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

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 3 and 4 . determinant and metrices1 Mark
MCQ
If $\left[ {\begin{array}{*{20}{c}}
1&1\\
0&1
\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}
1&2\\
0&1
\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}
1&3\\
0&1
\end{array}} \right]\,........\left[ {\begin{array}{*{20}{c}}
1&{n - 1}\\
0&1
\end{array}} \right]\, = \,\left[ {\begin{array}{*{20}{c}}
1&{78}\\
0&1
\end{array}} \right]$ then the inverse of $\left[ {\begin{array}{*{20}{c}}
1&n\\
0&1
\end{array}} \right]$ is
  • A
    $\left[ {\begin{array}{*{20}{c}}
    1&{ - 12}\\
    0&1
    \end{array}} \right]$
  • B
    $\left[ {\begin{array}{*{20}{c}}
    1&0\\
    {13}&1
    \end{array}} \right]$
  • C
    $\left[ {\begin{array}{*{20}{c}}
    1&0\\
    {12}&1
    \end{array}} \right]$
  • ✓
    $\left[ {\begin{array}{*{20}{c}}
    1&{ - 13}\\
    0&1
    \end{array}} \right]$
✓

Answer

Correct option: D.
$\left[ {\begin{array}{*{20}{c}}
1&{ - 13}\\
0&1
\end{array}} \right]$
d
$\left[ {\begin{array}{*{20}{c}}
1&1\\
0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&2\\
0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&2\\
0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&3\\
0&1
\end{array}} \right]........\left[ {\begin{array}{*{20}{c}}
1&{n - 1}\\
0&1
\end{array}} \right]$  $ = \left[ {\begin{array}{*{20}{c}}
1&{78}\\
0&1
\end{array}} \right]$

$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
1&{1 + 2 + 3 + ... + n - 1}\\
0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&{78}\\
0&1
\end{array}} \right]$

$ \Rightarrow \frac{{n\left( {n - 1} \right)}}{2} = 78 \Rightarrow n = 13, - 12$     (reject)

$\therefore $ we have to find inverse of $\left[ {\begin{array}{*{20}{c}}
1&{ - 13}\\
0&1
\end{array}} \right]$

$\therefore \left[ {\begin{array}{*{20}{c}}
1&{ - 13}\\
0&1
\end{array}} \right]$

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