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System of Linear Equations
Foundations of Linear Systems
A System of Equations consists of two or more Linear Equations sharing common variables. The Solution Set is the collection of values satisfying all equations simultaneously.
$$\begin{cases} a_1x + b_1y + c_1z = k_1 \\\\ a_2x + b_2y + c_2z = k_2 \\\\ a_3x + b_3y + c_3z = k_3 \end{cases}$$
A 3×3 non-homogeneous linear system in standard matrix form $AX = B$.
$a_i, b_i, c_i$=Coefficients of variables(unitless)
$x, y, z$=Unknown variables to solve for(units)
$k_i$=Constant terms on the right-hand side(units)
Homogeneous System: When all $k_i = 0$, the system always has at least the Trivial Solution $(0, 0, 0)$.
Non-Homogeneous System: When at least one $k_i \neq 0$, the system may have zero, one, or infinitely many solutions.
Matrix Form: The system can be written as $AX = B$ where $A$ is the Coefficient Matrix, $X$ is the variable column, and $B$ is the constants column.
A system is Consistent if it has at least one solution, and Inconsistent if it has none. The Rank of the coefficient matrix $A$ versus the Augmented Matrix $[A | B]$ determines which case applies.
Unique Solution: rank($A$) = rank($[A|B]$) = number of variables. The system has exactly one solution.
Infinitely Many Solutions: rank($A$) = rank($[A|B]$) < number of variables. Free variables exist, parameterised by arbitrary constants.
No Solution: rank($A$) ≠ rank($[A|B]$). The augmented matrix introduces a contradiction row like $[0 \ 0 \ 0 \ | \ k]$ where $k \neq 0$.
Elementary Row Operations
Elementary Row Operations are the building blocks for transforming any matrix into Echelon Form. They do not change the solution set of the system.
Swap Rows: $R_i \leftrightarrow R_j$ — interchange two rows. Used to get a leading 1 into position.
Scale a Row: $kR_i \to R_i'$ — multiply every entry by a non-zero scalar $k$. Used to normalise the leading entry to 1.
Row Replacement: $R_i + kR_j \to R_i'$ — add a scalar multiple of one row to another. The workhorse operation for creating zeros below (and above) pivots.
Key Property: Each operation is reversible, so the transformed system is equivalent to the original.
Echelon and Reduced Echelon Forms
The Echelon Form of a matrix has a staircase pattern of leading 1s, with all entries below each leading 1 being zero. The Reduced Echelon Form additionally requires all entries above each leading 1 to be zero.
Echelon Form Rules: (i) Each successive non-zero row has its leading entry further right than the row above. (ii) All non-zero rows precede zero rows. (iii) The leading entry in each non-zero row is 1.
Reduced Echelon Form: In addition to echelon form, if a leading 1 appears in column $j$, every other entry in column $j$ is 0.
Back-Substitution: From echelon form, solve from the bottom row upward by substituting known values into earlier equations.
Direct Reading: From reduced echelon form, the solution is read directly — each variable equals the constant in its row.
Echelon Form Example
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$\begin{bmatrix} 1 & -1 & 2 & | & -3 \\\\ 0 & 1 & -1 & | & 3 \\\\ 0 & 0 & 1 & | & -1 \end{bmatrix}$ — staircase of leading 1s, zeros below
Reduced Echelon Form Example
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$\begin{bmatrix} 1 & 0 & 0 & | & 1 \\\\ 0 & 1 & 0 & | & 2 \\\\ 0 & 0 & 1 & | & -1 \end{bmatrix}$ — identity on the left, solution on the right
Rank of a Matrix
The Rank of a matrix is the number of non-zero rows in its Echelon Form. It measures the number of truly independent equations in the system.
Computing Rank: Reduce to echelon form, then count non-zero rows.
Rank and Solutions: If rank($A$) = rank($[A|B]$) = $n$ (number of variables), unique solution. If rank < $n$, free variables exist ($n - \text{rank}$ free variables).
