Chapter 8.1, Problem 149E

a.

To determine

To find $A-B$ if possible.

Answer to Problem 149E

The difference is $A-B=\left[\begin{array}{c}2\\ 6\end{array}\right]$

Explanation of Solution

Given information:

The matrices are,

  $A=\left[\begin{array}{c}6\\ 3\end{array}\right]\text{and}B=\left[\begin{array}{c}4\\ -3\end{array}\right]$

Concept used:

Operations like addition and subtraction of matrices are only possible when they are of same dimension.

Dimension of matrix with a rows and b columns is given by $a\times b$ .

If a matrix is multiplied by a scalar then each element of the matrix is multiplied by the same scalar.

Calculation:

The difference of both matrices is,

  $\begin{array}{l}A-B=\left[\begin{array}{c}6\\ 3\end{array}\right]-\left[\begin{array}{c}4\\ -3\end{array}\right]\\ A-B=\left[\begin{array}{c}6-4\\ 3-\left(-3\right)\end{array}\right]\\ A-B=\left[\begin{array}{c}2\\ 6\end{array}\right]\end{array}$

b.

To determine

To find $2B-3A$ if possible.

Answer to Problem 149E

  $2B-3A=\left[\begin{array}{c}-10\\ -15\end{array}\right]$

Explanation of Solution

Given information:

The matrices are,

  $A=\left[\begin{array}{c}6\\ 3\end{array}\right]\text{and}B=\left[\begin{array}{c}4\\ -3\end{array}\right]$

Concept used:

Operations like addition and subtraction of matrices are only possible when they are of same dimension.

Dimension of matrix with a rows and b columns is given by $a\times b$ .

If a matrix is multiplied by a scalar then each element of the matrix is multiplied by the same scalar.

Calculation:

Calculate $2B-3A$ as shown:

  $\begin{array}{l}2B-3A=2\left[\begin{array}{c}4\\ -3\end{array}\right]-3\left[\begin{array}{c}6\\ 3\end{array}\right]\\ 2B-3A=\left[\begin{array}{c}8\\ -6\end{array}\right]-\left[\begin{array}{c}18\\ 9\end{array}\right]\\ 2B-3A=\left[\begin{array}{c}8-18\\ -6-9\end{array}\right]\\ 2B-3A=\left[\begin{array}{c}-10\\ -15\end{array}\right]\end{array}$

c.

To determine

To find $AB$ if possible.

Answer to Problem 149E

The product $AB$ of given matrices is not possible.

Explanation of Solution

Given:

The matrices:

  $A=\left[\begin{array}{c}6\\ 3\end{array}\right]\text{and}B=\left[\begin{array}{c}4\\ -3\end{array}\right]$

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 $c_{ij}=a_{i1}{b}_{1j}+a_{i2}{b}_{2j}+a_{i3}{b}_{3j}\mathrm{...}+a_{in}{b}_{nj}$ .

Calculation:

In order to find product of below matrices AB

  $A=\left[\begin{array}{c}6\\ 3\end{array}\right]\text{and}B=\left[\begin{array}{c}4\\ -3\end{array}\right]$

First check if this product is possible.

Note that here number of columns of A is 1 but the number of rows of B is 2.

Thus, product of given matrices is not possible.

d.

To determine

To find $BA$ if possible.

Answer to Problem 149E

The product $BA$ of given matrices is not possible.

Explanation of Solution

Given:

The matrices:

  $A=\left[\begin{array}{c}6\\ 3\end{array}\right]\text{and}B=\left[\begin{array}{c}4\\ -3\end{array}\right]$

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 $c_{ij}=a_{i1}{b}_{1j}+a_{i2}{b}_{2j}+a_{i3}{b}_{3j}\mathrm{...}+a_{in}{b}_{nj}$ .

Calculation:

In order to find product of below matrices BA.

  $A=\left[\begin{array}{c}6\\ 3\end{array}\right]\text{and}B=\left[\begin{array}{c}4\\ -3\end{array}\right]$

First check if this product is possible.

Note that here number of columns of Bis 1 but the number of rows of Ais 2.

Thus, product of given matrices is not possible.