Question

Let \( A=\begin{bmatrix}2 & -1 \\ 0 & 2\end{bmatrix} \). If \( B=I-{}^{3}C_{1}(\mathrm{adj}\,A)+{}^{3}C_{2}(\mathrm{adj}\,A)^{2}-{}^{3}C_{3}(\mathrm{adj}\,A)^{3} \), then the sum of all elements of the matrix \( B \) is

• A. -1
• B. -3
• C. -4
• D. -5

Hint:

Matrix Tip:Identity matrix operations mimic scalar algebra rules. $I^n = I$ and $I \times A = A$. This allows us to use standard binomial theorem expansions on matrix polynomials safely.Memorize the $2\times2$ adjoint shortcut: swap the principal diagonal, negate the secondary diagonal.

Answer 1

The Correct Option is D

Concept:
Matrices - Adjoint Matrix and Binomial Expansion.

Step 1: Recognize the polynomial pattern.
The given expression for matrix $B$ is $I - 3(\operatorname{adj} A) + 3(\operatorname{adj} A)^2 - (\operatorname{adj} A)^3$.
This matches the binomial expansion $(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3$.
Thus, $B = (I - \operatorname{adj} A)^3$.

Step 2: Calculate the adjoint of matrix $A$.
For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the adjoint is $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Therefore, $\operatorname{adj} A = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}$.

Step 3: Subtract from identity matrix.
\[ I - \operatorname{adj} A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} -1 & -1 \\ 0 & -1 \end{bmatrix} \]

Step 4: Cube the matrix.
First square: \[ \begin{bmatrix} -1 & -1 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & -1 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \] Now multiply again: \[ \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & -1 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} -1 & -3 \\ 0 & -1 \end{bmatrix} \]

Step 5: Sum of elements.
\[ (-1) + (-3) + 0 + (-1) = -5 \] $$ \therefore \text{Sum of elements of matrix } B = -5 $$