# Adding more difficult Matrices

I've already asked a question similar to this but this one is more difficult and plus want to make sure I'm understanding this correctly.

I've got this equation below:

$$\begin{bmatrix} 2 & 0 & -1 \\ 4 & -5 & 2 \\ \end{bmatrix} + 2* \begin{bmatrix} 7 & -5 & 1 \\ 1 & -4 & -3 \\ \end{bmatrix}$$

I believe it is solved like so:

$$\begin{bmatrix} 2 & 0 & -1 \\ 4 & -5 & 2 \\ \end{bmatrix} + \begin{bmatrix} 14 & -10 & 2 \\ 2 & -8 & -6 \\ \end{bmatrix}$$

$$\begin{bmatrix} 16 & -10 & 1 \\ 6 & -13 & -4 \\ \end{bmatrix}$$

Am I doing the addition right? Add both row 1's together then both row 2's together. For example, it would look something like this, when adding:

$$\begin{bmatrix} 2+14 & 0-10 & -1 + 2 \\ 4+2 & -5-8 & 2+-6 \\ \end{bmatrix}$$

EDIT: caught an addition error I made with the 2 + 14 for my final answer. Originally I had 28 down because I was multiplying but I think what I want is 16.

• Everything there is correct. – Doug M Feb 5 '18 at 19:51
• Thanks. That's what I thought but i'm still new to this and my teacher doesn't explain it too well. – Liath Feb 5 '18 at 19:57
• Would it help if I said that matrices are vectors and follow the identical rules for addition and scalar multiplication? If not, don't worry, I don't know how the subject has been introduced to you. – Doug M Feb 5 '18 at 20:01
• Addition and scalar multiplication are done coefficientwise. – Bernard Feb 5 '18 at 20:01
• Perhaps use a word other than complex in your title since complex numbers mean something else... – Michael Burr Feb 5 '18 at 20:25

## 1 Answer

Robert Howard, per your request:

Everything there is correct.

Matrices are vectors and follow the identical rules for addition and scalar multiplication.