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JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant

यदि \(\left[ {\begin{array}{*{20}{c}}
1&1\\
0&1
\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}
1&2\\
0&1
\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}
1&3\\
0&1
\end{array}} \right]\,........\left[ {\begin{array}{*{20}{c}}
1&{n - 1}\\
0&1
\end{array}} \right]\, = \,\left[ {\begin{array}{*{20}{c}}
1&{78}\\
0&1
\end{array}} \right]\) है तो \(\left[ {\begin{array}{*{20}{c}}
1&n\\
0&1
\end{array}} \right]\) व्युत्क्रम (inverse) है:

  1. A \(\left[ {\begin{array}{*{20}{c}}
    1&{ - 12}\\
    0&1
    \end{array}} \right]\)
  2. B \(\left[ {\begin{array}{*{20}{c}}
    1&0\\
    {13}&1
    \end{array}} \right]\)
  3. C \(\left[ {\begin{array}{*{20}{c}}
    1&0\\
    {12}&1
    \end{array}} \right]\)
  4. D \(\left[ {\begin{array}{*{20}{c}}
    1&{ - 13}\\
    0&1
    \end{array}} \right]\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\left[ {\begin{array}{*{20}{c}}
1&{ - 13}\\
0&1
\end{array}} \right]\)

Step-by-step Solution

Detailed explanation

\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[…

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