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 $A=\begin{bmatrix} 1 & 2 & 2 \\ 3 & 2 & 3 \\ 1 & 1 & 2 \end{bmatrix}$ and $A^{-1}=\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$, then $\sum_{1 \leq i, j \leq 3} a_{ij} =$

If $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]$,$10 B=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right]$ and $B$ is the inverse of $A$,then the value of $\alpha$ is

Suppose $n > 1$ and $A$ is a non-singular matrix of order $n$ such that $|\operatorname{adj} A| = |\operatorname{adj}(\operatorname{adj} A)|$. Then the matrix whose rank is $n$ is:

Let $A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}$. Verify that $A(\text{adj } A) = (\text{adj } A) A = |A| I$.

If $A = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$,then $\text{adj } A$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module