Question

If $A=\begin{bmatrix}2 & -1\\ -1 & 3\end{bmatrix}$, then the inverse of $(2A^{2}+5A)$ is

• A. $\frac{1}{95}\begin{bmatrix}7 & 3\\ 3 & 4\end{bmatrix}$
• B. $\frac{1}{95}\begin{bmatrix}-7 & 3\\ 3 & -4\end{bmatrix}$
• C. $\frac{1}{95}\begin{bmatrix}-7 & -3\\ 3 & 4\end{bmatrix}$
• D. $\frac{1}{95}\begin{bmatrix}4 & 3\\3 & 7\end{bmatrix}$

Hint:

Logic Tip: The Cayley-Hamilton theorem states that a matrix satisfies its own characteristic equation. $|A - \lambda I| = \lambda^2 - 5\lambda + 5 = 0 \implies A^2 - 5A + 5I = 0$. You can sometimes use this to reduce matrix polynomials ($A^2 = 5A - 5I$) before computing, trading matrix multiplication for simple addition.

Answer 1

The Correct Option is A

Concept:
Matrix polynomial evaluation followed by inversion. For a $2 \times 2$ matrix \[ M = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \] the inverse is: \[ M^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \] 
Step 1: Calculate $A^2$.
\[ A = \begin{bmatrix} 2 & -1 \\ -1 & 3 \end{bmatrix} \] \[ A^2 = A \cdot A = \begin{bmatrix} 2 & -1 \\ -1 & 3 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ -1 & 3 \end{bmatrix} \] \[ A^2 = \begin{bmatrix} (2)(2) + (-1)(-1) & (2)(-1) + (-1)(3) \\ (-1)(2) + (3)(-1) & (-1)(-1) + (3)(3) \end{bmatrix} \] \[ A^2 = \begin{bmatrix} 5 & -5 \\ -5 & 10 \end{bmatrix} \] 
Step 2: Compute $M = 2A^2 + 5A$.
\[ 2A^2 = \begin{bmatrix} 10 & -10 \\ -10 & 20 \end{bmatrix}, \quad 5A = \begin{bmatrix} 10 & -5 \\ -5 & 15 \end{bmatrix} \] \[ M = \begin{bmatrix} 10 & -10 \\ -10 & 20 \end{bmatrix} + \begin{bmatrix} 10 & -5 \\ -5 & 15 \end{bmatrix} = \begin{bmatrix} 20 & -15 \\ -15 & 35 \end{bmatrix} \] 
Step 3: Find $M^{-1}$.
\[ |M| = (20)(35) - (-15)(-15) = 700 - 225 = 475 \] \[ M^{-1} = \frac{1}{475} \begin{bmatrix} 35 & 15 \\ 15 & 20 \end{bmatrix} \] Factor out $5$: \[ M^{-1} = \frac{5}{475} \begin{bmatrix} 7 & 3 \\ 3 & 4 \end{bmatrix} = \frac{1}{95} \begin{bmatrix} 7 & 3 \\ 3 & 4 \end{bmatrix} \] 
Final Answer:
\[ \boxed{ \frac{1}{95} \begin{bmatrix} 7 & 3 \\ 3 & 4 \end{bmatrix} } \]