If A = begin{bmatrix} 1 & 2 -4 & -1 end{bmat | vedclass

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

Question

(A) Given matrix $A = \begin{bmatrix} 1 & 2 \ -4 & -1 \end{bmatrix}$.First, we calculate the determinant of $A$:$|A| = (1)(-1) - (2)(-4) = -1 + 8 = 7

Vedclass · Accepted Answer

$\frac{1}{7} \begin{bmatrix} -1 & -2 \ 4 & 1 \end{bmatrix}$

Vedclass · Answer

(A) Given matrix $A = \begin{bmatrix} 1 & 2 \ -4 & -1 \end{bmatrix}$.First, we calculate the determinant of $A$:$|A| = (1)(-1) - (2)(-4) = -1 + 8 = 7$.Since $|A| 
eq 0$, $A^{-1}$ exists.The adjoint of a $2 	imes 2$ matrix $\begin{bmatrix} a & b \ c & d \end{bmatrix}$ is given by $\begin{bmatrix} d & -b \ -c & a \end{bmatrix}$.Therefore, $	ext{adj}(A) = \begin{bmatrix} -1 & -2 \ 4 & 1 \end{bmatrix}$.The inverse is given by $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$.$A^{-1} = \frac{1}{7} \begin{bmatrix} -1 & -2 \ 4 & 1 \end{bmatrix}$.

If $A = \begin{bmatrix} 1 & 2 \\ -4 & -1 \end{bmatrix}$, then $A^{-1}$ is

Similar Questions

If $A = \begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix}$ and $A \cdot \text{adj}(A) = AA^T$,then $5a + b =$

Find the inverse of the matrix,if it exists: $\left[\begin{array}{cc}6 & -3 \\ -2 & 1\end{array}\right]$

If $A = \begin{bmatrix} 8 & -2 \\ -4 & 1 \end{bmatrix}$ is given,then $A^{-1}$ is:

If $A = \begin{bmatrix} 1 & -1 & 1 \\ 0 & 2 & -3 \\ 2 & 1 & 3 \end{bmatrix}$ and $B = \operatorname{adj} A$,$C = 5A$,then $\frac{|\operatorname{adj} B|}{|C|} = $

If $A = \begin{bmatrix} 2 & 3 \\ -3 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$,then $(B^{-1} A^{-1})^{-1} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module