Question

If
\[ \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} \cdot A \cdot \begin{pmatrix} -3 & 2 \\ 5 & -3 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \]
then \(A\) is:

• A. \(\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}\)
• B. \(\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\)
• C. \(\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}\)
• D. \(\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}\)

Hint:

When solving matrix equations, break them down into individual equations for each element, and solve the system systematically.

Answer 1

The Correct Option is A

The given matrix equation is:

\[\begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} \cdot A \cdot \begin{pmatrix} -3 & 2 \\ 5 & -3 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.\]

Let \[A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}.\] Substituting A into the equation, we compute step-by-step.

Step 1: Simplify the right product \[A \cdot \begin{pmatrix} -3 & 2 \\ 5 & -3 \end{pmatrix}.\]

\[A \cdot \begin{pmatrix} -3 & 2 \\ 5 & -3 \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot \begin{pmatrix} -3 & 2 \\ 5 & -3 \end{pmatrix}.\]

Multiply the matrices:

\[\begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot \begin{pmatrix} -3 & 2 \\ 5 & -3 \end{pmatrix} = \begin{pmatrix} -3a + 5b & 2a - 3b \\ -3c + 5d & 2c - 3d \end{pmatrix}.\]

Step 2: Multiply by \[\begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}.\]

Now multiply:

\[\begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} \cdot \begin{pmatrix} -3a + 5b & 2a - 3b \\ -3c + 5d & 2c - 3d \end{pmatrix}.\]

Perform the matrix multiplication:

\[\begin{pmatrix} 2(-3a + 5b) + 1(-3c + 5d) & 2(2a - 3b) + 1(2c - 3d) \\ 3(-3a + 5b) + 2(-3c + 5d) & 3(2a - 3b) + 2(2c - 3d) \end{pmatrix}.\]

Simplify each term: - First row, first column:

\[2(-3a + 5b) + (-3c + 5d) = -6a + 10b - 3c + 5d.\]

- First row, second column:

\[2(2a - 3b) + (2c - 3d) = 4a - 6b + 2c - 3d.\]

- Second row, first column:

\[3(-3a + 5b) + 2(-3c + 5d) = -9a + 15b - 6c + 10d.\]

- Second row, second column:

\[3(2a - 3b) + 2(2c - 3d) = 6a - 9b + 4c - 6d.\]

Thus, the resulting matrix is:

\[\begin{pmatrix} -6a + 10b - 3c + 5d & 4a - 6b + 2c - 3d \\ -9a + 15b - 6c + 10d & 6a - 9b + 4c - 6d \end{pmatrix}.\]

Step 3: Equate to identity matrix.

We equate:

\[\begin{pmatrix} -6a + 10b - 3c + 5d & 4a - 6b + 2c - 3d \\ -9a + 15b - 6c + 10d & 6a - 9b + 4c - 6d \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.\]

From this, solve the system of equations:

1. \(-6a + 10b - 3c + 5d = 1\)
2. \(4a - 6b + 2c - 3d = 0\)
3. \(-9a + 15b - 6c + 10d = 0\)
4. \(6a - 9b + 4c - 6d = 1\)

Solve this system, and you find:

\[A = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}.\]

Conclusion: The matrix A is:

\[\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}.\]