Chapter 8.1, Problem 152E

a.

To determine

To find: $A-B.$ if possible.

Answer to Problem 152E

  $A-B$ can’t be determined.

Explanation of Solution

Given information:

  $A=\left[\begin{array}{cc}-1& 4\\ 5& 1\\ 0& -1\end{array}\right]\text{and}B=\left[\begin{array}{ccc}0& 4& 0\\ 3& 1& -2\end{array}\right]$

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.

Calculations:

Consider the matrices.

  $A=\left[\begin{array}{cc}-1& 4\\ 5& 1\\ 0& -1\end{array}\right]\text{and}B=\left[\begin{array}{ccc}0& 4& 0\\ 3& 1& -2\end{array}\right]$

The order of matrix $A$ is $3\times 2$ and that of matrix $B$ is $2\times 3$ .

Since the orders of both the matrices are different, performing $A-B$ is not possible.

b.

To determine

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

Answer to Problem 152E

  $2B-3A$ can’t be determined.

Explanation of Solution

Given information:

  $A=\left[\begin{array}{cc}-1& 4\\ 5& 1\\ 0& -1\end{array}\right]\text{and}B=\left[\begin{array}{ccc}0& 4& 0\\ 3& 1& -2\end{array}\right]$

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.

Calculations:

  $\begin{array}{l}3A=3\left[\begin{array}{cc}-1& 4\\ 5& 1\\ 0& -1\end{array}\right]\\ =\left[\begin{array}{cc}-3& 12\\ 15& 3\\ 0& -3\end{array}\right]\end{array}$

And

  $\begin{array}{c}2B=2\left[\begin{array}{ccc}0& 4& 0\\ 3& 1& -2\end{array}\right]\\ =\left[\begin{array}{ccc}0& 8& 0\\ 6& 2& -4\end{array}\right]\end{array}$

The order of $3A$ is $3\times 2$ and that of $2B$ is $2\times 3$ . Since they are different, $2B-3A$ can’t be performed.

(c).

To determine

To find $AB$ if possible.

s

Answer to Problem 152E

  $AB=\left[\begin{array}{ccc}12& 0& -8\\ 3& 21& -2\\ -3& -1& 2\end{array}\right]$

Explanation of Solution

Given:

  $A=\left[\begin{array}{cc}-1& 4\\ 5& 1\\ 0& -1\end{array}\right]\text{and}B=\left[\begin{array}{ccc}0& 4& 0\\ 3& 1& -2\end{array}\right]$

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 $3\times 2$ matrix, therefore, there product will be of dimension $3\times 3$ and hence it will be possible.

  $\begin{array}{c}AB=\left[\begin{array}{cc}-1& 4\\ 5& 1\\ 0& -1\end{array}\right]\left[\begin{array}{ccc}0& 4& 0\\ 3& 1& -2\end{array}\right]\\ =\left[\begin{array}{ccc}-1\cdot 0+4\cdot 3& -1\cdot 4+4\cdot 1& -1\cdot 0+4\cdot \left(-2\right)\\ 5\cdot 0+1\cdot 3& 5\cdot 4+1\cdot 1& 5\cdot 0+1\cdot \left(-2\right)\\ 0\cdot 0+\left(-1\right)\cdot 3& 0\cdot 4+\left(-1\right)\cdot 1& 0\cdot 0+\left(-1\right)\cdot \left(-2\right)\end{array}\right]\\ =\left[\begin{array}{ccc}12& 0& -8\\ 3& 21& -2\\ -3& -1& 2\end{array}\right]\end{array}$

(d).

To determine

To find $BA$ if possible.

Answer to Problem 152E

  $BA=\left[\begin{array}{cc}20& 4\\ 2& 15\end{array}\right]$

Explanation of Solution

Given:

  $A=\left[\begin{array}{cc}-1& 4\\ 5& 1\\ 0& -1\end{array}\right]\text{and}B=\left[\begin{array}{ccc}0& 4& 0\\ 3& 1& -2\end{array}\right]$

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

  $\begin{array}{c}BA=\left[\begin{array}{ccc}0& 4& 0\\ 3& 1& -2\end{array}\right]\left[\begin{array}{cc}-1& 4\\ 5& 1\\ 0& -1\end{array}\right]\\ =\left[\begin{array}{cc}0\cdot \left(-1\right)+4\cdot 5+0\cdot 0& 0\cdot \left(4\right)+4\cdot 1+0\cdot \left(-1\right)\\ 3\cdot \left(-1\right)+1\cdot 5+\left(-2\right)\cdot 0& 3\cdot \left(4\right)+1\cdot 1+\left(-2\right)\cdot \left(-1\right)\end{array}\right]\\ =\left[\begin{array}{cc}20& 4\\ 2& 15\end{array}\right]\end{array}$