Worked examples · Sheet 1

Matrix operations

Theory used: Sum and product by a scalar Row-by-column product Transpose matrix Properties of the product

Exercise 1

Given the matrices

A = \begin{pmatrix} 2 & -1 \\ 0 & 3 \end{pmatrix}, \qquad B = \begin{pmatrix} 1 & 4 \\ -2 & 1 \end{pmatrix},

compute $3A - 2B^{T}$ .

B^{T} = \begin{pmatrix} 1 & -2 \\ 4 & 1 \end{pmatrix}

3A = \begin{pmatrix} 6 & -3 \\ 0 & 9 \end{pmatrix}, \qquad 2B^{T} = \begin{pmatrix} 2 & -4 \\ 8 & 2 \end{pmatrix}

3A - 2B^{T} = \begin{pmatrix} 6 - 2 & -3 + 4 \\ 0 - 8 & 9 - 2 \end{pmatrix} = \begin{pmatrix} 4 & 1 \\ -8 & 7 \end{pmatrix}

The transpose of $B$ is obtained by writing the rows of $B$ as columns, so the first row $(1, 4)$ becomes the first column and the second row $(-2, 1)$ becomes the second column. The product by a scalar multiplies every entry, hence $3A$ has the entries of $A$ multiplied by $3$ and $2B^{T}$ those of $B^{T}$ multiplied by $2$ . The two matrices are both $2 \times 2$ , so the difference is defined and is computed entry by entry. The result is $3A - 2B^{T} = \begin{pmatrix} 4 & 1 \\ -8 & 7 \end{pmatrix}$ .

Exercise 2

Given the matrices

A = \begin{pmatrix} 1 & 0 & 2 \\ -1 & 3 & 1 \end{pmatrix}, \qquad B = \begin{pmatrix} 2 & 1 \\ 0 & -1 \\ 1 & 4 \end{pmatrix},

compute, if they are defined, the products $AB$ and $BA$ .

AB = \begin{pmatrix} 1 \cdot 2 + 0 \cdot 0 + 2 \cdot 1 & 1 \cdot 1 + 0 \cdot (-1) + 2 \cdot 4 \\ -1 \cdot 2 + 3 \cdot 0 + 1 \cdot 1 & -1 \cdot 1 + 3 \cdot (-1) + 1 \cdot 4 \end{pmatrix} = \begin{pmatrix} 4 & 9 \\ -1 & 0 \end{pmatrix}

BA = \begin{pmatrix} 2 - 1 & 0 + 3 & 4 + 1 \\ 0 + 1 & 0 - 3 & 0 - 1 \\ 1 - 4 & 0 + 12 & 2 + 4 \end{pmatrix} = \begin{pmatrix} 1 & 3 & 5 \\ 1 & -3 & -1 \\ -3 & 12 & 6 \end{pmatrix}

The matrix $A$ is $2 \times 3$ and the matrix $B$ is $3 \times 2$ . The product $AB$ is defined because the number of columns of $A$ , that is $3$ , equals the number of rows of $B$ , and the result is a $2 \times 2$ matrix. Each entry of $AB$ is obtained with the row-by-column product; for example the entry in position $(1, 2)$ is the first row of $A$ , $(1, 0, 2)$ , times the second column of $B$ , $(1, -1, 4)$ , that is $1 + 0 + 8 = 9$ .

$BA$ is also defined, because $B$ has $2$ columns and $A$ has $2$ rows, but it is a $3 \times 3$ matrix. The entry in position $(3, 2)$ , for example, is the third row of $B$ , $(1, 4)$ , times the second column of $A$ , $(0, 3)$ , that is $0 + 12 = 12$ . The two products have different dimensions, so in this case $AB$ and $BA$ cannot even be compared.

Exercise 3

Find all matrices $X$ of order $2$ that commute with

A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix},

that is, such that $AX = XA$ .

X = \begin{pmatrix} a & b \\ c & d \end{pmatrix}

AX = \begin{pmatrix} a + 2c & b + 2d \\ c & d \end{pmatrix}, \qquad XA = \begin{pmatrix} a & 2a + b \\ c & 2c + d \end{pmatrix}

AX = XA \iff \begin{cases} a + 2c = a \\[4pt] b + 2d = 2a + b \\[4pt] c = c \\[4pt] d = 2c + d \end{cases} \iff \begin{cases} c = 0 \\[4pt] d = a \end{cases}

X = \begin{pmatrix} a & b \\ 0 & a \end{pmatrix}, \qquad a, b \in \mathbb{R}

We write the unknown matrix with four entries $a$ , $b$ , $c$ , $d$ and compute the two products with the row-by-column rule. Two matrices are equal when all their corresponding entries are equal, so the equality $AX = XA$ is equivalent to a system of four equations. The first equation gives $2c = 0$ , that is, $c = 0$ ; the second gives $2d = 2a$ , that is, $d = a$ ; the third is always true and the fourth again gives $c = 0$ . The entries $a$ and $b$ remain free, hence the matrices that commute with $A$ are exactly those of the form $\begin{pmatrix} a & b \\ 0 & a \end{pmatrix}$ . For example $I_2$ and $A$ itself are of this form, whereas $\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$ is not, and indeed it does not commute with $A$ .