Q : 4         If A =\begin{bmatrix} 1 &0 &1 \\ 0& 1& 2\\ 0& 0 &4 \end{bmatrix}  then show that  |3A|=27|A|

Answers (1)

Given MatrixA =\begin{bmatrix} 1 &0 &1 \\ 0& 1& 2\\ 0& 0 &4 \end{bmatrix}

Calculating 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, \left | 3A \right | = 3(3(12) - 6(0) ) - 0(0(12)-0(6)) + 3(0-0) = 3(36) = 108

 calculating 27|A|,

|A| = \begin{vmatrix} 1 & 0 &1 \\ 0 & 1 & 2\\ 0& 0 &4 \end{vmatrix} = 1\begin{vmatrix} 1 &2 \\ 0 & 4 \end{vmatrix} - 0\begin{vmatrix} 0 &2 \\ 0& 4 \end{vmatrix} + 1\begin{vmatrix} 0 &1 \\ 0& 0 \end{vmatrix} = 4 -0 + 0 = 4

So, 27|A| = 27(4) = 108

Therefore |3A|=27|A|.

Hence proved.

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