If AB = begin{bmatrix} -6 & 26 -1 & 19 end{b | vedclass

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

Question

(C) Given $AB = \begin{bmatrix} -6 & 26 \ -1 & 19 \end{bmatrix}$ and $11B^{-1} = \begin{bmatrix} 5 & -3 \ 2 & 1 \end{bmatrix}$.We know that $A = (AB

Vedclass · Accepted Answer

$\begin{bmatrix} 2 & 4 \ 3 & 2 \end{bmatrix}$

Vedclass · Answer

(C) Given $AB = \begin{bmatrix} -6 & 26 \ -1 & 19 \end{bmatrix}$ and $11B^{-1} = \begin{bmatrix} 5 & -3 \ 2 & 1 \end{bmatrix}$.We know that $A = (AB)B^{-1}$.From the second equation,$B^{-1} = \frac{1}{11} \begin{bmatrix} 5 & -3 \ 2 & 1 \end{bmatrix}$.Substituting these into the expression for $A$:$A = \begin{bmatrix} -6 & 26 \ -1 & 19 \end{bmatrix} 	imes \frac{1}{11} \begin{bmatrix} 5 & -3 \ 2 & 1 \end{bmatrix}$$A = \frac{1}{11} \begin{bmatrix} (-6)(5) + (26)(2) & (-6)(-3) + (26)(1) \ (-1)(5) + (19)(2) & (-1)(-3) + (19)(1) \end{bmatrix}$$A = \frac{1}{11} \begin{bmatrix} -30 + 52 & 18 + 26 \ -5 + 38 & 3 + 19 \end{bmatrix}$$A = \frac{1}{11} \begin{bmatrix} 22 & 44 \ 33 & 22 \end{bmatrix}$$A = \begin{bmatrix} 2 & 4 \ 3 & 2 \end{bmatrix}$.Thus,the correct option is $C$.

If $AB = \begin{bmatrix} -6 & 26 \\ -1 & 19 \end{bmatrix}$ and $11B^{-1} = \begin{bmatrix} 5 & -3 \\ 2 & 1 \end{bmatrix}$,then $A = $ . . . . . . .

Similar Questions

If $(BA)^{-1} = C$ where $B = \begin{bmatrix} 2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1 \end{bmatrix}$ and $C = \begin{bmatrix} -1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{bmatrix}$,then $A^{-1}$ is given by

If $A$ and $B$ are square matrices of the same order such that $(A+B)(A-B)=A^{2}-B^{2}$,then $(A B A^{-1})^{2}$ is equal to

If $A^{-1}=\left[\begin{array}{lll}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 5 & 5\end{array}\right]$,then $A=$

If $A$ is a symmetric matrix with real entries, then

If $A = \begin{bmatrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{bmatrix}$,then $|A| |adj A|$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module