Zero Row in Augmented Matrix: A row $[0 \ 0 \ 0 \ | \ k]$ with $k \neq 0$ means the system is inconsistent. A row $[0 \ 0 \ 0 \ | \ 0]$ is simply redundant.
Dimensional Check: For a 3×3 system: rank 3 → unique solution, rank 2 → one free variable (line of solutions), rank 1 → two free variables (plane of solutions).
Solving by Matrix Inverse Method
When the Coefficient Matrix $A$ is non-singular ($|A| \neq 0$), the system $AX = B$ has the unique solution $X = A^{-1}B$.
$$X = A^{-1}B = \frac{1}{|A|} (\text{adj}\ A) \cdot B$$
The solution vector obtained by multiplying the inverse of the coefficient matrix with the constants vector.
$A^{-1}$=Inverse of the coefficient matrix
$|A|$=Determinant of $A$
$\text{adj}\ A$=Adjoint (transpose of cofactor matrix)
$B$=Column vector of constants $[k_1, k_2, k_3]^T$
Procedure: (1) Compute $|A|$. If zero, this method fails — use echelon form instead. (2) Find cofactors, form adjoint. (3) Compute $A^{-1} = \frac{1}{|A|} \text{adj}\ A$. (4) Multiply $X = A^{-1}B$.
Row Operations Alternative: Augment $[A | I]$ and row-reduce to $[I | A^{-1}]$. This avoids computing cofactors.
When It Fails: $|A| = 0$ means the system has either no solution or infinitely many — you must use echelon form to distinguish.
Cramer's Rule
Cramer's Rule expresses each variable as a ratio of two Determinants: the numerator determinant is formed by replacing the variable's column in $A$ with the constants column $B$.
$$x_i = \frac{|A_i|}{|A|}$$
Each variable equals a fraction whose denominator is $|A|$ and whose numerator is the determinant of $A$ with column $i$ replaced by $B$.
$|A|$=Determinant of the coefficient matrix
$|A_i|$=Determinant of $A$ with column $i$ replaced by the constants column $B$
$x_i$=The $i$-th variable being solved for
For 3 Variables: $x_1 = \frac{|A_1|}{|A|}$, $x_2 = \frac{|A_2|}{|A|}$, $x_3 = \frac{|A_3|}{|A|}$ where $A_i$ is $A$ with column $i$ replaced by $B$.
Prerequisite: $|A| \neq 0$ (non-singular system). If $|A| = 0$, Cramer's Rule is inapplicable.
Efficiency Note: Cramer's Rule requires computing 4 determinants for a 3×3 system. For larger systems, echelon form is more efficient.
Homogeneous Linear Systems
A Homogeneous system has the form $AX = O$ (all constants zero). It always has the Trivial Solution $X = O$. The key question is whether Non-Trivial Solutions exist.
$$|A| = 0 \iff \text{non-trivial solutions exist}$$
The determinant test: a homogeneous system has solutions beyond $(0,0,0)$ if and only if the coefficient matrix is singular.
$|A|$=Determinant of the coefficient matrix
$= 0$=Matrix is singular, rows are linearly dependent
Trivial Solution: $x_1 = x_2 = x_3 = 0$ always works because $A \cdot O = O$.
Non-Trivial Condition: $|A| = 0$ means the system has infinitely many solutions. Solve by setting one variable as parameter $t$ and expressing others in terms of $t$.
$|A| \neq 0$ Case: The inverse exists, so $X = A^{-1}O = O$ — only the trivial solution.
Finding Solutions: When $|A| = 0$, solve any two equations simultaneously, express variables in terms of one free variable, then verify the third equation is automatically satisfied.
Solving Techniques Comparison
Three main methods solve 3×3 non-homogeneous systems: the matrix inverse method, Echelon Form reduction, and Cramer's Rule. Each has distinct advantages.
