If A=\(\begin{bmatrix}1 & 0 & 1\\0 & 1 & 2\\0 & 0 & 4\end{bmatrix}\), then show that |3A| = 27|A|.
The given matrix is
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 (C1) for easier calculation.
| 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(4 – 0) – 0 + 0 = 4
So 27 |A| = 27 (4) = 108 ……. (i)
Now 3A = 3\(\begin{bmatrix}1 & 0 & 1\\0 & 1 & 2\\0 & 0 & 4\end{bmatrix}\)=\(\begin{bmatrix}3 & 0 & 3\\0 & 3 & 6\\0 & 0 & 12\end{bmatrix}\)
So |3A| = 3\(\begin{vmatrix}3 & 6\\0 & 12\end{vmatrix}\) - 0\(\begin{vmatrix}0 & 3\\0 & 12\end{vmatrix}\) + 0\(\begin{vmatrix}0 &3\\0 & 6\end{vmatrix}\)
= 3 (36 – 0) = 3(36) 108 ……….. (ii)
From equations (i) and (ii), we have:
|3A| = 27|A|
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