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TS EAMCET 5-Aug-2021 Shift 1 Question Paper

Section: Mathematics

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Question : 5 of 160

Marks: +1, -0

If

A

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

A^5

=

$A$
Identity Matrix
Null Matrix
$A^{-1}$

Solution: 👈: Video Solution

A=\begin{pmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{pmatrix}

A^{2}=\begin{pmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{pmatrix}\begin{pmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{pmatrix}

A^{2}=\begin{pmatrix} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{pmatrix}

and

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

=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}

A^{3}=I

\therefore \; \; A^{5}=A^{2} \cdot A^{3}=A^{2} \cdot I

=A^{2}(A A^{-1})

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

=I \cdot A^{-1}

\therefore \; \; A^{5}=A^{-1}

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