Chapter 3.6, Problem 43PPS

To determine

To find: $X\left(ZZ\right)$

Answer to Problem 43PPS

  $\left(XX\right)Z=\left[\begin{array}{c}-12+69y+15x\\ 42.75y-18xy-18y^2+20.25x\end{array}\right]$ .

Explanation of Solution

Given information:

  $X=\left[\begin{array}{cc}2& -6\\ 3y& -4.5\end{array}\right]$

  $Y=\left[\begin{array}{cc}-5& -1.5\\ x+2& y\\ 13& 1.2\end{array}\right]$

  $Z=\left[\begin{array}{c}-3\\ x+y\end{array}\right]$

Calculation:

In order for matrix multiplication to e defined, the number of Columns in the first matrix must be equal to number of rows of second matrix.

Let Matrix A be in form $\left(m\times n\right)$ .

Matrix B be in the form $\left(n\times k\right)$ .

Then the product matrix will be in the form $\left(m\times k\right)$ .

Here for $\left(XX\right)$

  $X=\left[\begin{array}{cc}2& -6\\ 3y& -4.5\end{array}\right]$ in the form $\left(2\times 2\right)$

  $X=\left[\begin{array}{cc}2& -6\\ 3y& -4.5\end{array}\right]$ in the form $\left(2\times 2\right)$

As the number of Columns of matrix $X$ is equal to number of rows of Matrix $X$ .

So, the Product $XX$ is defined.

  $\begin{array}{l}XX=\left[\begin{array}{cc}2& -6\\ 3y& -4.5\end{array}\right]\left[\begin{array}{cc}2& -6\\ 3y& -4.5\end{array}\right]\\ \Rightarrow XX=\left[\begin{array}{cc}\left(2\times 2\right)+\left(\left(-6\right)\times 3y\right)& \left(2\times \left(-6\right)\right)+\left(\left(-6\right)\times \left(-4.5\right)\right)\\ \left(3y\times 2\right)+\left(\left(-4.5\right)\times 3y\right)& \left(3y\times \left(-6\right)\right)+\left(\left(-4.5\right)\times \left(-4.5\right)\right)\end{array}\right]\\ \Rightarrow XX=\left[\begin{array}{cc}4-18y& -12+27\\ 6y-13.5y& -18y+20.25\end{array}\right]\\ \Rightarrow XX=\left[\begin{array}{cc}4-18y& 15\\ -7.5y& -18y+20.25\end{array}\right]\end{array}$

Here for $\left(XX\right)Z$

  $XX=\left[\begin{array}{cc}4-18y& 15\\ -7.5y& -18y+20.25\end{array}\right]$ in the form $\left(2\times 2\right)$

  $Z=\left[\begin{array}{c}-3\\ x+y\end{array}\right]$ in the form $\left(2\times 1\right)$

As the number of Columns of matrix $XX$ is equal to number of rows of Matrix $Z$ .

So, the Product $\left(XX\right)Z$ is defined.

  $\begin{array}{l}\left(XX\right)Z=\left[\begin{array}{cc}4-18y& 15\\ -7.5y& -18y+20.25\end{array}\right]\left[\begin{array}{c}-3\\ x+y\end{array}\right]\\ \Rightarrow \left(XX\right)Z=\left[\begin{array}{c}\left(4-18y\right)\left(-3\right)+15\left(x+y\right)\\ \left(-7.5y\right)\left(-3\right)+\left(-18y+20.25\right)\left(x+y\right)\end{array}\right]\\ \Rightarrow \left(XX\right)Z=\left[\begin{array}{c}-12+54y+15x+15y\\ 22.5y-18xy-18y^2+20.25x+20.25y\end{array}\right]\\ \Rightarrow \left(XX\right)Z=\left[\begin{array}{c}-12+69y+15x\\ 42.75y-18xy-18y^2+20.25x\end{array}\right]\end{array}$