Chapter 8, Problem 42RE

a.

To determine

Find theaddition of the given matrices.

Answer to Problem 42RE

The addition of the matrices is $\left[\begin{array}{cc}7& 14\\ 9& 26\\ 30& 30\end{array}\right]$ .

Explanation of Solution

Given information:

The givenmatricesare

  $A=\left[\begin{array}{cc}4& 3\\ -6& 1\\ 10& 1\end{array}\right]$ , $B=\left[\begin{array}{cc}3& 11\\ 15& 25\\ 20& 29\end{array}\right]$

Calculation:

If $A=\left[a_{ij}\right]$ and $B=\left[b_{ij}\right]$ are matrices of order $m\times n$ , then their sum is the $m\times n$ matrix given by

  $A+B=\left[a_{ij}+b_{ij}\right]$

The sum of two matrices of different orders is undefined.

Now

  $\begin{array}{l}A+B\\ =\left[\begin{array}{cc}4& 3\\ -6& 1\\ 10& 1\end{array}\right]+\left[\begin{array}{cc}3& 11\\ 15& 25\\ 20& 29\end{array}\right]\end{array}$

  $\begin{array}{l}=\left[\begin{array}{cc}4+3& 3+11\\ -6+15& 1+25\\ 10+20& 1+29\end{array}\right]\\ =\left[\begin{array}{cc}7& 14\\ 9& 26\\ 30& 30\end{array}\right]\end{array}$

Hence theaddition of the matricesis $\left[\begin{array}{cc}7& 14\\ 9& 26\\ 30& 30\end{array}\right]$ .

b.

To determine

Find the subtraction of the given matrices.

Answer to Problem 42RE

The difference of the matrices is $\left[\begin{array}{cc}1& -8\\ -21& -24\\ -10& -28\end{array}\right]$ .

Explanation of Solution

Given information:

The given matrices are

  $A=\left[\begin{array}{cc}4& 3\\ -6& 1\\ 10& 1\end{array}\right]$ , $B=\left[\begin{array}{cc}3& 11\\ 15& 25\\ 20& 29\end{array}\right]$

Calculation:

If $A=\left[a_{ij}\right]$ and $B=\left[b_{ij}\right]$ are matrices of order $m\times n$ , then their difference is the $m\times n$ matrix given by

  $A-B=\left[a_{ij}-b_{ij}\right]$

The sum of two matrices of different orders is undefined.

Now

  $\begin{array}{l}A-B\\ =\left[\begin{array}{cc}4& 3\\ -6& 1\\ 10& 1\end{array}\right]-\left[\begin{array}{cc}3& 11\\ 15& 25\\ 20& 29\end{array}\right]\end{array}$

  $\begin{array}{l}=\left[\begin{array}{cc}4-3& 3-11\\ -6-15& 1-25\\ 10-20& 1-29\end{array}\right]\\ =\left[\begin{array}{cc}1& -8\\ -21& -24\\ -10& -28\end{array}\right]\end{array}$

Hence the difference of the matrices is $\left[\begin{array}{cc}1& -8\\ -21& -24\\ -10& -28\end{array}\right]$ .

c.

To determine

Find the scalar multiplication of the given matrix.

Answer to Problem 42RE

The scalar multiplication of the matrix is $\left[\begin{array}{cc}16& 12\\ -24& 4\\ 40& 4\end{array}\right]$ .

Explanation of Solution

Given information:

The given matrixis

  $A=\left[\begin{array}{cc}4& 3\\ -6& 1\\ 10& 1\end{array}\right]$

and the scalar quantity is $4$ .

Calculation:

If $A=\left[a_{ij}\right]$ is an $m\times n$ matrix and $c$ is a scalar, then the scalar multiple of $A$ by $c$ is the $m\times n$ matrix given by

  $cA=\left[c{a}_{ij}\right]$

Now

  $\begin{array}{l}4A\\ =4\left[\begin{array}{cc}4& 3\\ -6& 1\\ 10& 1\end{array}\right]\\ =\left[\begin{array}{cc}4\cdot 4& 3\cdot 4\\ -6\cdot 4& 1\cdot 4\\ 10\cdot 4& 1\cdot 4\end{array}\right]\\ =\left[\begin{array}{cc}16& 12\\ -24& 4\\ 40& 4\end{array}\right]\end{array}$

Hence the scalar multiplication of the matrixis $\left[\begin{array}{cc}16& 12\\ -24& 4\\ 40& 4\end{array}\right]$ .

d.

To determine

Find the addition of matrices with scalar multiplication.

Answer to Problem 42RE

The addition of matrices is $\left[\begin{array}{cc}13& 36\\ 39& 76\\ 70& 58\end{array}\right]$ .

Explanation of Solution

Given information:

The given matrices are

  $A=\left[\begin{array}{cc}4& 3\\ -6& 1\\ 10& 1\end{array}\right]$ , $B=\left[\begin{array}{cc}3& 11\\ 15& 25\\ 20& 29\end{array}\right]$

and the given expression is $A+3B$ .

Calculation:

If $A=\left[a_{ij}\right]$ is an $m\times n$ matrix and $c$ is a scalar, then the scalar multiple of $A$ by $c$ is the $m\times n$ matrix given by

  $cA=\left[c{a}_{ij}\right]$

If $A=\left[a_{ij}\right]$ and $B=\left[b_{ij}\right]$ are matrices of order $m\times n$ , then their sum is the $m\times n$ matrix given by

  $A+B=\left[a_{ij}+b_{ij}\right]$

The sum of two matrices of different orders is undefined.

Now

  $\begin{array}{l}3B\\ =3\left[\begin{array}{cc}3& 11\\ 15& 25\\ 20& 29\end{array}\right]\\ =\left[\begin{array}{cc}3\cdot 3& 11\cdot 3\\ 15\cdot 3& 25\cdot 3\\ 20\cdot 3& 29\cdot 3\end{array}\right]\\ =\left[\begin{array}{cc}9& 33\\ 45& 75\\ 60& 57\end{array}\right]\end{array}$

  $\begin{array}{l}A+3B\\ =\left[\begin{array}{cc}4& 3\\ -6& 1\\ 10& 1\end{array}\right]+\left[\begin{array}{cc}9& 33\\ 45& 75\\ 60& 57\end{array}\right]\\ =\left[\begin{array}{cc}4+9& 3+33\\ -6+45& 1+75\\ 10+60& 1+57\end{array}\right]\\ =\left[\begin{array}{cc}13& 36\\ 39& 76\\ 70& 58\end{array}\right]\end{array}$

Hence the addition of matrices is $\left[\begin{array}{cc}13& 36\\ 39& 76\\ 70& 58\end{array}\right]$ .