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  • Help me please, - Matrices and Determinants - JEE Main-5

If

\begin{bmatrix} 1& 1\\ 0& 1 \end{bmatrix}\begin{bmatrix} 1& 2\\ 0& 1 \end{bmatrix}\begin{bmatrix} 1 & 3\\ 0 & 1 \end{bmatrix}..................\begin{bmatrix} 1 & n-1\\ 0& 1 \end{bmatrix}=\begin{bmatrix} 1 & 78\\ 0& 1 \end{bmatrix} ,

then the inverse of \begin{bmatrix} 1 & n\\ 0 & 1 \end{bmatrix} is :

  • Option 1)

      \begin{bmatrix} 1 & 0\\ 12 & 1 \end{bmatrix}

  • Option 2)

     \begin{bmatrix} 1& -13\\ 0& 1\end{bmatrix}

  • Option 3)

     \begin{bmatrix} 1 & -12\\ 0 & 1 \end{bmatrix}

  • Option 4)

     \begin{bmatrix} 1 & 0\\ 13 &1 \end{bmatrix}

 

Answers (1)

best_answer

\begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1& 2\\ 0& 1 \end{bmatrix}=\begin{bmatrix} 1 & 3\\ 0 & 1 \end{bmatrix}

and \begin{bmatrix} 1 & 3\\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 3\\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 6\\ 0 & 1 \end{bmatrix}

and so on

So,\begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1& 2\\ 0& 1 \end{bmatrix}\begin{bmatrix} 1 & 3\\ 0 & 1 \end{bmatrix}...........\begin{bmatrix} 1 & n-1\\ 0& 1 \end{bmatrix}=\begin{bmatrix} 1& 1+2+3+.....n-1\\ 0 & 1 \end{bmatrix}

\Rightarrow \begin{bmatrix} 1& \frac{n\left ( n-1 \right )}{2}\\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 78\\ 0& 1 \end{bmatrix}

     \frac{n\left ( n-1 \right )}{2}=78

     n=13

     Inverse of \begin{bmatrix} 1 & 13\\ 0& 1 \end{bmatrix}  is  \begin{bmatrix} 1 & -13\\ 0& 1 \end{bmatrix}


Option 1)

  \begin{bmatrix} 1 & 0\\ 12 & 1 \end{bmatrix}

Option 2)

 \begin{bmatrix} 1& -13\\ 0& 1\end{bmatrix}

Option 3)

 \begin{bmatrix} 1 & -12\\ 0 & 1 \end{bmatrix}

Option 4)

 \begin{bmatrix} 1 & 0\\ 13 &1 \end{bmatrix}

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