Let A be a 3 3 matrix such that A [ * 20 c 1&2&3 0&2&3 0&1&1 ] = … - Vidyadip

MCQ

Let $A$ be a $3\times3$ matrix such that

$A\left[ {\begin{array}{*{20}{c}}
1&2&3 \\
0&2&3 \\
0&1&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0&0&1 \\
1&0&0 \\
0&1&0
\end{array}} \right]$ Then $A^{-1}$ is

✓
$\left[ {\begin{array}{*{20}{c}}
3&1&2 \\
3&0&2 \\
1&0&1
\end{array}} \right]$
B
$\left[ {\begin{array}{*{20}{c}}
3&2&1 \\
3&2&0 \\
1&1&0
\end{array}} \right]$
C
$\left[ {\begin{array}{*{20}{c}}
0&1&3 \\
0&2&3 \\
1&1&1
\end{array}} \right]$
D
$\left[ {\begin{array}{*{20}{c}}
1&2&3 \\
0&1&1 \\
0&2&3
\end{array}} \right]$

✓

Answer

Correct option: A.

$\left[ {\begin{array}{*{20}{c}}
3&1&2 \\
3&0&2 \\
1&0&1
\end{array}} \right]$

a
Given $A\left[ {\begin{array}{*{20}{c}}
1&2&3\\
0&2&3\\
0&1&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0&0&1\\
1&0&0\\
0&1&0
\end{array}} \right]$

Applying ${C_1} \leftrightarrow {C_3}$

$A\left[ {\begin{array}{*{20}{c}}
3&2&1\\
3&2&0\\
1&1&0
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&0&0\\
0&0&1\\
0&1&0
\end{array}} \right]$

Again Applying ${C_2} \leftrightarrow {C_3}$

$A\left[ {\begin{array}{*{20}{c}}
3&1&2\\
3&0&2\\
1&0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&0&0\\
0&1&0\\
0&0&1
\end{array}} \right]$

pre-multiplying both sides by ${A^{ - 1}}$

${A^{ - 1}}A\left[ {\begin{array}{*{20}{c}}
3&1&2\\
3&0&2\\
1&0&1
\end{array}} \right] = {A^{ - 1}}\left[ {\begin{array}{*{20}{c}}
1&0&0\\
0&1&0\\
0&0&1
\end{array}} \right]$

$I\left[ {\begin{array}{*{20}{c}}
3&1&2\\
3&0&2\\
1&0&1
\end{array}} \right] = {A^{ - 1}}I = {A^{ - 1}}$

($\because$ ${A^{ - 1}}A = I$ and $I=$ Identity matrix)

Hence, ${A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
3&1&2\\
3&0&2\\
1&0&1
\end{array}} \right]$

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