Matrices Question 12
Question 12 - 31 January - Shift 1
Let $A= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{bmatrix}$ Then the sum of the diagonal elements of the matrix $(A+I)^{11}$ is equal to:
(1) 6144
(2) 4094
(3) 4097
(4) 2050
Show Answer
Answer: (3)
Solution:
Formula: Properties of Matrix Multiplication, Properties of Positive Integral Powers Of A SQUARE MATRIX, Properties of Trace of matrix
$\begin{aligned} & A^2=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{array}\right] \\ & =\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{array}\right]=\mathrm{A} \\ & \Rightarrow \mathrm{A}^3=\mathrm{A}^4=\ldots \ldots=\mathrm{A} \\ & (\mathrm{A}+\mathrm{I})^{11}={ }^{11} \mathrm{C}0 \mathrm{~A}^{11}+{ }^{11} \mathrm{C}1 \mathrm{~A}^{10}+ \\ & \text {….. }{ }^{11} \mathrm{C}{10} \mathrm{~A}+{ }^{11} \mathrm{C}{11} \mathrm{I} \\ & =\left({ }^{11} \mathrm{C}_0+{ }^{11} \mathrm{C}1+\ldots . .{ }^{11} \mathrm{C}{10}\right) \mathrm{A}+\mathrm{I} \\ & =\left(2^{11}-1\right) \mathrm{A}+\mathrm{I}=2047 \mathrm{~A}+\mathrm{I} \\ & \therefore \text { Sum of diagonal elements }=2047(1+4-3)+3 \\ & =4094+3=4097 \\ & \end{aligned}$
