If $A$ is any matrix and $p$ is any scalar (real number), then the product $pA$ is the matrix obtained by multiplying each entry of matrix $A$ by the scalar $p$. The matrix $pA$ is said to be a scalar multiple of $A$.

Example:

Let

Then

Properties of scalar multiplication

Let $r$ and $s$ be real numbers (scalars), and let $A$ and $B$ be matrices of the same order. Then the following properties hold:

1. $r(A \pm B) = rA \pm rB$
2. $(r \pm s)A = rA \pm sA$
3. $r(sA) = (rs)A$

Example 10.6

Let

a. Find $A + B$:

b. Find $4(A + B)$:

Alternatively, using the distributive property:

Then:

Conclusion: $4(A + B) = 4A + 4B$, as expected.

Two matrices can be multiplied if and only if the number of columns in the first matrix is equal to the number of rows in the second matrix. If matrix $A$ is of order $m \times a$ and matrix $B$ is of order $b \times n$, then the multiplication $AB$ is possible if and only if $a = b$. The resulting matrix $AB$ will be of order $m \times n$.

Procedure for matrix multiplication

Let:

Then, to compute $AB = A \cdot B$, follow these steps:

1. Multiply the first row of $A$ by the first column of $B$: $$$$

(a ;; b) \cdot \begin{bmatrix} r \ u \end{bmatrix} = ar + bu \quad \text{(1st element)}

(a ;; b) \cdot \begin{bmatrix} s \ v \end{bmatrix} = as + bv \quad \text{(2nd element)}

(c ;; d) \cdot \begin{bmatrix} r \ u \end{bmatrix} = cr + du \quad \text{(3rd element)}

(c ;; d) \cdot \begin{bmatrix} s \ v \end{bmatrix} = cs + dv \quad \text{(4th element)}

AB = \begin{bmatrix} ar + bu & as + bv \ cr + du & cs + dv \end{bmatrix}

A = \begin{bmatrix} 2 & 2 \ 3 & 5 \end{bmatrix}, \quad B = \begin{bmatrix} 3 & 2 \ 1 & 4 \end{bmatrix}

AB = \begin{bmatrix} 2 & 2 \ 3 & 5 \end{bmatrix} \cdot \begin{bmatrix} 3 & 2 \ 1 & 4 \end{bmatrix}

\begin{bmatrix} (2 \cdot 3 + 2 \cdot 1) & (2 \cdot 2 + 2 \cdot 4) \ (3 \cdot 3 + 5 \cdot 1) & (3 \cdot 2 + 5 \cdot 4) \end{bmatrix}

\begin{bmatrix} 6 + 2 & 4 + 8 \ 9 + 5 & 6 + 20 \end{bmatrix}

\begin{bmatrix} 8 & 12 \ 14 & 26 \end{bmatrix}

BA = \begin{bmatrix} 3 & 2 \ 1 & 4 \end{bmatrix} \cdot \begin{bmatrix} 2 & 2 \ 3 & 5 \end{bmatrix}

\begin{bmatrix} (3 \cdot 2 + 2 \cdot 3) & (3 \cdot 2 + 2 \cdot 5) \ (1 \cdot 2 + 4 \cdot 3) & (1 \cdot 2 + 4 \cdot 5) \end{bmatrix}

\begin{bmatrix} 6 + 6 & 6 + 10 \ 2 + 12 & 2 + 20 \end{bmatrix}

\begin{bmatrix} 12 & 16 \ 14 & 22 \end{bmatrix}

AB = \begin{bmatrix} 8 & 12 \ 14 & 26 \end{bmatrix} \ne BA = \begin{bmatrix} 12 & 16 \ 14 & 22 \end{bmatrix}

A = \begin{bmatrix} 2 & 5 & 7 \ 8 & 10 & 9 \end{bmatrix}, \quad B = \begin{bmatrix} 7 \ 3 \ 2 \end{bmatrix}

AB = \begin{bmatrix} 2 & 5 & 7 \ 8 & 10 & 9 \end{bmatrix} \cdot \begin{bmatrix} 7 \ 3 \ 2 \end{bmatrix}

\begin{bmatrix} 2 \cdot 7 + 5 \cdot 3 + 7 \cdot 2 \ 8 \cdot 7 + 10 \cdot 3 + 9 \cdot 2 \end{bmatrix}

\begin{bmatrix} 14 + 15 + 14 \ 56 + 30 + 18 \end{bmatrix}

\begin{bmatrix} 43 \ 104 \end{bmatrix}

Let $A$ be an $m \times n$ matrix. The transpose of matrix $A$, denoted by $A^T$, is the $n \times m$ matrix obtained by interchanging the rows and columns of $A$.

Examples:

If

If

If

Example

Given:

Verify that $(AB)^T = B^T A^T$

Step 1: Compute $AB$

Step 2: Compute $(AB)^T$

Step 3: Compute $B^T$ and $A^T$

Step 4: Compute $B^T A^T$

Conclusion:

Hence, verified that $(AB)^T = B^T A^T$.