Inverse of the matrix left[begin{array}{cc}0. | vedclass

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

Question

(A) Let $A = \left[\begin{array}{cc}0.8 & -0.6 \ 0.6 & 0.8\end{array}ight]$.The determinant of $A$ is given by $|A| = (0.8)(0.8) - (-0.6)(0.6) = 0.

Vedclass · Accepted Answer

$\left[\begin{array}{cc}0.8 & 0.6 \ -0.6 & 0.8\end{array}ight]$

Vedclass · Answer

(A) Let $A = \left[\begin{array}{cc}0.8 & -0.6 \ 0.6 & 0.8\end{array}ight]$.The determinant of $A$ is given by $|A| = (0.8)(0.8) - (-0.6)(0.6) = 0.64 + 0.36 = 1$.Since $|A| 
eq 0$,the inverse $A^{-1}$ exists.For a $2 	imes 2$ matrix $A = \left[\begin{array}{cc}a & b \ c & d\end{array}ight]$,the inverse is given by $A^{-1} = \frac{1}{|A|} \left[\begin{array}{cc}d & -b \ -c & a\end{array}ight]$.Substituting the values,we get $A^{-1} = \frac{1}{1} \left[\begin{array}{cc}0.8 & 0.6 \ -0.6 & 0.8\end{array}ight] = \left[\begin{array}{cc}0.8 & 0.6 \ -0.6 & 0.8\end{array}ight]$.

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

Similar Questions

If $A = \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix}$ and $A + A^{-1} = I$,then $\alpha =$

The inverse of matrix $A = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ is

Let $A$ be a $3 \times 3$ matrix such that $A \begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 3 \\ 0 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$. Then $A^{-1}$ is

If $A = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$ such that $XA = B^T$ and $A^{-1}Y = B$, then find the value of $XY$.

If $A$ is a matrix of order $2$ and $I$ is the identity matrix of order $2$ such that $A^2 - 4A + 3I = 0$,then $(A + 3I)^{-1} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module