If B=left[begin{array}{ll}1 & 3 1 & alphaend | vedclass

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

Question

(B) We know that for any square matrix $A$,the property relating the adjoint and the determinant is $|\operatorname{adj}(A)| = |A|^{n-1}$,where $n$ is

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) We know that for any square matrix $A$,the property relating the adjoint and the determinant is $|\operatorname{adj}(A)| = |A|^{n-1}$,where $n$ is the order of the matrix.Here,$n=2$ and $|A|=2$.So,$|\operatorname{adj}(A)| = |A|^{2-1} = |A|^1 = 2$.Given $\operatorname{adj}(A) = B = \left[\begin{array}{ll}1 & 3 \ 1 & \alpha\end{array}ight]$,we calculate the determinant of $B$:$|B| = (1)(\alpha) - (3)(1) = \alpha - 3$.Since $|\operatorname{adj}(A)| = |B|$,we have $\alpha - 3 = 2$.Therefore,$\alpha = 2 + 3 = 5$.

If $B=\left[\begin{array}{ll}1 & 3 \\ 1 & \alpha\end{array}\right]$ is the adjoint of a matrix $A$ and $|A|=2$,then the value of $\alpha$ is

Similar Questions

Let $F(\alpha ) = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$,where $\alpha \in \mathbb{R}$. Then $[F(\alpha )]^{-1}$ is equal to

Which of the following is true for the matrix product $AB$?

If $A$ is a singular matrix,then $\text{adj } A$ is

If $A = \begin{bmatrix} 5 & -2 \\ 4 & 3 \end{bmatrix}$,then $A(\operatorname{adj} A) = $ . . . . . . .

Find $adj$ $A$ for $A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module