Find the inverse of the matrix A = left[begin | vedclass

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

Question

(A) Let $A = \left[\begin{array}{ll}2 & 1 \ 7 & 4\end{array}ight]$.First,we find the determinant of $A$:$|A| = (2 	imes 4) - (1 	imes 7) = 8 - 7

Vedclass · Accepted Answer

$A^{-1} = \left[\begin{array}{cc}4 & -1 \ -7 & 2\end{array}ight]$

Vedclass · Answer

(A) Let $A = \left[\begin{array}{ll}2 & 1 \ 7 & 4\end{array}ight]$.First,we find the determinant of $A$:$|A| = (2 	imes 4) - (1 	imes 7) = 8 - 7 = 1$.Since $|A| 
eq 0$,the inverse $A^{-1}$ exists.The formula for the inverse of a $2 	imes 2$ matrix $\left[\begin{array}{ll}a & b \ c & d\end{array}ight]$ is $\frac{1}{|A|} \left[\begin{array}{cc}d & -b \ -c & a\end{array}ight]$.Substituting the values:$A^{-1} = \frac{1}{1} \left[\begin{array}{cc}4 & -1 \ -7 & 2\end{array}ight] = \left[\begin{array}{cc}4 & -1 \ -7 & 2\end{array}ight]$.

Find the inverse of the matrix $A = \left[\begin{array}{ll}2 & 1 \\ 7 & 4\end{array}\right]$,if it exists.

Similar Questions

Let $A = \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix}$ satisfy $A^2 + \alpha(adj(adj(A))) + \beta(adj(A)(adj(adj(A)))) = \begin{bmatrix} 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix}$ for some $\alpha, \beta \in R$. Then $(\alpha - \beta)^2$ is equal to . . . . . . .

If $P = \begin{bmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{bmatrix}$ is the adjoint of a $3 \times 3$ matrix $A$ and $|A| = 4$,then the value of $\alpha$ is:

Inverse of the matrix $\left[\begin{array}{cc}0.8 & -0.6 \\ 0.6 & 0.8\end{array}\right]$ is

If $A = \begin{bmatrix} 3 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 7 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{bmatrix}$, then find $(2A + 3B)^{-1}$.

If $|\operatorname{Adj} A|=x$ and $|\operatorname{Adj} B|=y$, then $\left|(\operatorname{Adj}(AB))^{-1}\right|=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module