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STD 12 - 3 and 4 . determinant and metrices — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3 and 4 . determinant and metrices4 Marks
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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