Chapter B, Problem 1E

(a)

To determine

To find: The matrix of $A+B$.

Answer to Problem 1E

The result after performing the indicated operation of addition is $\underset{\_}{\left[\begin{array}{cc}2& 11\\ 2& -1\end{array}\right]}$.

Explanation of Solution

Given that the matrix A is $\left[\begin{array}{cc}4& 5\\ -6& 9\end{array}\right]$ and B is $\left[\begin{array}{cc}-2& 6\\ 8& -10\end{array}\right]$.

It is known that two or more matrices can be added when they have same order.

Compute the matrix $A+B$ as follows.

$\begin{array}{c}A+B=\left[\begin{array}{cc}4& 5\\ -6& 9\end{array}\right]+\left[\begin{array}{cc}-2& 6\\ 8& -10\end{array}\right]\\ =\left[\begin{array}{cc}4+\left(-2\right)& 5+6\\ \left(-6\right)+8& 9+\left(-10\right)\end{array}\right]\\ =\left[\begin{array}{cc}2& 11\\ 2& -1\end{array}\right]\end{array}$

Thus, the result after performing the indicated operation of addition is $\underset{\_}{\left[\begin{array}{cc}2& 11\\ 2& -1\end{array}\right]}$.

(b)

To determine

To find: The matrix of $B-A$.

Answer to Problem 1E

The result after performing the indicated operation of addition is $\left[\begin{array}{cc}-6& 1\\ 14& -19\end{array}\right]$.

Explanation of Solution

Given that the matrix A is $\left[\begin{array}{cc}4& 5\\ -6& 9\end{array}\right]$ and B is $\left[\begin{array}{cc}-2& 6\\ 8& -10\end{array}\right]$.

It is known that two or more matrices can be subtracted when they have same order.

Compute the matrix $B-A$ as follows.

$\begin{array}{c}B-A=\left[\begin{array}{cc}-2& 6\\ 8& -10\end{array}\right]-\left[\begin{array}{cc}4& 5\\ -6& 9\end{array}\right]\\ =\left[\begin{array}{cc}\left(-2\right)-4& 6-5\\ 8-\left(-6\right)& \left(-10\right)-9\end{array}\right]\\ =\left[\begin{array}{cc}-6& 1\\ 14& -19\end{array}\right]\end{array}$

Thus the result after performing the indicated operation is $\left[\begin{array}{cc}-6& 1\\ 14& -19\end{array}\right]$.

(c)

To determine

To find: The matrix of $2A+3B$.

Answer to Problem 1E

The matrix of $2A+3B$ is $\left[\begin{array}{cc}2& 28\\ 12& -12\end{array}\right]$.

Explanation of Solution

Given that the matrix A is $\left[\begin{array}{cc}4& 5\\ -6& 9\end{array}\right]$ and B is $\left[\begin{array}{cc}-2& 6\\ 8& -10\end{array}\right]$.

Multiply each term of the matrix A by a scalar value $2$ to get the matrix $2A$.

$\begin{array}{c}2\left[\begin{array}{cc}4& 5\\ -6& 9\end{array}\right]=\left[\begin{array}{cc}2\cdot 4& 2\cdot 5\\ 2\left(-6\right)& 2\cdot 9\end{array}\right]\\ =\left[\begin{array}{cc}8& 10\\ -12& 18\end{array}\right]\\ \end{array}$

Similarly, multiply each term of the matrix B by a scalar value $3$ to get the matrix $3B$.

$\begin{array}{c}3\left[\begin{array}{cc}-2& 6\\ 8& -10\end{array}\right]=\left[\begin{array}{cc}3\left(-2\right)& 3\cdot 6\\ 3\cdot 8& 3\left(-10\right)\end{array}\right]\\ =\left[\begin{array}{cc}-6& 18\\ 24& -30\end{array}\right]\end{array}$

It is known that two or more matrices can be added when they have same order.

Now, add $2A$ and $3B$ to get the result of matrix $2A+3B$ as follows.

$\begin{array}{c}2A+3B=\left[\begin{array}{cc}8& 10\\ -12& 18\end{array}\right]+\left[\begin{array}{cc}-6& 18\\ 24& -30\end{array}\right]\\ =\left[\begin{array}{cc}8+\left(-6\right)& 10+18\\ \left(-12\right)+24& 18+\left(-30\right)\end{array}\right]\\ =\left[\begin{array}{cc}2& 28\\ 12& -12\end{array}\right]\end{array}$

Thus, the matrix of $2A+3B$ is $\left[\begin{array}{cc}2& 28\\ 12& -12\end{array}\right]$.