Worked examples · Sheet 2

Row-by-column products

Theory used: Row-by-column product The column-by-column product and linear systems Properties of the row-by-column product Commuting matrices Special products of matrices

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

Given the matrices

A = (10 - 1 3 21), B = 21 - 1 104,

compute, if defined, the products $A B$ and $B A$ .

(2 \times 3) \cdot (3 \times 2) = 2 \times 2, (3 \times 2) \cdot (2 \times 3) = 3 \times 3

A B = (2 - 1 - 2 0 + 3 - 1 1 + 0 + 8 0 + 0 + 4) = (- 1 2 94)

B A = 2 + 0 1 + 0 - 1 + 0 - 2 + 3 - 1 + 0 1 + 12 4 + 1 2 + 0 - 2 + 4 = 21 - 1 1 - 1 13 522

The product $A B$ is defined because the number of columns of $A$ , that is $3$ , equals the number of rows of $B$ , and the result has as many rows as $A$ and as many columns as $B$ . The product $B A$ is defined for the same reason, since $B$ has $2$ columns and $A$ has $2$ rows, and the result is $3 \times 3$ . Each entry is obtained by multiplying a row of the first factor by a column of the second: for example the entry in position $(1, 2)$ of $A B$ is the product of the first row $(1, - 1, 2)$ of $A$ by the second column $(1, 0, 4)$ of $B$ , that is, $1 \cdot 1 + (- 1) \cdot 0 + 2 \cdot 4 = 9$ . The two matrices $A B$ and $B A$ have different dimensions, therefore they are different.

Exercise 2

Given the matrices

A = (1 - 1 20), B = (03 1 - 2),

determine whether $A$ and $B$ commute and compare $(A + B)^{2}$ with $A^{2} + 2 A B + B^{2}$ .

A B = (0 + 6 0 + 0 1 - 4 - 1 + 0) = (60 - 3 - 1), B A = (0 - 1 3 + 2 0 + 0 6 + 0) = (- 1 5 06)

A + B = (12 3 - 2), (A + B)^{2} = (1 + 6 2 - 4 3 - 6 6 + 4) = (7 - 2 - 3 10)

A^{2} = (- 1 - 1 2 - 2), B^{2} = (3 - 6 - 2 7)

A^{2} + 2 A B + B^{2} = (- 1 + 12 + 3 - 1 + 0 - 6 2 - 6 - 2 - 2 - 2 + 7) = (14 - 7 - 6 3) \neq = (A + B)^{2}

The products $A B$ and $B A$ are both $2 \times 2$ but have different entries, for example $(A B)_{11} = 6$ and $(B A)_{11} = - 1$ , therefore $A$ and $B$ do not commute. By the distributive property $(A + B)^{2} = A^{2} + A B + B A + B^{2}$ , and this expression coincides with $A^{2} + 2 A B + B^{2}$ if and only if $B A = A B$ . Since the two matrices do not commute, the formula for the square of a binomial valid for numbers gives a result different from $(A + B)^{2}$ , as the computations confirm. The value of $(A + B)^{2}$ is checked with the correct formula: the sum $A^{2} + A B + B A + B^{2}$ has first entry $- 1 + 6 - 1 + 3 = 7$ and, proceeding in the same way, gives back the matrix $(7 - 2 - 3 10)$ .

Exercise 3

Given the symmetric matrix and the vector

A = 210 1 - 1 3 031, x = 1 - 2 1,

compute $A x$ as a linear combination of the columns of $A$ and the number $x^{T} A x$ .

A x = 1210 - 2 1 - 1 3 + 1031 = 2 - 2 + 0 1 + 2 + 3 0 - 6 + 1 = 06 - 5

x^{T} A x = (1 - 2 1) 06 - 5 = 0 - 12 - 5 = - 17

The product of $A$ , which is $3 \times 3$ , by the column vector $x$ , which is $3 \times 1$ , is a $3 \times 1$ column vector. By the reading of the column-by-column product, $A x$ is the linear combination of the columns of $A$ with coefficients the components $1$ , $- 2$ , $1$ of $x$ ; adding the three columns multiplied in this way gives $(0, 6, - 5)$ . By the associative property $x^{T} A x = x^{T} (A x)$ , so it suffices to multiply the row vector $x^{T}$ , which is $1 \times 3$ , by the column vector $A x$ just found. The result is a $1 \times 1$ matrix, that is, the number $- 17$ . Expressions of this type are quadratic forms, which appear in the study of conics and quadrics.