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

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(BC) = (AB)C

Answers (1)

We have to prove that: A(BC) = (AB)C

\begin{array}{l} \text { LHS = } \mathrm{A}(\mathrm{BC})=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]\left(\left[\begin{array}{cc} 4 & 0 \\ 1 & 5 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]\right) \\ \Rightarrow \mathrm{LHS}=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]\left(\begin{array}{cc} 4 \times 2+0 \times 1 & 4 \times 0+0 \times(-2) \\ 1 \times 2+5 \times 1 & 1 \times 0+5 \times(-2)]) \end{array}\right. \\ \Rightarrow \mathrm{LHS}=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]\left[\begin{array}{cc} 8 & 0 \\ 7 & -10 \end{array}\right] \\ LHS= {\left[\begin{array}{cc} 22 & -20 \\ 13 & -30 \end{array}\right]} \end{array}

\begin{aligned} &\mathrm{RHS}=(\mathrm{AB}) \mathrm{C}=\left(\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]\left[\begin{array}{cc} 4 & 0 \\ 1 & 5 \end{array}\right]\right)\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]\\ &\text { By matrix multiplication as done for LHS }\\ &\Rightarrow \mathrm{RHS}=\left[\begin{array}{cc} 6 & 10 \\ -1 & 15 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]\\ &\Rightarrow \mathrm{RHS}=\left[\begin{array}{ll} 22 & -20 \\ 13 & -30 \end{array}\right]\\ &\text { Evidently, LHS = RHS }=\left[\begin{array}{rr} 22 & -20 \\ 13 & -30 \end{array}\right]\\ &\therefore \mathrm{A}(\mathrm{BC})=(\mathrm{AB}) \mathrm{C} \ldots \text { .proved } \end{aligned}

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