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  • Let A = [1 2 - 1 3] and a = 4, b = -2. Show that: a(C - A) = aC -aA

Let A=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right], B=\left[\begin{array}{ll} 4 & 0 \\ 1 & 5 \end{array}\right], C=\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right] and a = 4, b = -2.

Show that:
a(C - A) = aC -aA

Answers (1)

We have to prove: a(C - A) = aC -aA

As,

\\LHS =a(C-A)=4\left(\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right]-\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]\right)$ \\$\Rightarrow \mathrm{LHS}=4\left(\left[\begin{array}{cc}2-1 & 0-2 \\ 1-(-1) & -2-3\end{array}\right]\right)=4\left[\begin{array}{cc}1 & -2 \\ 2 & -5\end{array}\right]=\left[\begin{array}{cc}4 & -8 \\ 8 & -20\end{array}\right]$ \\$\mathrm{RHS}=\mathrm{aC}-\mathrm{aA}={ }^{4}\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right]-4\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]$ \\$\Rightarrow a C-a A=\left[\begin{array}{cc}8 & 0 \\ 4 & -8\end{array}\right]-\left[\begin{array}{cc}4 & 8 \\ -4 & 12\end{array}\right]=\left[\begin{array}{cc}4 & -8 \\ 8 & -20\end{array}\right]
Clearly LHS =\mathrm{RHS}=\left[\begin{array}{cc}4 & -8 \\ 8 & -20\end{array}\right]
Hence, we have
$a(C-A)=a C-a A \ldots$proved

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