Question

If $A=\left[\begin{array}{rr}2 & 3 \\ 5 & -2\end{array}\right]$ and $A^{-1}=K A$, then $K$ is

• A. 19
• B. $-\frac{1}{19}$
• C. $-19$
• D. $\frac{1}{19}$

Hint:

Whenever a $2 \times 2$ matrix features a trace of zero (sum of main diagonal elements $= 2 + (-2) = 0$), it follows the identity property $A^2 = -|A| \cdot I$. Substituting this into $I = KA^2$ yields $I = K(-|A|I) \implies K = -\frac{1}{|A|}$. Since $|A| = -19$, $K = -\frac{1}{-19} = \frac{1}{19}$. This relationship works instantly for any zero-trace matrix!

Answer 1

The Correct Option is D

Step 1: Understand the relation.
We are told $A^{-1} = KA$ and must find the scalar $K$ for $A = \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$.
Step 2: Compute the determinant.
$|A| = (2)(-2) - (3)(5) = -4 - 15 = -19$.
Step 3: Write the adjugate.
For a $2\times 2$ matrix, swap the diagonal entries and negate the off-diagonal: $\text{adj }A = \begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix}$.
Step 4: Form the inverse.
$A^{-1} = \dfrac{1}{|A|}\,\text{adj }A = \dfrac{1}{-19}\begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix}$.
Step 5: Simplify the sign.
Pulling the minus inside flips every entry: $A^{-1} = \dfrac{1}{19}\begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$, which is exactly $\dfrac{1}{19}A$.
Step 6: Read off K.
Comparing $A^{-1} = \dfrac{1}{19}A$ with $A^{-1} = KA$ gives $K = \dfrac{1}{19}$, option 4, matching the key.
\[ \boxed{K = \dfrac{1}{19}} \]