GATE Electrical 2022 | Question: 32

Question

Consider a matrix $A = \begin{bmatrix} 1 & 0 & 0\\ 0 & 4 & -2\\ 0 & 1 & 1 \end{bmatrix}$.

The matrix $A$ satisfies the equation $6A^{-1} = A^{2} + cA + dl$, where $c$ and $d$ are scalars and $I$ is the identity matrix.

Then $\left ( c + d \right )$ is equal to

• A. $5$
• B. $17$
• C. $-6$
• D. $11$

Answer 1

Step1: Find the Characteristic Equation:
The characteristic equation is given by $|A - \lambda I| = 0$:
$$
\begin{vmatrix}
1-\lambda & 0 & 0 \\
0 & 4-\lambda & -2 \\
0 & 1 & 1-\lambda
\end{vmatrix} = 0
$$

step 2. Solve the Determinant.
Expanding along the first row:
$$ (1-\lambda) \left[ (4-\lambda)(1-\lambda) - (-2)(1) \right] = 0 $$
$$ (1-\lambda) \left[ \lambda^2 - 5\lambda + 4 + 2 \right] = 0 $$
$$ (1-\lambda)(\lambda^2 - 5\lambda + 6) = 0 $$
$$ \lambda^3 - 5\lambda^2 + 6\lambda - \lambda^2 + 5\lambda - 6 = 0 $$
$$ \lambda^3 - 6\lambda^2 + 11\lambda - 6 = 0 $$

step 3: Apply Cayley-Hamilton Theorem:
Every square matrix satisfies its own characteristic equation. Replacing $\lambda$ with $A$:
$$ A^3 - 6A^2 + 11A - 6I = 0 $$

Multiply the entire equation by $A^{-1}$:
$$ A^2 - 6A + 11I - 6A^{-1} = 0 $$
Rearrange to isolate $6A^{-1}$:
$$ 6A^{-1} = A^2 - 6A + 11I $$

step 5: Comparison and Final Calculation:
Comparing with the given form $6A^{-1} = A^2 + cA + dI$:

The value of $(c + d)$ is:
$$ -6 + 11 = 5 $$
Option A is correct.