Question

If \( A=\begin{bmatrix}2 & 3 \\ 5 & -2\end{bmatrix} \), then:

• A. \( A^{-1}=\frac{1}{11}A \)
• B. \( A^{-1}=\frac{1}{19}A \)
• C. \( A^{-1}=-\frac{1}{19}A \)
• D. \( A^{-1}=\frac{1}{7}A \)

Hint:

For a \(2\times2\) matrix: \[ A= \begin{bmatrix} a & b c & d \end{bmatrix} \] always remember: \[ A^{-1}= \frac{1}{ad-bc} \begin{bmatrix} d & -b -c & a \end{bmatrix} \] A shortcut for objective questions:
• Compute the determinant first.
• Check whether the adjoint matrix is proportional to the original matrix.
• This often allows very quick matching with the options. Also note: \[ AA^{-1}=I \] where \(I\) is the identity matrix.

Answer 1

The Correct Option is C

To find the inverse of the matrix \( A = \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix} \), we use the formula for the inverse of a 2x2 matrix.

The formula for the inverse of a 2x2 matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is given by:

\(A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\)

First, we calculate the determinant of matrix \( A \):

\(det(A) = ad - bc = (2)(-2) - (3)(5) = -4 - 15 = -19\)

Since the determinant is non-zero, the matrix is invertible.

Next, apply the formula to find \( A^{-1} \):

\(A^{-1} = \frac{1}{-19} \begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix}\)

Simplify by distributing the scalar:

\(A^{-1} = \begin{bmatrix} \frac{2}{19} & \frac{3}{19} \\ \frac{5}{19} & -\frac{2}{19} \end{bmatrix}\)

In the problem, it's given that \( A^{-1} \) is \(-\frac{1}{19}A\). If we verify this:

\(-\frac{1}{19} A = -\frac{1}{19} \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix} = \begin{bmatrix} \frac{-2}{19} & \frac{-3}{19} \\ \frac{-5}{19} & \frac{2}{19} \end{bmatrix}\)

Thus, \( A^{-1} = -\frac{1}{19} A \) matches with our calculated inverse matrix.

Therefore, the correct answer is \( A^{-1} = -\frac{1}{19} A \).