If the matrix A = begin{bmatrix} 1 & 2 -5 & | vedclass

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

Question

(B) Given $A = \begin{bmatrix} 1 & 2 \ -5 & 1 \end{bmatrix}$.First,we find the determinant $|A| = (1)(1) - (2)(-5) = 1 + 10 = 11$.Next,we find the in

Vedclass · Accepted Answer

$\frac{4}{11}$

Vedclass · Answer

(B) Given $A = \begin{bmatrix} 1 & 2 \ -5 & 1 \end{bmatrix}$.First,we find the determinant $|A| = (1)(1) - (2)(-5) = 1 + 10 = 11$.Next,we find the inverse $A^{-1} = \frac{1}{|A|} 	ext{adj}(A) = \frac{1}{11} \begin{bmatrix} 1 & -2 \ 5 & 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{11} & \frac{-2}{11} \ \frac{5}{11} & \frac{1}{11} \end{bmatrix}$.Given $A^{-1} = xA + yI$,we have:$\begin{bmatrix} \frac{1}{11} & \frac{-2}{11} \ \frac{5}{11} & \frac{1}{11} \end{bmatrix} = x \begin{bmatrix} 1 & 2 \ -5 & 1 \end{bmatrix} + y \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} x+y & 2x \ -5x & x+y \end{bmatrix}$.Comparing the elements,we get $2x = \frac{-2}{11} \Rightarrow x = \frac{-1}{11}$.Also,$x+y = \frac{1}{11} \Rightarrow \frac{-1}{11} + y = \frac{1}{11} \Rightarrow y = \frac{2}{11}$.Finally,$2x + 3y = 2(\frac{-1}{11}) + 3(\frac{2}{11}) = \frac{-2}{11} + \frac{6}{11} = \frac{4}{11}$.

If the matrix $A = \begin{bmatrix} 1 & 2 \\ -5 & 1 \end{bmatrix}$ and $A^{-1} = xA + yI$,where $I$ is a unit matrix of order $2$,then the value of $2x + 3y$ is

Similar Questions

If $K \in R_0$,then $\det(adj(KI_n))$ is equal to:

${\left[ {\begin{array}{*{20}{c}}{ - 6}&5\\{ - 7}&6\end{array}} \right]^{ - 1}}$ =

If $A$ and $B$ are invertible matrices,which one of the following statements is not correct?

If $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 1 & 1 \\ 1 & 3 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 3 & 4 \\ 3 & 2 & 2 \\ 2 & 4 & 2 \end{bmatrix}$,then $\sqrt{|\operatorname{Adj}(AB)|} = $

The adjoint of the matrix $A = \begin{bmatrix} 2 & -3 \\ 3 & 5 \end{bmatrix}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module