Linear algebra

Systems of linear equations

Row-reducing the augmented matrix solves any linear system systematically, and the rank tells you whether solutions exist.

Basic, Standard, Advanced: University Year 1 · 1st semester

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Gaussian elimination

The three elementary row operations do not change the solutions:

swap two rows;
multiply a row by a nonzero number;
add a multiple of one row to another.

Reduce the augmented matrix $(A \mid \boldsymbol{b})$ until each pivot is $1$ with zeros above and below it (reduced row echelon form); then read off the solution.

\left(\begin{array}{ccc|c} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 3 \end{array}\right) \quad\Rightarrow\quad x = 2,\ y = -1,\ z = 3

Rank and solvability

The number of nonzero rows in an echelon form is the rank. With $n$ unknowns:

Solvability

$\operatorname{rank} A = \operatorname{rank}(A \mid \boldsymbol{b}) = n$ : exactly one solution
$\operatorname{rank} A = \operatorname{rank}(A \mid \boldsymbol{b}) < n$ : infinitely many solutions ( $n - \operatorname{rank} A$ free parameters)
$\operatorname{rank} A < \operatorname{rank}(A \mid \boldsymbol{b})$ : no solution

A homogeneous system $A\boldsymbol{x} = \boldsymbol{0}$ always has $\boldsymbol{x} = \boldsymbol{0}$ ; for square $A$ it has a nontrivial solution exactly when $|A| = 0$ .

Parametric solutions and solution spaces

When there are infinitely many solutions, set each free variable (a column without a pivot) equal to a parameter such as $t$ and express the others in terms of it. If the reduced form is $x + 2z = 2$ and $y - 3z = -3$ , then with $z = t$ :

x = 2 - 2t,\quad y = -3 + 3t,\quad z = t

The solution space of $A\boldsymbol{x} = \boldsymbol{0}$ has dimension $n - \operatorname{rank} A$ .

Q1BasicUniv Y1 · Sem 1Systems of linear equations

Solve the system of equations.

\begin{cases} -x - y + z = -1 \\ 3x + 2y = 4 \\ 2x + 3y - 3z = 1 \end{cases}

Answer

x = 2,\ y = -1,\ z = 0

Solution

The augmented matrix is
$\left(\begin{array}{ccc|c}-1 & -1 & 1 & -1 \\ 3 & 2 & 0 & 4 \\ 2 & 3 & -3 & 1\end{array}\right)$
Row-reducing,
$\left(\begin{array}{ccc|c}1 & 0 & 0 & 2 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 0\end{array}\right)$
Hence
$x = 2,\ y = -1,\ z = 0$

Q2StandardUniv Y1 · Sem 1Systems of linear equations

This system has infinitely many solutions. Setting $z = t$ ( $t$ any real number), express $x$ and $y$ in terms of $t$ .

\begin{cases} -x - 2y - 4z = -2 \\ -2x - 2y - 2z = 0 \\ x - 2y - 8z = -6 \end{cases}

(1) $x$
(2) $y$

Answer

(1) $x = -2 + 2t$
(2) $y = 2 - 3t$

Solution

Reducing the augmented matrix,
$\left(\begin{array}{ccc|c}1 & 0 & -2 & -2 \\ 0 & 1 & 3 & 2 \\ 0 & 0 & 0 & 0\end{array}\right)$
This means
$x - 2z = -2,\quad y + 3z = 2$
Setting $z = t$ ,
$x = -2 + 2t,\ y = 2 - 3t,\ z = t$

Q3AdvancedUniv Y1 · Sem 1Systems of linear equations

Find the constants $a$ and $b$ for which the system has infinitely many solutions.

\begin{cases} 3x - 3y - 3z = -3 \\ 2x - y - 3z = -1 \\ 4x - 5y + az = b \end{cases}

Answer

a = -3,\ b = -5

Solution

Comparing the $x$ and $y$ coefficients, the third equation matches $2\times$ (first) $+\ (-1)\times$ (second).
For infinitely many solutions, the whole third equation must be this combination:
$a = 2 \cdot (-3) + (-1) \cdot (-3) = -3,\quad b = 2 \cdot (-3) + (-1) \cdot (-1) = -5$
(If $a$ differs, the rank is 3 and the solution is unique; if $a$ matches but $b$ differs, there is no solution.)

Q1BasicUniv Y1 · Sem 1Systems of linear equations

Use elementary row operations to bring the matrix to reduced row echelon form.

\begin{pmatrix}-3 & 2 & -2 & -7 \\ -1 & 1 & -1 & -3 \\ -3 & 2 & -1 & -4\end{pmatrix}

Q2StandardUniv Y1 · Sem 1Systems of linear equations

Find the values of the constant $k$ for which $\begin{cases} (7 - k)x - 5y = 0 \\ 10x + (-8 - k)y = 0 \end{cases}$ has a solution other than $x = y = 0$ .

Q3AdvancedUniv Y1 · Sem 1Systems of linear equations

Find the constants $a$ and $b$ for which the system has infinitely many solutions.

\begin{cases} 3x - y + 2z = 4 \\ -3x - y + 2z = -4 \\ 2y + az = b \end{cases}

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