix} 0 & -1 & 1 \\ \frac{1}{5} & \frac{2}{5} & -\frac{2}{5} \\ 0 & 3 & -2 \end{pmatrix} A \)
Applying \( R_3 \rightarrow R_3 - 10R_2 \):
\( \begin{pmatrix} 1 & -4 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & -5 \end{pmatrix} = \begin{pmatrix} 0 & -1 & 1 \\ \frac{1}{5} & \frac{2}{5} & -\frac{2}{5} \\ -2 & -1 & 2 \end{pmatrix} A \)
Applying \( R_3 \rightarrow -\frac{1}{5}R_3 \):
\( \begin{pmatrix} 1 & -4 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & -1 & 1 \\ \frac{1}{5} & \frac{2}{5} & -\frac{2}{5} \\ \frac{2}{5} & \frac{1}{5} & -\frac{2}{5} \end{pmatrix} A \)
Applying \( R_1 \rightarrow R_1 + 4R_2 \):
\( \begin{pmatrix} 1 & 0 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} \frac{4}{5} & \frac{3}{5} & -\frac{3}{5} \\ \frac{1}{5} & \frac{2}{5} & -\frac{2}{5} \\ \frac{2}{5} & \frac{1}{5} & -\frac{2}{5} \end{pmatrix} A \)
Applying \( R_1 \rightarrow R_1 - 3R_3 \) and \( R_2 \rightarrow R_2 - R_3 \):
\( \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} -\frac{2}{5} & 0 & \frac{3}{5} \\ -\frac{1}{5} & \frac{1}{5} & 0 \\ \frac{2}{5} & \frac{1}{5} & -\frac{2}{5} \end{pmatrix} A \)
Hence, \( A^{-1} = \begin{pmatrix} -\frac{2}{5} & 0 & \frac{3}{5} \\ -\frac{1}{5} & \frac{1}{5} & 0 \\ \frac{2}{5} & \frac{1}{5} & -\frac{2}{5} \end{pmatrix} \).
In simple words: We used a longer series of row operations to change the 3x3 matrix into an identity matrix. Each step was also applied to the identity matrix on the right, which eventually gave us the inverse.

Exam Tip: For 3x3 matrices, systematize your operations: first get 1s on the main diagonal, then zeros below them, then zeros above them. This reduces errors and makes the process more manageable.

Question 16. \( \begin{pmatrix} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{pmatrix} \)
Answer: Let \( A = \begin{pmatrix} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{pmatrix} \). We use the initial setup \( A = IA \).
So, \( \begin{pmatrix} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} A \).
Applying \( R_2 \rightarrow R_2 + 3R_1 \) and \( R_3 \rightarrow R_3 - 2R_1 \):
\( \begin{pmatrix} 1 & 3 & -2 \\ 0 & 9 & -11 \\ 0 & -1 & 4 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 0 & 1 \end{pmatrix} A \)
Applying \( R_2 \leftrightarrow R_3 \):
\( \begin{pmatrix} 1 & 3 & -2 \\ 0 & -1 & 4 \\ 0 & 9 & -11 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ -2 & 0 & 1 \\ 3 & 1 & 0 \end{pmatrix} A \)
Applying \( R_2 \rightarrow (-1)R_2 \):
\( \begin{pmatrix} 1 & 3 & -2 \\ 0 & 1 & -4 \\ 0 & 9 & -11 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 0 & -1 \\ 3 & 1 & 0 \end{pmatrix} A \)
Applying \( R_3 \rightarrow R_3 - 9R_2 \):
\( \begin{pmatrix} 1 & 3 & -2 \\ 0 & 1 & -4 \\ 0 & 0 & 25 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 0 & -1 \\ -15 & 1 & 9 \end{pmatrix} A \)
Applying \( R_3 \rightarrow \frac{1}{25}R_3 \):
\( \begin{pmatrix} 1 & 3 & -2 \\ 0 & 1 & -4 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 0 & -1 \\ -\frac{15}{25} & \frac{1}{25} & \frac{9}{25} \end{pmatrix} A = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 0 & -1 \\ -\frac{3}{5} & \frac{1}{25} & \frac{9}{25} \end{pmatrix} A \)
Applying \( R_1 \rightarrow R_1 + 2R_3 \) and \( R_2 \rightarrow R_2 + 4R_3 \):
\( \begin{pmatrix} 1 & 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1-\frac{6}{5} & \frac{2}{25} & 0+\frac{18}{25} \\ 2-\frac{12}{5} & \frac{4}{25} & -1+\frac{36}{25} \\ -\frac{3}{5} & \frac{1}{25} & \frac{9}{25} \end{pmatrix} A = \begin{pmatrix} -\frac{1}{5} & \frac{2}{25} & \frac{18}{25} \\ -\frac{2}{5} & \frac{4}{25} & \frac{11}{25} \\ -\frac{3}{5} & \frac{1}{25} & \frac{9}{25} \end{pmatrix} A \)
Applying \( R_1 \rightarrow R_1 - 3R_2 \):
\( \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} -\frac{1}{5}-3(-\frac{2}{5}) & \frac{2}{25}-3(\frac{4}{25}) & \frac{18}{25}-3(\frac{11}{25}) \\ -\frac{2}{5} & \frac{4}{25} & \frac{11}{25} \\ -\frac{3}{5} & \frac{1}{25} & \frac{9}{25} \end{pmatrix} A = \begin{pmatrix} -\frac{1}{5}+\frac{6}{5} & \frac{2-12}{25} & \frac{18-33}{25} \\ -\frac{2}{5} & \frac{4}{25} & \frac{11}{25} \\ -\frac{3}{5} & \frac{1}{25} & \frac{9}{25} \end{pmatrix} A \)
\( \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} \frac{5}{5} & -\frac{10}{25} & -\frac{15}{25} \\ -\frac{2}{5} & \frac{4}{25} & \frac{11}{25} \\ -\frac{3}{5} & \frac{1}{25} & \frac{9}{25} \end{pmatrix} A = \begin{pmatrix} 1 & -\frac{2}{5} & -\frac{3}{5} \\ -\frac{2}{5} & \frac{4}{25} & \frac{11}{25} \\ -\frac{3}{5} & \frac{1}{25} & \frac{9}{25} \end{pmatrix} A \)
Therefore, the inverse matrix \( A^{-1} = \begin{pmatrix} 1 & -\frac{2}{5} & -\frac{3}{5} \\ -\frac{2}{5} & \frac{4}{25} & \frac{11}{25} \\ -\frac{3}{5} & \frac{1}{25} & \frac{9}{25} \end{pmatrix} \).
In simple words: We systematically used row operations to change the given matrix into an identity matrix. The same transformations applied to the identity matrix on the right produced the inverse matrix.

