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

Concept: For any square matrix \[ A=\begin{bmatrix}a & b c & d\end{bmatrix} \] its inverse exists only when: \[ |A|=ad-bc\neq 0 \] The inverse of a \(2\times2\) matrix is given by: \[ A^{-1}=\frac{1}{|A|} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \] Another important idea used in this question is recognizing whether the adjoint matrix becomes a scalar multiple of the original matrix itself. We now solve the question carefully step-by-step.

Step 1: Writing the given matrix We are given: \[ A= \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix} \] We first compute the determinant of \(A\).

Step 2: Finding the determinant of the matrix For a matrix \[ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \] the determinant is: \[ |A|=ad-bc \] Here, \[ a=2,\quad b=3,\quad c=5,\quad d=-2 \] Therefore, \[ |A|=(2)(-2)-(3)(5) \] \[ =-4-15 \] \[ =-19 \] Since: \[ |A|=-19\neq 0 \] the inverse exists.

Step 3: Finding the inverse using the standard formula Using: \[ A^{-1}=\frac{1}{|A|} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \] we get: \[ A^{-1} = \frac{1}{-19} \begin{bmatrix} -2 & -3 -5 & 2 \end{bmatrix} \] Now take negative sign common from the matrix: \[ = -\frac{1}{19} \begin{bmatrix} 2 & 3 5 & -2 \end{bmatrix} \] But the matrix inside the bracket is exactly \(A\). Hence, \[ A^{-1}=-\frac{1}{19}A \]

Step 4: Comparing with the given options We obtained: \[ A^{-1}=-\frac{1}{19}A \] This matches option (C). Final Answer: \[ \boxed{A^{-1}=-\frac{1}{19}A} \] Therefore, the correct option is: \[ \boxed{(C)} \]