Matrices and Determinants 2 Question 4
4. 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
(2019 Main, 9 April I)
(a) $\begin{bmatrix}1 & 0 \\ 12 & 1\end{bmatrix}$
(b) $\begin{bmatrix}1 & -13 \\ 0 & 1\end{bmatrix}$
(c) $\begin{bmatrix}1 & 0 \\ 13 & 1\end{bmatrix}$
(d) $\begin{bmatrix}1 & -12 \\ 0 & 1\end{bmatrix}$
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Answer:
Correct Answer: 4. (b)
Solution:
- Given
$\begin{bmatrix}1 & 1 \\ 0 & 1 \end{bmatrix}$ $\begin{bmatrix}1 & 2 \\ 0 & 1 \end{bmatrix}$ $\begin{bmatrix}1 & 3 \\ 0 & 1 \end{bmatrix}$ $\begin{bmatrix}1 & n-1 \\ 0 & 1 \end{bmatrix}$ $=\begin{bmatrix} 1 & 78 \\ 0 & 1 \end{bmatrix} $
$\begin{bmatrix}1 & 1 \\ 0 & 1 \end{bmatrix}$ $\begin{bmatrix}1 & 2 \\ 0 & 1 \end{bmatrix}$ $=\begin{bmatrix} 1 & 2+1\\ 0 & 1 \end{bmatrix} $
$ \begin{bmatrix} 1 & 2+1 \\ 0 & 1 \end{bmatrix}$ $ \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}$ $ = \begin{bmatrix} 1 & 3+2+1 \\ 0 & 1 \end{bmatrix}$
$\begin{bmatrix}1 & 1 \\ 0 & 1 \end{bmatrix}$ $\begin{bmatrix}1 & 2 \\ 0 & 1 \end{bmatrix}$ $\begin{bmatrix}1 & 3 \\ 0 & 1 \end{bmatrix}$ $\begin{bmatrix}1 & n-1 \\ 0 & 1 \end{bmatrix}$
$=\begin{bmatrix} 1 & (n-1)+(n-2)+\ldots+3+2+1 \\ 0 & 1 \end{bmatrix} $
$ =\begin{bmatrix} 1 & \frac{n(n-1)}{2}\\ 0 & 1 \end{bmatrix}$ $=\begin{bmatrix} 1 & 78 \\ 0 & 1 \end{bmatrix} $
Since, both matrices are equal, so equating corresponding element, we get
$ \begin{aligned} \frac{n(n-1)}{2} & =78 \Rightarrow n(n-1)=156 \\ & =13 \times 12=13(13-1) \end{aligned} $
$ \Rightarrow \quad n=13 $
So, $\quad A=\begin{bmatrix}1 & 13 \\ 0 & 1\end{bmatrix}=A^{-1}=\begin{bmatrix}1 & -13 \\ 0 & 1\end{bmatrix}$
$\because$ if $|A|=1$ and $A=\begin{bmatrix}a & b \\ c & d\end{bmatrix}$, then $A^{-1}=\begin{bmatrix}d & -b \\ -c & a\end{bmatrix}$
