Question

If A $\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 2\\ 0& 0& 4\end{array}\right]$, then show that |3 A| = 27 |A|.

Answer 1

$$\begin{gathered} A=\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 2\\ 0& 0& 4\end{array}\right] \\ \Rightarrow 3A=\left[\begin{array}{ccc}3& 0& 3\\ 0& 3& 6\\ 0& 0& 12\end{array}\right]\left[\mathrm{Multiplying}\mathrm{each}\mathrm{element}\mathrm{of}\mathrm{A}\mathrm{by}3\right] \\ \Rightarrow \left|\begin{array}{c}3A\end{array}\right|={\left(-1\right)}^{1+1}3\left(36-0\right)+{\left(-1\right)}^{1+2}0\left(0-0\right)+{\left(-1\right)}^{1+3}3\left(0-0\right)=3\left(36-0\right)-0\left(0-0\right)+3\left(0-0\right)\left[\mathrm{Expanding}\mathrm{along}{R}_1\right] \\ =3\times 36=108...\left(1\right) \\ \Rightarrow \left|\begin{array}{c}A\end{array}\right|={\left(-1\right)}^{1+1}1\left(4-0\right)+{\left(-1\right)}^{1+2}0\left(0-0\right)+{\left(-1\right)}^{1+3}1\left(0-0\right)=1\left(4-0\right)-0\left(0-0\right)+1\left(0-0\right)=4\left[\mathrm{Expanding}\mathrm{along}{R}_1\right] \\ \Rightarrow 27\left|A\right|=27\times 4=108...\left(2\right) \\ \therefore \left|3A\right|=27\left|A\right|\left[\mathrm{From}\mathrm{eqs}.\left(1\right)\mathrm{and}\left(2\right)\right] \end{gathered}$$