Exam Tip: Keep your fractions simplified at each step to make the arithmetic easier and reduce the chance of errors, especially with larger matrices.

Question 17. \( \begin{pmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{pmatrix} \)
Answer: Let \( A = \begin{pmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{pmatrix} \). We use the starting point \( A = IA \).
So, \( \begin{pmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} A \).
Applying \( R_1 \leftrightarrow R_2 \):
\( \begin{pmatrix} 5 & 1 & 0 \\ 2 & 0 & -1 \\ 0 & 1 & 3 \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} A \)
Applying \( R_1 \rightarrow R_1 - 2R_2 \):
\( \begin{pmatrix} 1 & 1 & 2 \\ 2 & 0 & -1 \\ 0 & 1 & 3 \end{pmatrix} = \begin{pmatrix} -2 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} A \)
Applying \( R_2 \rightarrow R_2 - 2R_1 \):
\( \begin{pmatrix} 1 & 1 & 2 \\ 0 & -2 & -5 \\ 0 & 1 & 3 \end{pmatrix} = \begin{pmatrix} -2 & 1 & 0 \\ 5 & -2 & 0 \\ 0 & 0 & 1 \end{pmatrix} A \)
Applying \( R_2 \leftrightarrow R_3 \):
\( \begin{pmatrix} 1 & 1 & 2 \\ 0 & 1 & 3 \\ 0 & -2 & -5 \end{pmatrix} = \begin{pmatrix} -2 & 1 & 0 \\ 0 & 0 & 1 \\ 5 & -2 & 0 \end{pmatrix} A \)
Applying \( R_3 \rightarrow R_3 + 2R_2 \):
\( \begin{pmatrix} 1 & 1 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 & 0 \\ 0 & 0 & 1 \\ 5 & -2 & 2 \end{pmatrix} A \)
Applying \( R_1 \rightarrow R_1 - R_2 \):
\( \begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 & -1 \\ 0 & 0 & 1 \\ 5 & -2 & 2 \end{pmatrix} A \)
Applying \( R_1 \rightarrow R_1 + R_3 \) and \( R_2 \rightarrow R_2 - 3R_3 \):
\( \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{pmatrix} A \)
Therefore, the inverse matrix \( A^{-1} = \begin{pmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{pmatrix} \).
In simple words: We used a step-by-step method of row operations to change the given matrix into an identity matrix. These same operations, applied to the identity matrix on the right, produced the inverse.

Exam Tip: Practice makes perfect for 3x3 matrices. Focus on one column at a time (e.g., get the first column as \( \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \), then the second as \( \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \), and so on) to keep track of your steps.

Question 18. B will be inverse of each other only, if
(A) AB = BA
(B) AB = BA = O
(C) AB = O, BA = I
(D) AB = BA = I
Answer: (D) AB = BA = I
In simple words: If two matrices are inverses of each other, then multiplying them in any order always gives the identity matrix.

Exam Tip: This is a fundamental definition in matrix algebra. Know the properties of the identity matrix and inverse matrices thoroughly.