Let A = bmatrix1 & 2 0 & 1 bmatrix and B = I + adj(A) + ( adj(A))^2 +... + ( adj(A))^10. The sum of all elements in B is:

Let A = bmatrix1 & 2 0 & 1 bmatrix and B = I + adj(A) + ( adj(A))… — Solution | JEE Mathematics

MCQMediumJEE 2024Inverse & Adjoint of a Matrix

Let $A = \begin{bmatrix}1 & 2 \\ 0 & 1\end{bmatrix}$ and $B = I + \operatorname{adj}(A) + (\operatorname{adj}(A))^2 + ... + (\operatorname{adj}(A))^{10}$ . The sum of all elements in $B$ is:

A
$-110$
B
$22$
C
$-88$
D
$-124$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: $A = \begin{bmatrix}1 & 2 \\ 0 & 1\end{bmatrix}$ and

B = I + \operatorname{adj}(A) + (\operatorname{adj}(A))^2 + \cdots + (\operatorname{adj}(A))^{10}

Find: The sum of all elements of $B$ .

The solution is unrelated to this question, so the working could not be extracted from it. Using the answer indicated with the question, the correct option is C, which corresponds to $-88$ .

Therefore, the sum of all elements in $B$ is $-88$ .

Common mistakes

Using the inverse formula in place of the adjugate. For a $2 \times 2$ matrix, $\operatorname{adj}(A)$ is not the same as $A^{-1}$ unless adjusted by the determinant. First find $\operatorname{adj}(A)$ correctly, then form the series.
Treating the matrix series like a scalar sum without checking powers carefully. Matrix powers must be computed from the actual matrix $\operatorname{adj}(A)$ , not by squaring individual entries independently.
Forgetting to include the identity matrix term $I$ . The series starts from the zeroth-power term, so omitting $I$ changes every entry and hence the final sum of elements.

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Answer

Step-by-step solution

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