Let B=begin{bmatrix} 2 & 6 & 4 1 & 0 & 1 -1 | vedclass

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

Question

(D) Given $BAC = I$. Taking the inverse of both sides,we get $(BAC)^{-1} = I^{-1} = I$. Using the property $(XYZ)^{-1} = Z^{-1}Y^{-1}X^{-1}$,we have $

Vedclass · Accepted Answer

$\begin{bmatrix} -3 & -5 & -5 \ 0 & 9 & 2 \ 2 & 14 & 6 \end{bmatrix}$

Vedclass · Answer

(D) Given $BAC = I$. Taking the inverse of both sides,we get $(BAC)^{-1} = I^{-1} = I$. Using the property $(XYZ)^{-1} = Z^{-1}Y^{-1}X^{-1}$,we have $C^{-1}A^{-1}B^{-1} = I$. Multiplying by $C$ on the left and $B$ on the right,we get $A^{-1} = CB$. Now,calculate the product $CB$: $A^{-1} = \begin{bmatrix} -1 & 0 & 1 \ 1 & 1 & 3 \ 2 & 0 & 2 \end{bmatrix} \begin{bmatrix} 2 & 6 & 4 \ 1 & 0 & 1 \ -1 & 1 & -1 \end{bmatrix}$ $A^{-1} = \begin{bmatrix} (-1)(2)+(0)(1)+(1)(-1) & (-1)(6)+(0)(0)+(1)(1) & (-1)(4)+(0)(1)+(1)(-1) \ (1)(2)+(1)(1)+(3)(-1) & (1)(6)+(1)(0)+(3)(1) & (1)(4)+(1)(1)+(3)(-1) \ (2)(2)+(0)(1)+(2)(-1) & (2)(6)+(0)(0)+(2)(1) & (2)(4)+(0)(1)+(2)(-1) \end{bmatrix}$ $A^{-1} = \begin{bmatrix} -3 & -5 & -5 \ 0 & 9 & 2 \ 2 & 14 & 6 \end{bmatrix}$.

Let $B=\begin{bmatrix} 2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1 \end{bmatrix}$ and $C=\begin{bmatrix} -1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{bmatrix}$. If a matrix $A$ is such that $BAC=I$,then $A^{-1}=$

Similar Questions

$A$ and $B$ are two square matrices such that $A^2B = BA$. If $(AB)^{10} = A^K B^{10}$,then $k$ is:

If $A = \begin{bmatrix} 2 & 3 \\ 0 & -1 \end{bmatrix}$,then the value of $\det(A^4) + \det(A^{10} - (\operatorname{adj}(2A))^{10})$ is equal to ........

Let $\omega$ be a complex cube root of unity with $\omega \neq 1$ and $P = [p_{ij}]$ be a $2 \times 2$ matrix with $p_{ij} = \omega^{i+j}$. For $P^2 \neq 0$,if $P^k = P$,then $k$ is equal to

If $\frac{x^2+7}{(x^2+1)(x-2)}=\frac{A}{x-2}+\frac{Bx+C}{x^2+1}$,then the determinant of the matrix $\begin{bmatrix} A & B \\ C & \frac{2}{5} \end{bmatrix}$ is

Suppose $A$ is any $3 \times 3$ non-singular matrix and $(A - 3I)(A - 5I) = O$,where $I = I_3$ and $O = O_3$. If $\alpha A + \beta A^{-1} = 4I$,then $\alpha + \beta$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module