Linear Algebra · Book or PDF

A First Course in Linear Algebra

A.G. Hamilton · Study material generated by AI from this edition. Not affiliated with the publisher.

Chapter 8 · pp. 90–95

Solutions to Simultaneous Equations 2

This chapter revisits solving linear equations using matrix ideas: rank, invertibility and the augmented matrix. The main question is how to tell, from ranks alone, whether a system has no solution, exactly one, or infinitely many. You will also see how many free parameters a solution needs. (p. 90)

Writing a system as a matrix equation

A set of $p$ linear equations in $q$ unknowns can be written as one equation, $Ax = h$ . Here $A$ is the $p \times q$ matrix of coefficients, $x$ is the $q$ -vector of unknowns, and $h$ is the $p$ -vector of right-hand sides. (p. 90)

Example: the system $2x - y = 1$ , $x + 3y = 11$ has $A$ with rows $(2, -1)$ and $(1, 3)$ , unknowns $x = (x, y)$ , and $h = (1, 11)$ .

Every such system falls into exactly one of three cases: no solution, exactly one solution, or infinitely many solutions. (p. 90)

The augmented matrix $[A|h]$ is $A$ with $h$ added as an extra last column. Gaussian elimination (GE) is done on this matrix. The first $q$ columns change in exactly the same way as when GE is applied to $A$ alone. This is why we can compare the two matrices and their ranks. (p. 90)

When does a solution exist?

A system is inconsistent when, after GE, the last non-zero row is all zeros except for its final entry, for example $(0\ 0\ 0\ |\ 1)$ . That row would say $0 = 1$ .

In terms of rank, such a row counts as a non-zero row of $[A|h]$ , but its left part is a zero row of $A$ . So $[A|h]$ has a higher rank than $A$ . This gives the key rule:

$Ax = h$ has a solution if and only if $\text{rank}[A|h] = \text{rank}\,A$ . (p. 90)

Example 8.3(i) has a $3 \times 3$ singular matrix $A$ . After GE, $A$ has rank 2, but $[A|h]$ ends with a row $(0\ 0\ 0\ |\ 1)$ and has rank 3. The ranks differ, so there is no solution. (p. 91)

Homogeneous systems. If $h = 0$ , the system $Ax = 0$ is always consistent, because setting every unknown to zero works. In rank terms, adding a column of zeros to $A$ cannot create a new non-zero row, so $\text{rank}[A|0] = \text{rank}\,A$ automatically. (p. 90)

Square matrices: invertible or singular

Suppose a solution exists. Is it unique? For a square matrix $A$ , the answer depends on whether $A$ is invertible.

Invertible $A$ . Multiply both sides of $Ax = h$ on the left by $A^{-1}$ . This gives $x = A^{-1}h$ . Any solution must equal this, so there is exactly one. An invertible $p \times p$ matrix has rank $p$ , so no parameters are needed. (p. 92)

Example: for $2x - y = 1$ , $x + 3y = 11$ , the determinant of $A$ is $2\cdot 3 - (-1)\cdot 1 = 7$ , which is not zero, so $A$ is invertible. Here $A^{-1} = \frac{1}{7}\begin{pmatrix}3 & 1\\ -1 & 2\end{pmatrix}$ , and $A^{-1}h = (2, 3)$ . Check: $2(2) - 3 = 1$ and $2 + 3(3) = 11$ .

Singular $A$ . If $A$ is singular, GE on $A$ leaves a last row of all zeros. For $[A|h]$ there are two possibilities for the last row:

It is $(0 \dots 0 \mid \text{non-zero})$ . Then there is no solution.
It is all zeros. Then you check the last non-zero row, and the system has free parameters, so there are infinitely many solutions.

So a singular $A$ gives either no solutions or infinitely many, never exactly one. (p. 92)

Example 8.3(ii) has a singular $3 \times 3$ matrix. After GE, both $A$ and $[A|h]$ have rank 2, which is less than 3. The last row of $[A|h]$ is all zeros, so there are infinitely many solutions. (p. 91)

Counting parameters

For a $p \times p$ matrix $A$ of rank $r$ , suppose $\text{rank}[A|h]$ is also $r$ . Then the solutions need

$p - r$

parameters. The reason is that each unit of rank fixes one unknown through a leading 1, and the remaining $p - r$ unknowns can be chosen freely. If $r = p$ , there are 0 parameters, so the solution is unique. (pp. 92–93)

A question to think about: could a $4 \times 4$ system have a solution set with three parameters? Yes, if the rank is 1.

Writing out solutions that have parameters

Process for a consistent system with free unknowns:

Apply GE to $[A|h]$ .
Give each unknown that has no leading 1 its own parameter.
Work upward from the last non-zero equation (back-substitution) to express the other unknowns.

Example 8.4 (two parameters). GE reduces the system to a single equation, $x_1 - x_2 + 3x_3 = -2$ , because the other rows become zero. Let $x_2 = t$ and $x_3 = u$ . Then $x_1 = t - 3u - 2$ . The rank is 1 and $p = 3$ , so $3 - 1 = 2$ parameters. (pp. 91, 93)

Example 8.5 ( $4 \times 4$ , rank 2). After GE, only two non-zero rows remain:

$x_1 + 2x_2 + x_3 = 1$
$x_2 + x_3 - x_4 = 1$

Let $x_3 = t$ and $x_4 = u$ . The second equation gives $x_2 = 1 - t + u$ . Substituting into the first gives $x_1 = -1 + t - 2u$ . Two parameters, matching $4 - 2 = 2$ . (p. 93)

Non-square systems and uniqueness

Now let $A$ be $p \times q$ . The equation $Ax = h$ has a unique solution if and only if $\text{rank}\,A = \text{rank}[A|h] = q$ , the number of columns. (pp. 92, 94)

The reasoning: a unique solution needs a leading 1 in every column. So after GE, the first $q$ rows have 1s on the diagonal positions $(1,1), (2,2), \dots, (q,q)$ with zeros below. Any further rows must be all zeros. This is exactly the situation where both ranks equal $q$ . (p. 94)

Example 8.6. Here $A$ is $4 \times 3$ , so $q = 3$ . Four possible GE outcomes are:

(i) Both ranks are 3, so there is a unique solution.
(ii) The ranks differ, because a row like $(0\ 0\ 0\ |\ 1)$ appears, so there is no solution.
(iii) Again the ranks differ, so there is no solution.
(iv) Both ranks are 2, which is less than 3, so there are infinitely many solutions. (p. 93)

Fewer equations than unknowns. If $p < q$ , then $\text{rank}\,A \le p < q$ . The rank can never reach $q$ , so a unique solution is impossible. The system has either no solutions or infinitely many. For example, two equations in three unknowns can never pin down a single point. (p. 94)

Summary of the rank test

Find $\text{rank}\,A$ and $\text{rank}[A|h]$ using GE.
If they differ, there is no solution.
If they are equal to $r$ and $r$ equals the number of unknowns $q$ , there is exactly one solution.
If they are equal to $r$ and $r < q$ , there are infinitely many solutions, with $q - r$ parameters. (For square systems this is $p - r$ .) (pp. 90–94)

Check yourself

A system has $\text{rank}\,A = 2$ and $\text{rank}[A|h] = 3$ . What does this tell you, and what would the last row of the reduced augmented matrix look like?
Why is $Ax = 0$ never inconsistent?
A $5 \times 5$ system has $\text{rank}\,A = \text{rank}[A|h] = 3$ . How many parameters does the solution need?
Explain why a system of 3 equations in 4 unknowns can never have exactly one solution.
If $A$ is singular, can $Ax = h$ have exactly one solution? Explain.