Chapter 12.2, Problem 47PPSE

Solution

Whether the two expressions $\left(P+Q\right)\left(R+S\right)$ and $\left(P+Q\right)R+\left(P+Q\right)S$ are equal or not.

Yes, both expressions are equal that is $\left(P+Q\right)\left(R+S\right)=\left(P+Q\right)R+\left(P+Q\right)S$ .

Given information: Matrices are $P=\left[\begin{array}{cc}3& 4\\ 1& 2\end{array}\right],Q=\left[\begin{array}{cc}1& 0\\ 3& -2\end{array}\right],R=\left[\begin{array}{cc}1& 4\\ -2& 1\end{array}\right]\text{ and }S=\left[\begin{array}{cc}0& 1\\ 2& 1\end{array}\right]$ .

Formula used:

Matrix multiplication is,

  $\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]=\left[\begin{array}{cc}{a}_{11}{b}_{11}+a_{12}{b}_{21}& a_{11}{b}_{12}+a_{12}{b}_{22}\\ a_{21}{b}_{11}+a_{22}{b}_{21}& a_{21}{b}_{12}+a_{22}{b}_{22}\end{array}\right]$ .

Calculation:

Consider the matrices as $P=\left[\begin{array}{cc}3& 4\\ 1& 2\end{array}\right],Q=\left[\begin{array}{cc}1& 0\\ 3& -2\end{array}\right],R=\left[\begin{array}{cc}1& 4\\ -2& 1\end{array}\right]\text{ and }S=\left[\begin{array}{cc}0& 1\\ 2& 1\end{array}\right]$ .

Now,

  $\begin{array}{c}\left(P+Q\right)\left(R+S\right)=\left(\left[\begin{array}{cc}3& 4\\ 1& 2\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 3& -2\end{array}\right]\right)\left(\left[\begin{array}{cc}1& 4\\ -2& 1\end{array}\right]+\left[\begin{array}{cc}0& 1\\ 2& 1\end{array}\right]\right)\\ =\left[\begin{array}{cc}3+1& 4+0\\ 1+3& 2+\left(-2\right)\end{array}\right]\left[\begin{array}{cc}1+0& 4+1\\ -2+2& 1+1\end{array}\right]\\ =\left[\begin{array}{cc}4& 4\\ 4& 0\end{array}\right]\left[\begin{array}{cc}1& 5\\ 0& 2\end{array}\right]\\ =\left[\begin{array}{cc}4\times 1+4\times 0& 4\times 5+4\times 2\\ 4\times 1+0\times 0& 4\times 5+0\times 2\end{array}\right]\end{array}$

Further,

  $\begin{array}{c}\left(P+Q\right)\left(R+S\right)=\left[\begin{array}{cc}4+0& 20+8\\ 4+0& 20+0\end{array}\right]\\ =\left[\begin{array}{cc}4& 28\\ 4& 20\end{array}\right]\end{array}$

Also,

  $\left(P+Q\right)R+\left(P+Q\right)S=\left(\left[\begin{array}{cc}3& 4\\ 1& 2\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 3& -2\end{array}\right]\right)\left[\begin{array}{cc}1& 4\\ -2& 1\end{array}\right]+\left(\left[\begin{array}{cc}3& 4\\ 1& 2\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 3& -2\end{array}\right]\right)\left[\begin{array}{cc}0& 1\\ 2& 1\end{array}\right]$

  $=\left[\begin{array}{cc}3+1& 4+0\\ 1+3& 2+\left(-2\right)\end{array}\right]\left[\begin{array}{cc}1& 4\\ -2& 1\end{array}\right]+\left[\begin{array}{cc}3+1& 4+0\\ 1+3& 2+\left(-2\right)\end{array}\right]\left[\begin{array}{cc}0& 1\\ 2& 1\end{array}\right]$

  $=\left[\begin{array}{cc}4& 4\\ 4& 0\end{array}\right]\left[\begin{array}{cc}1& 4\\ -2& 1\end{array}\right]+\left[\begin{array}{cc}4& 4\\ 4& 0\end{array}\right]\left[\begin{array}{cc}0& 1\\ 2& 1\end{array}\right]$

  $=\left[\begin{array}{cc}4\times 1+4\times \left(-2\right)& 4\times 4+4\times 1\\ 4\times 1+0\times \left(-2\right)& 4\times 4+0\times 1\end{array}\right]+\left[\begin{array}{cc}4\times 0+4\times 2& 4\times 1+4\times 1\\ 4\times 0+0\times 2& 4\times 1+0\times 1\end{array}\right]$

Further,

  $\begin{array}{c}\left(P+Q\right)R+\left(P+Q\right)S=\left[\begin{array}{cc}4-8& 16+4\\ 4+0& 16+0\end{array}\right]+\left[\begin{array}{cc}0+8& 4+4\\ 0+0& 4+0\end{array}\right]\\ =\left[\begin{array}{cc}-4& 20\\ 4& 16\end{array}\right]+\left[\begin{array}{cc}8& 8\\ 0& 4\end{array}\right]\\ =\left[\begin{array}{cc}-4+8& 20+8\\ 4+0& 16+4\end{array}\right]\\ =\left[\begin{array}{cc}4& 28\\ 4& 20\end{array}\right]\end{array}$

Therefore, $\left(P+Q\right)\left(R+S\right)=\left(P+Q\right)R+\left(P+Q\right)S$ .