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Question 4

# 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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