The element of the second row and third colum | vedclass

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

Question

(B) Let $A = \begin{bmatrix} 1 & 2 & 1 \ 2 & 1 & 0 \ -1 & 0 & 1 \end{bmatrix}$.First,we calculate the determinant $|A| = 1(1-0) - 2(2-0) + 1(0 - (-1

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(B) Let $A = \begin{bmatrix} 1 & 2 & 1 \ 2 & 1 & 0 \ -1 & 0 & 1 \end{bmatrix}$.First,we calculate the determinant $|A| = 1(1-0) - 2(2-0) + 1(0 - (-1)) = 1 - 4 + 1 = -2$.The inverse is given by $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$.The element in the $2^{nd}$ row and $3^{rd}$ column of $A^{-1}$ is $\frac{1}{|A|} 	imes C_{32}$,where $C_{32}$ is the cofactor of the element in the $3^{rd}$ row and $2^{nd}$ column of $A$.The cofactor $C_{32} = (-1)^{3+2} M_{32} = -1 	imes \begin{vmatrix} 1 & 1 \ 2 & 0 \end{vmatrix} = -1 	imes (0 - 2) = 2$.Thus,the required element is $\frac{C_{32}}{|A|} = \frac{2}{-2} = -1$.

The element of the second row and third column in the inverse of $\begin{bmatrix} 1 & 2 & 1 \\ 2 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}$ is

Similar Questions

If $A = \begin{bmatrix} 2 & 3 \\ -4 & 1 \end{bmatrix}$,then $\text{adj}(3A^2 + 12A)$ is equal to

If $A = \begin{bmatrix} 1 & 2 & -2 \\ 2 & -1 & 2 \\ -1 & 1 & -2 \end{bmatrix}$,then $A + 2A^{-1} =$

If ${A^2} - A + I = 0$,then ${A^{-1}} = $

If $A=\begin{bmatrix} 2a & -3b \\ 3 & 2 \end{bmatrix}$ and $A \cdot \operatorname{adj} A = A A^{T}$,then $2a + 3b$ is

If $A$ is a square matrix satisfying the equation $A^2 - 5A + 7I = 0$,where $I$ is the identity matrix and $0$ is the null matrix of the same order,then $A^{-1} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module