If $A=\begin{bmatrix}2&3\ 5&-2\end{bmatrix}$ such that $A^{-1}=kA$ then $k=$

Q: If $A=\begin{bmatrix}2&3\\ 5&-2\end{bmatrix}$ such that $A^{-1}=kA$ then $k=$

If $A=\begin{bmatrix}2&3\\ 5&-2\end{bmatrix}$ such that $A^{-1}=kA$ then $k=$

Question

Accepted Answer

To determine the value of $ k $ such that $ A^{-1} = kA $ for the matrix $ A = \begin{bmatrix} 2 & 3 \ 5 & -2 \end{bmatrix} $, we need to find the inverse of matrix $ A $ and compare it to $ kA $.

### Step 1: Calculate the Inverse of Matrix $ A $

The inverse of a 2x2 matrix $ A = \begin{bmatrix} a & b \ c & d \end{bmatrix} $ is given by:

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

For the given matrix $ A = \begin{bmatrix} 2 & 3 \ 5 & -2 \end{bmatrix} $:

- $ a = 2 $, $ b = 3 $, $ c = 5 $, $ d = -2 $

The determinant $ ad - bc $ is:

$$
ad - bc = (2)(-2) - (3)(5) = -4 - 15 = -19
$$

Thus, the inverse $ A^{-1} $ is:

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

### Step 2: Compare $ A^{-1} $ to $ kA $

We know $ A^{-1} = kA $. Therefore, we need to express $ kA $ and compare it to the calculated $ A^{-1} $:

$$
kA = k \begin{bmatrix} 2 & 3 \ 5 & -2 \end{bmatrix} = \begin{bmatrix} 2k & 3k \ 5k & -2k \end{bmatrix}
$$

For $ A^{-1} = kA $, we equate the corresponding elements:

$$
\begin{bmatrix} \frac{2}{19} & \frac{3}{19} \ \frac{5}{19} & -\frac{2}{19} \end{bmatrix} = \begin{bmatrix} 2k & 3k \ 5k & -2k \end{bmatrix}
$$

This gives us the equations:

1. $ 2k = \frac{2}{19} $
2. $ 3k = \frac{3}{19} $
3. $ 5k = \frac{5}{19} $
4. $ -2k = -\frac{2}{19} $

Solving any of these equations for $ k $:

$$
2k = \frac{2}{19} \implies k = \frac{2}{19} \div 2 = \frac{1}{19}
$$

Thus, $ k = \frac{1}{19} $.

### Conclusion

The correct value of $ k $ is $\frac{1}{19}$, which corresponds to Option 2. This is consistent across all elements of the matrix, confirming that Option 2 is indeed correct.

### Explanation of Incorrect Options

1. **Option 1 (19):** If $ k = 19 $, then $ 2k = 38 $, which does not match $\frac{2}{19}$.
2. **Option 3 (-19):** If $ k = -19 $, then $ 2k = -38 $, which does not match $\frac{2}{19}$.
3. **Option 4 (-$\frac{1}{19}$):** If $ k = -\frac{1}{19} $, then $ 2k = -\frac{2}{19} $, which does not match $\frac{2}{19}$.

Therefore, Option 2 is the only correct choice, as it satisfies the condition $ A^{-1} = kA $.