Matrices & Determinants
Adjoint of a matrix
Grade Class 12
Question:
Let $A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$ and $B = I + \text{adj}(A) + (\text{adj } A)^2 + \dots + (\text{adj } A)^{10}$. Then, the sum of all the elements of the matrix $B$ is :
(1) -110
(2) 22
(3) -88
(4) -124
Step-by-Step Solution
Key Concept: Calculate adj(A) for the given matrix A, then use the formula for the sum of a geometric series of matrices: B = (I - (adj A)^11)(I - adj A)^-1.
Given $A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$. Then $\text{adj}(A) = \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix}$. Let $M = \text{adj}(A)$. Then $M^k = \begin{bmatrix} 1 & -2k \\ 0 & 1 \end{bmatrix}$. $B = \sum_{k=0}^{10} M^k = \sum_{k=0}^{10} \begin{bmatrix} 1 & -2k \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 11 & -2 \sum_{k=0}^{10} k \\ 0 & 11 \end{bmatrix} = \begin{bmatrix} 11 & -2 \cdot \frac{10 \cdot 11}{2} \\ 0 & 11 \end{bmatrix} = \begin{bmatrix} 11 & -110 \\ 0 & 11 \end{bmatrix}$. The sum of elements is $11 - 110 + 0 + 11 = -88$.
Correct Answer: 3