Question

If \(\begin{pmatrix} 2 & 1 \\ 3 & 2 \\ \end{pmatrix}A=\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}\), then the matrix A is

• A. \(\begin{pmatrix} 2 & 1 \\ 3 & 2 \\ \end{pmatrix}\)
• B. \(\begin{pmatrix} 2 & -1 \\ -3 & 2 \\ \end{pmatrix}\)
• C. \(\begin{pmatrix} -2 & 1 \\ 3 & -2 \\ \end{pmatrix}\)
• D. \(\begin{pmatrix} 2 & -1 \\ 3 & 2 \\ \end{pmatrix}\)

Answer 1

The Correct Option is B

Given equation:

$$ \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $$

This means that matrix $ A $ is the inverse of the matrix $ \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} $.

Step 1: Find the inverse of the matrix $ \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} $.

The formula to find the inverse of a 2x2 matrix $ \begin{pmatrix} a & b \\ c & d \end{pmatrix} $ is:

$$ \text{Inverse} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} $$

For the matrix $ \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} $, we have:

$ a = 2 $, $ b = 1 $, $ c = 3 $, $ d = 2 $

The determinant $ ad - bc = (2)(2) - (1)(3) = 4 - 3 = 1 $.

Thus, the inverse is:

$$ \frac{1}{1} \begin{pmatrix} 2 & -1 \\ -3 & 2 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -3 & 2 \end{pmatrix} $$

Step 2: Conclusion

The matrix $ A $ is:

$$ A = \begin{pmatrix} 2 & -1 \\ -3 & 2 \end{pmatrix} $$

Answer: (B) $ \begin{pmatrix} 2 & -1 \\ -3 & 2 \end{pmatrix} $.

Answer 2

If \(\begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\), then we need to find the matrix A.

Let \(M = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}\). 

Then we have \(MA = I\), where I is the identity matrix. This implies that \(A = M^{-1}\).

To find the inverse of M, we use the formula:

\(M^{-1} = \frac{1}{det(M)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}\), where \(M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\).

The determinant of M is \(det(M) = (2)(2) - (1)(3) = 4 - 3 = 1\).

Therefore, \(A = M^{-1} = \frac{1}{1} \begin{pmatrix} 2 & -1 \\ -3 & 2 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -3 & 2 \end{pmatrix}\).

Thus, the correct option is (B) \(\begin{pmatrix} 2 & -1 \\ -3 & 2 \end{pmatrix}\).