Chapter 8.2, Problem 50E

(a).

To determine

To find $AB$ , where $A=\left[\begin{array}{ccc}1& 3& -2\\ -5& 10& 1\end{array}\right]$ and $B=\left[\begin{array}{c}3\\ 3\\ 3\end{array}\right]$ .

Answer to Problem 50E

  $AB=\left[\begin{array}{c}6\\ 18\end{array}\right]$

Explanation of Solution

Given:

Consider $A=\left[\begin{array}{ccc}1& 3& -2\\ -5& 10& 1\end{array}\right]$ and $B=\left[\begin{array}{c}3\\ 3\\ 3\end{array}\right]$ .

Concept used:

If $A=\left[a_{ij}\right]$ is an $m\times n$ matrix and $B=\left[b_{ij}\right]$ is an $n\times p$ matrix, then the product $AB$ is an $m\times p$ matrix given by $AB=\left[c_{ij}\right]$ , where $c_{ij}=a_{i1}{b}_{1j}+a_{i2}{b}_{2j}+a_{i3}{b}_{3j}+\cdots +a_{in}{b}_{nj}$ .

Calculation:

Here, Since A is a $2\times 3$ and B is a $3\times 1$ matrix, therefore, there product will be of dimension $2\times 1$ and hence it will be possible.

  $\begin{array}{c}AB=\left[\begin{array}{ccc}1& 3& -2\\ -5& 10& 1\end{array}\right]\left[\begin{array}{c}3\\ 3\\ 3\end{array}\right]\\ =\left[\begin{array}{c}3+9-6\\ -15+30+3\end{array}\right]\\ =\left[\begin{array}{c}6\\ 18\end{array}\right]\end{array}$

(b).

To determine

To find $BA$ , where $A=\left[\begin{array}{ccc}1& 3& -2\\ -5& 10& 1\end{array}\right]$ and $B=\left[\begin{array}{c}3\\ 3\\ 3\end{array}\right]$ .

Answer to Problem 50E

  $BA=$ Not possible

Explanation of Solution

Given:

Consider $A=\left[\begin{array}{ccc}1& 3& -2\\ -5& 10& 1\end{array}\right]$ and $B=\left[\begin{array}{c}3\\ 3\\ 3\end{array}\right]$ .

Concept used:

If $A=\left[a_{ij}\right]$ is an $m\times n$ matrix and $B=\left[b_{ij}\right]$ is an $n\times p$ matrix, then the product $AB$ is an $m\times p$ matrix given by $AB=\left[c_{ij}\right]$ , where $c_{ij}=a_{i1}{b}_{1j}+a_{i2}{b}_{2j}+a_{i3}{b}_{3j}+\cdots +a_{in}{b}_{nj}$ .

Calculation:

Since B is a $3\times 1$ and A is a $2\times 3$ matrix, therefore, there the product $BA$ will not be possible according to the definition of the matrix multiplication which states that number of columns in matrix B should be equal to number of rows in matrix A .

(c).

To determine

To find $A^2$ , where $A=\left[\begin{array}{ccc}1& 3& -2\\ -5& 10& 1\end{array}\right]$ .

Answer to Problem 50E

  $A^2$ is not possible.

Explanation of Solution

Given:

Consider $A=\left[\begin{array}{ccc}1& 3& -2\\ -5& 10& 1\end{array}\right]$ .

Concept used:

If $A=\left[a_{ij}\right]$ is an $m\times n$ matrix and $B=\left[b_{ij}\right]$ is an $n\times p$ matrix, then the product $AB$ is an $m\times p$ matrix given by $AB=\left[c_{ij}\right]$ , where $c_{ij}=a_{i1}{b}_{1j}+a_{i2}{b}_{2j}+a_{i3}{b}_{3j}+\cdots +a_{in}{b}_{nj}$ .

Calculation:

Square of a matrix is only possible if it is a square matrix. Here, matrix A is of dimension $2\times 3$ , which is not a square matrix, therefore, product $A^2$ will not be possible.