If $\left[ {\begin{array}{*{20}{c}} 1&1\\ 0&1 \end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}} 1&2\\ 0&1 \end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}} 1&3\\ 0&1 \end{array}} \right] \cdot \cdot \cdot \cdot \left[ {\begin{array}{*{20}{c}} 1&{n - 1}\\ 0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&{78}\\ 0&1 \end{array}} \right],$ then the inverse of $\left[ {\begin{array}{*{20}{c}} 1&n\\ 0&1 \end{array}} \right]$ is

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

$\left[ {\begin{array}{*{20}{c}} 1&0\\ {12}&1 \end{array}} \right]$

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

$\left[ {\begin{array}{*{20}{c}} 1&{ - 12}\\ 0&1 \end{array}} \right]$

JEE MATHEMATICS MCQ: MATRICES

1 Easy

1 of 1

Application +4 | -1

\left[ {\begin{array}{cccccccccccccccccccc} 1&1\\ 0&1 \end{array}} \right] \cdot \left[ {\begin{array}{cccccccccccccccccccc} 1&2\\ 0&1 \end{array}} \right] \cdot \left[ {\begin{array}{cccccccccccccccccccc} 1&3\\ 0&1 \end{array}} \right] \cdot \cdot \cdot \cdot \left[ {\begin{array}{cccccccccccccccccccc} 1&{n - 1}\\ 0&1 \end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc} 1&{78}\\ 0&1 \end{array}} \right],

then the inverse of

\left[ {\begin{array}{cccccccccccccccccccc} 1&n\\ 0&1 \end{array}} \right]

View answer and explanation

Correct option: B

\left[ {\begin{array}{cccccccccccccccccccc} 1&{ - 13}\\ 0&1 \end{array}} \right]

Option B (

\left[ {\begin{array}{cccccccccccccccccccc} 1&{ - 13}\\ 0&1 \end{array}} \right]

) is correct because, by multiplying the matrices and solving for n, we can find the inverse of the given matrix.