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

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 - B)C = AC - BC

Answers (2)

c) To prove: (A - B)C = AC - BC

A s, L H S=(A-B) C
Substituting the values of \mathrm{A} . \mathrm{B} and \mathrm{C} and multiplying according to the rule of matrix multiplication.
(A-B)=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]-\left[\begin{array}{cc} 4 & 0 \\ 1 & 5 \end{array}\right]=\left[\begin{array}{cc} 1-4 & 2-0 \\ -1-1 & 3-5 \end{array}\right]=\left[\begin{array}{cc} -3 & 2 \\ -2 & -2 \end{array}\right]

LHS=(A-B) C=\left[\begin{array}{cc} -3 & 2 \\ -2 & -2 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]=\left[\begin{array}{cc} -4 & -4 \\ -6 & 4 \end{array}\right]


A C=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]=\left[\begin{array}{cc} 4 & -4 \\ 1 & -6 \end{array}\right]

B C=\left[\begin{array}{ll} 4 & 0 \\ 1 & 5 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]=\left[\begin{array}{cc} 8 & 0 \\ 7 & -10 \end{array}\right]

RHS=A C-B C=\left[\begin{array}{cc} 4-8 & -4-0 \\ 1-7 & -6+10 \end{array}\right]=\left[\begin{array}{cc} -4 & -4 \\ -6 & 4 \end{array}\right]=LHS

Hence (A-B) C=A C-B C...proved

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c) To prove: (A - B)C = AC - BC

A s, L H S=(A-B) C
Substituting the values of \mathrm{A} . \mathrm{B} and \mathrm{C} and multiplying according to the rule of matrix multiplication.
\\LHS =\left(\left[\begin{array}{cc}1 & -1 \\ 2 & 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]$ \\$\Rightarrow \mathrm{LHS}=\left[\begin{array}{cc}-3 & -1 \\ 1 & -2\end{array}\right]\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right]=\left[\begin{array}{cc}-7 & 2 \\ 0 & 4\end{array}\right]$ $\mathrm{RHS}=\mathrm{AC} \cdot \mathrm{BC}=\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right]-\left[\begin{array}{cc}4 & 0 \\ 1 & 5\end{array}\right]\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right]
\\ \Rightarrow \mathrm{RHS}=\left[\begin{array}{cc}1 & 2 \\ 7 & -6\end{array}\right]-\left[\begin{array}{cc}8 & 0 \\ 7 & -10\end{array}\right]=\left[\begin{array}{cc}1-8 & 2-0 \\ 7-7 & -6-(-10)\end{array}\right]=\left[\begin{array}{cc}-7 & 2 \\ 0 & 4\end{array}\right]

We have, L H S=R H S=\left[\begin{array}{cc}-7 & 2 \\ 0 & 4\end{array}\right]
Hence (A-B) C=A C-B C...proved

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