Chapter 7, Problem 6RE

Solution

a.

To calculate: The matrix multiplication $AB$ by using the definition of the matrix multiplication.

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

The multiplication is not possible.

Given Information:

The matrix are,

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

Formula Used:

The matrix multiplication is possible if and only if order of the first matrix column equal to the order of the second matrix row.

Calculation:

Consider the given matrix,

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

It can be observed that the number of column of matrix $A$ is not equal to the number of row of the matrix $B$ . As $A$ have 2 column and $B$ have 4 rows.

Thus, the multiplication $AB$ is not possible.

Hence, the multiplication does not exists.

b.

To calculate: The matrix multiplication $BA$ by using the definition of the matrix multiplication.

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

The multiplication is $\left[\begin{array}{cc}-3& -1\\ -1& 3\\ -3& 4\\ -1& 2\end{array}\right]$ .

Given Information:

The matrix are,

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

Formula Used:

The matrix multiplication is possible if and only if order of the first matrix column equal to the order of the second matrix row.

Calculation:

Consider the given matrix,

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

It can be observed that the number of column of matrix $B$ is equal to the number of row of the matrix $A$ .

Now, find the value of $BA$ .

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

Hence, the obtained matrix after multiplication is $\left[\begin{array}{cc}-3& -1\\ -1& 3\\ -3& 4\\ -1& 2\end{array}\right]$ .