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: Understanding the Question:
We are given a $2 \times 2$ matrix $A$. The inverse matrix $A^{-1}$ satisfies the relationship $A^{-1} = KA$. We need to find the value of the scalar constant $K$.

Step 2: Key Formula or Approach:
1. The matrix inverse formula for a $2 \times 2$ matrix is: $$A^{-1} = \frac{1}{|A|} \text{adj } A$$ 2. Alternatively, multiply both sides of the given equation $A^{-1} = KA$ by $A$ from the left or right side: $$A \cdot A^{-1} = A \cdot (KA) \implies I = K A^2$$

Step 3: Detailed Explanation:
Let's compute the determinant of the matrix $A$: $$|A| = \begin{vmatrix} 2 & 3 \\ 5 & -2 \end{vmatrix} = (2)(-2) - (3)(5) = -4 - 15 = -19$$ Now, find the adjugate matrix ($\text{adj } A$) by swapping the main diagonal elements and changing the signs of the off-diagonal elements: $$\text{adj } A = \left[\begin{array}{rr} -2 & -3 \\ -5 & 2\end{array}\right]$$ Substitute these into the standard matrix inverse formula: $$A^{-1} = \frac{1}{-19} \left[\begin{array}{rr} -2 & -3 \\ -5 & 2\end{array}\right] = \frac{1}{19} \left[\begin{array}{rr} 2 & 3 \\ 5 & -2\end{array}\right]$$ Notice that the resulting matrix is exactly our original matrix $A$: $$A^{-1} = \frac{1}{19} A$$ Comparing this with the problem statement $A^{-1} = KA$, we find that: $$K = \frac{1}{19}$$

Step 4: Final Answer:
The scalar value $K$ is equal to $\frac{1}{19}$, which matches option (D).