Chapter 3.6, Problem 4E

To determine

To find: $6B+DC$

Answer to Problem 4E

  $6B+DC$ is not possible.

Explanation of Solution

Given information:

  $A=\left[\begin{array}{cc}4.5& -9.0\\ -7.4& 9.4\end{array}\right]$

  $B=\left[\begin{array}{cc}1.9& -5.9\\ 2.9& 5.0\end{array}\right]$

  $C=\left[\begin{array}{ccc}7.0& 5.5& -1.9\\ 7.6& -9.9& 0.5\end{array}\right]$

  $D=\left[\begin{array}{cc}8.5& 8.0\\ -1.2& -5.9\\ 0.7& 8.9\end{array}\right]$

Calculation:

  $\begin{array}{l}B=\left[\begin{array}{cc}1.9& -5.9\\ 2.9& 5.0\end{array}\right]\\ \Rightarrow 6B=6\times \left[\begin{array}{cc}1.9& -5.9\\ 2.9& 5.0\end{array}\right]\\ \Rightarrow 6B=\left[\begin{array}{cc}6\times 1.9& 6\times \left(-5.9\right)\\ 6\times 2.9& 6\times 5.0\end{array}\right]\\ \Rightarrow 6B=\left[\begin{array}{cc}11.4& -35.4\\ 17.4& 30.0\end{array}\right]\end{array}$

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

  $D=\left[\begin{array}{cc}8.5& 8.0\\ -1.2& -5.9\\ 0.7& 8.9\end{array}\right]$ in the form $\left(3\times 2\right)$

  $C=\left[\begin{array}{ccc}7.0& 5.5& -1.9\\ 7.6& -9.9& 0.5\end{array}\right]$ in the form $\left(2\times 3\right)$

As the number of Columns of matrix D is equal to number of rows of Matrix C .

So, the Product $DC$ is defined.

And the product $DC$ in the form $\left(3\times 3\right)$ .

  $6B=\left[\begin{array}{cc}11.4& -35.4\\ 17.4& 30.0\end{array}\right]$ is in the form $\left(2\times 2\right)$

As Matrix $6B$ is in the form $\left(2\times 2\right)$ and Matrix $DC$ in the form $\left(3\times 3\right)$

So, we cannot add those two matrices as their dimensions are different.

So,

  $6B+DC$ is not possible.