Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS3 Marks
Question
If $\text{A}=\begin{bmatrix}1&0&1\\0&1&2\\0&0&4\end{bmatrix}$, then show that $\left|3\text{A}\right|=27\left|\text{A}\right|$
✓
Answer
The given matrix is $\text{A}=\begin{bmatrix}1&0&1\\0&1&2\\0&0&4\end{bmatrix}$.
It can be observed that in the first column, two entries are zero. Thus, we expand along the first column $(C_1)$ for easier calculation.
$|\text{A}|=1\begin{vmatrix}1&2\\0&4\end{vmatrix}-0\begin{vmatrix}0&1\\0&4\end{vmatrix}+0\begin{vmatrix}0&1\\1&2\end{vmatrix}=1\left(4-0\right)-0+0=4$
$ \therefore27|\text{A}|=27(4)=108 \dots\dots(1)$
$ \text{Now},3\text{A}=3\begin{bmatrix}1&0&1\\0&1&2\\0&0&4\end{bmatrix}=\begin{bmatrix}3&0&3\\0&3&6\\0&0&12\end{bmatrix}$
$\therefore|3\text{A}|=3\begin{vmatrix}3&6\\0&12\end{vmatrix}-0\begin{vmatrix}0&3\\0&12\end{vmatrix}+0\begin{vmatrix}0&3\\3&6\end{vmatrix}$
$=3\left(36-0\right)=3\left(36\right)=108 \dots\dots(2)$
From equations (1) and (2), we have:
$\left|3A\right|=27\left|A\right|$
Hence, the given result is proved.
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