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AP EAMCET Engineering 2018 Apr 22 Shift 1 Paper

Section: Mathematics

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Question : 4 of 160

Marks: +1, -0

For the matrix

A=\begin{bmatrix}3 & -3 & 4\\2 & -3 & 4\\0 & -1 & 1\end{bmatrix},A^{-1}=

$A$
$A^2$
$A^3$
$A^4$

Solution:

The determinant of A comes out to be,

\left|A\right|=\begin{bmatrix}3 & -3 & 4\\2 & -3 & 4\\0 & -1 & 1\end{bmatrix}

\left|A\right|=3(-3+4)+3(2)+4(-2)=1\neq0

Taking the adjacent of

A

we get.

\operatorname{adj}(A)=\begin{bmatrix}1 & -2 & -2\\-2 & 3 & 3\\0 & -4 & -3\end{bmatrix}

=\begin{bmatrix}1 & -1 & 0\\-2 & 3 & -4\\-2 & 3 & -3\end{bmatrix}

A^{-1}=\frac{\operatorname{adj}(A)}{\left|A\right|}=\begin{bmatrix}1 & -1 & 0\\-2 & 3 & -4\\-2 & 3 & -3\end{bmatrix}

On squaring the given determinant.

A^{2}=A\cdot A=\begin{bmatrix}3 & -3 & 4\\2 & -3 & 4\\0 & -1 & 1\end{bmatrix}

\begin{bmatrix}3 & -3 & 4\\2 & -3 & 4\\0 & -1 & 1\end{bmatrix}

A^2=\begin{bmatrix}3 & -4 & 4\\0 & -1 & 0\\-2 & 2 & -3\end{bmatrix}

A^3=\begin{bmatrix}3 & -4 & 4\\0 & -1 & 0\\-2 & 2 & -3\end{bmatrix}

\begin{bmatrix}3 & -3 & 4\\2 & -3 & 4\\0 & -1 & 1\end{bmatrix}=\begin{bmatrix}1 & -1 & 0\\-2 & 3 & -4\\-2 & 3 & -3\end{bmatrix}

Thus, the inverse of

A

comes out to be

A^{-1}=A^{3}

.

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