Chapter 12.1, Problem 1LC

To determine

To find: the sum or differences of the given matrix.

Answer to Problem 1LC

  $\left[\begin{array}{l}07\\ -45\end{array}\right]+\left[\begin{array}{l}-32\\ 4-1\end{array}\right]=\left[\begin{array}{l}-39\\ 04\end{array}\right]$ .

Explanation of Solution

Given:

The given matrix is $\left[\begin{array}{l}07\\ -45\end{array}\right]+\left[\begin{array}{l}-32\\ 4-1\end{array}\right]$ .

Concept used:

The sum or differences of the any matrix is possible, when the number of elements of both matrices are same. i.e. number of columns are same and number of rows are same.

Calculation:

The given matrix is $\left[\begin{array}{l}07\\ -45\end{array}\right]+\left[\begin{array}{l}-32\\ 4-1\end{array}\right]$ .

Here number of elements in both matrices are same.

Therefore, add the matrix. Now,

  $\begin{array}{l}\left[\begin{array}{l}07\\ -45\end{array}\right]+\left[\begin{array}{l}-32\\ 4-1\end{array}\right]=\left[\begin{array}{l}0+\left(-3\right)7+2\\ -4+45+\left(-1\right)\end{array}\right]\\ =\left[\begin{array}{l}0-37+2\\ -4+45-1\end{array}\right]\\ =\left[\begin{array}{l}-39\\ 04\end{array}\right]\end{array}$

Hence, the sum of the given matrix is $\left[\begin{array}{l}07\\ -45\end{array}\right]+\left[\begin{array}{l}-32\\ 4-1\end{array}\right]=\left[\begin{array}{l}-39\\ 04\end{array}\right]$ .