M - Multiplying Matrices Lesson

Multiplying Matrices

Before we learn about how to multiply a matrix by another matrix - let's first talk about scalar multiplication. This just means to multiply a matrix by a constant.

For instance, let equation image indicator $LaTeX: A=\begin{bmatrix} 1 & -2 & 5\\ 0 & 4 & -1 \end{bmatrix}$ $A=\begin{bmatrix} 1 & -2 & 5\\ 0 & 4 & -1 \end{bmatrix}$ then if we want to find 2A, we would multiply each entry by 2.

So, equation image indicator $LaTeX: 2A=\begin{bmatrix} 2 & -4 & 10\\ 0 & 8 & -2 \end{bmatrix}$ $2A=\begin{bmatrix} 2 & -4 & 10\\ 0 & 8 & -2 \end{bmatrix}$

Try a few problems to see if you've got it:

equation image indicator $LaTeX: A=\begin{bmatrix} -4 & 7 & 9\\ 0 & 3 & 2 \end{bmatrix} B= \begin{bmatrix} 3 & 6 & 8\\ -1 & -2 & 0 \end{bmatrix} C=\begin{bmatrix} 1 & 0 & 5\\ 4 & 2 & -2\\ 7 & -3 & 1 \end{bmatrix} D= \begin{bmatrix} 7 & -2 & 0\\ 5 & 3 & 3\\ 6 & -1 & 4 \end{bmatrix}$ $A=\begin{bmatrix} -4 & 7 & 9\\ 0 & 3 & 2 \end{bmatrix} B= \begin{bmatrix} 3 & 6 & 8\\ -1 & -2 & 0 \end{bmatrix} C=\begin{bmatrix} 1 & 0 & 5\\ 4 & 2 & -2\\ 7 & -3 & 1 \end{bmatrix} D= \begin{bmatrix} 7 & -2 & 0\\ 5 & 3 & 3\\ 6 & -1 & 4 \end{bmatrix}$

Try these problems to see if you get the correct solution.

1. text annotation indicator Problem: -3B

Solution: $LaTeX: -3B=\begin{bmatrix} -9& -18 & -24\\ 3 & 6 & 0 \end{bmatrix}$ $-3B=\begin{bmatrix} -9& -18 & -24\\ 3 & 6 & 0 \end{bmatrix}$

2. Problem: (1/2)C text annotation indicator

Solution: $LaTeX: \frac{1}{2}C=\begin{bmatrix} \frac{1}{2}& 0 & \frac{5}{2}\\ 2 & 1 & -1\\ \frac{7}{2} & -\frac{3}{2} & \frac{1}{2} \end{bmatrix}$ $\frac{1}{2}C=\begin{bmatrix} \frac{1}{2}& 0 & \frac{5}{2}\\ 2 & 1 & -1\\ \frac{7}{2} & -\frac{3}{2} & \frac{1}{2} \end{bmatrix}$

3. Problem: 3C+2D text annotation indicator

Solution: $LaTeX: 3C+2D=\begin{bmatrix} 17& -4 & 15\\ 22 & 12 & 0 \\ 33 & -11 & 11 \end{bmatrix}$ $3C+2D=\begin{bmatrix} 17& -4 & 15\\ 22 & 12 & 0 \\ 33 & -11 & 11 \end{bmatrix}$

4. Problem: 4B - A text annotation indicator

Solution: $LaTeX: 4B-A=\begin{bmatrix} 16 & 17 & 23\\ -4 & -11 & -2 \end{bmatrix}$ $4B-A=\begin{bmatrix} 16 & 17 & 23\\ -4 & -11 & -2 \end{bmatrix}$

So far, matrices have behaved fairly predictably and with the same properties as real numbers. But, matrix multiplication is unusual!

In order to multiply two matrices A and B, the number of columns in matrix A must be equal to the number of rows in matrix B. If matrix AB is possible, the resulting matrix has the same number of rows as matrix A and columns as matrix B.

matrix A * matrix B = AB

(4 x 2) * (2 x 3) = 4 x 3

equation image indicator

4x2 2x3
Since the inner values are equal, the matrices can be multiplied.
The green values are the dimensions of the answer 4x3 $LaTeX: = \begin{bmatrix} (a\times i)+(b\times l) & (a\times j)+(b\times m) & (a\times k)+(b\times n)\\ (c\times i)+(d\times l) & (c\times j)+(d\times m) & (c\times k)+(d\times n) \\ (e\times i)+(f\times l) & (e\times j)+(f\times m) & (e\times k)+(f\times n) \\ (g\times i)+(h\times l) & (g\times j)+(h\times m) & (g\times k)+(h\times n) & \end{bmatrix}$ $= \begin{bmatrix} (a\times i)+(b\times l) & (a\times j)+(b\times m) & (a\times k)+(b\times n)\\ (c\times i)+(d\times l) & (c\times j)+(d\times m) & (c\times k)+(d\times n) \\ (e\times i)+(f\times l) & (e\times j)+(f\times m) & (e\times k)+(f\times n) \\ (g\times i)+(h\times l) & (g\times j)+(h\times m) & (g\times k)+(h\times n) & \end{bmatrix}$

Watch this video to try a problem:

Try a few problems to see if you've got it:

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