A and B are two 3 times 3 non-singular matric | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given, $\operatorname{adj} A = |A| B$. Since $A$ is non-singular, $\operatorname{adj} A = |A| A^{-1}$. Thus, $|A| B = |A| A^{-1}$, which implies $

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Given, $\operatorname{adj} A = |A| B$. Since $A$ is non-singular, $\operatorname{adj} A = |A| A^{-1}$. Thus, $|A| B = |A| A^{-1}$, which implies $B = A^{-1}$. Therefore, $A B = A A^{-1} = I$, where $I$ is the identity matrix. Now, consider the sum $S = \sum_{k=1}^{\infty} \operatorname{tr}\left(\frac{1}{3^k}(A B)^k Cight)$. Since $(A B)^k = I^k = I$, the expression becomes $S = \sum_{k=1}^{\infty} \operatorname{tr}\left(\frac{1}{3^k} I Cight) = \sum_{k=1}^{\infty} \frac{1}{3^k} \operatorname{tr}(C)$. The trace of $C$ is $\operatorname{tr}(C) = 4 + (-2) + 6 = 8$. So, $S = 8 \sum_{k=1}^{\infty} \left(\frac{1}{3}ight)^k$. This is an infinite geometric series with first term $a = \frac{1}{3}$ and common ratio $r = \frac{1}{3}$. The sum is $S = 8 \left( \frac{1/3}{1 - 1/3} ight) = 8 \left( \frac{1/3}{2/3} ight) = 8 	imes \frac{1}{2} = 4$.

Similar Questions

Let $A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$ and $B$ be two matrices such that $A^{100} = 100B + I$. Then the sum of all the elements of $B^{100}$ is . . . . . . .

For $M=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right]$ and for any $n \in N$,the matrix $M^{n+1}-M^n=$

Let $A=[a_{ij}]$ be a matrix of order $3 \times 3$,with $a_{ij}=(\sqrt{2})^{i+j}$. If the sum of all the elements in the third row of $A^2$ is $\alpha+\beta \sqrt{2}$,where $\alpha, \beta \in Z$,then $\alpha+\beta$ is equal to

Let $A = [a_{ij}]_{2 \times 2}$ where $a_{ij} \neq 0$ for all $i, j$ and $A^2 = I$. Let $a$ be the sum of all diagonal elements of $A$ and $b = |A|$. Then $3a^2 + 4b^2$ is equal to:

Let $A=\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right],$ show that $(a \mathrm{I}+b \mathrm{A})^{n}=a^{n} \mathrm{I}+n a^{n-1} b \mathrm{A},$ where $\mathrm{I}$ is the identity matrix of order $2$ and $n \in \mathrm{N}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module