Matrix Inverse ($X = A^{-1}B$): Best when $A^{-1}$ is already known or when solving multiple systems with the same $A$ but different $B$.
Echelon Form: Most general method — works even when $|A| = 0$. Reveals whether the system is consistent, and gives parametric solutions for infinite cases.
Cramer's Rule: Quick for 3×3 when you only need one specific variable. Requires $|A| \neq 0$.
Limitation Check: Both matrix inverse and Cramer's Rule require $|A| \neq 0$. Only echelon form handles all cases (unique, infinite, no solution).
Application Strategy
Worked Examples
Problem Statement
Solve the system by reducing the augmented matrix to echelon form: $x_1 + 3x_2 + 2x_3 = 3$, $4x_1 + 5x_2 - 3x_3 = -3$, $3x_1 - 2x_2 + 17x_3 = 42$.
Augmented matrixRow operationsBack-substitution
Strategic Toolkit
$$[A|B] = \begin{bmatrix} 1 & 3 & 2 & | & 3 \\\\ 4 & 5 & -3 & | & -3 \\\\ 3 & -2 & 17 & | & 42 \end{bmatrix}$$
Step 1
Form the Augmented Matrix
Write the system as $[A|B]$ with coefficients and constants.
$$\begin{bmatrix} 1 & 3 & 2 & | & 3 \\\\ 4 & 5 & -3 & | & -3 \\\\ 3 & -2 & 17 & | & 42 \end{bmatrix}$$
Step 2
Eliminate Below First Pivot
$R_2 + (-4)R_1 \to R_2'$ and $R_3 + (-3)R_1 \to R_3'$. This creates zeros in column 1 below the pivot.
$$\begin{bmatrix} 1 & 3 & 2 & | & 3 \\\\ 0 & -7 & -11 & | & -15 \\\\ 0 & -11 & 11 & | & 33 \end{bmatrix}$$
Step 3
Swap to Position Better Pivot
Swap $R_2 \leftrightarrow R_3$ to bring $-11$ into the second pivot position (easier to normalise).
$$\begin{bmatrix} 1 & 3 & 2 & | & 3 \\\\ 0 & -11 & 11 & | & 33 \\\\ 0 & -7 & -11 & | & -15 \end{bmatrix}$$
Step 4
Normalise Second Row
$(-\frac{1}{11})R_2 \to R_2'$ to get a leading 1 in position (2,2).
$$\begin{bmatrix} 1 & 3 & 2 & | & 3 \\\\ 0 & 1 & -1 & | & -3 \\\\ 0 & -7 & -11 & | & -15 \end{bmatrix}$$
Step 5
Eliminate Below Second Pivot
$R_3 + 7R_2 \to R_3'$ creates a zero in column 2 of row 3.
$$\begin{bmatrix} 1 & 3 & 2 & | & 3 \\\\ 0 & 1 & -1 & | & -3 \\\\ 0 & 0 & -18 & | & -36 \end{bmatrix}$$
Step 6
Normalise Third Row
$(-\frac{1}{18})R_3 \to R_3'$ to get the leading 1 in position (3,3).
$$\begin{bmatrix} 1 & 3 & 2 & | & 3 \\\\ 0 & 1 & -1 & | & -3 \\\\ 0 & 0 & 1 & | & 2 \end{bmatrix}$$
Step 7
Back-Substitute
From row 3: $x_3 = 2$. Row 2: $x_2 - 2 = -3 \Rightarrow x_2 = -1$. Row 1: $x_1 + 3(-1) + 2(2) = 3 \Rightarrow x_1 = 2$.
$$x_1 = 2, \quad x_2 = -1, \quad x_3 = 2$$
Problem Statement
Use Cramer's Rule to solve: $3x_1 + x_2 - x_3 = -4$, $x_1 + x_2 - 2x_3 = -4$, $-x_1 + 2x_2 - x_3 = 1$.
Cramer's RuleDeterminant computationColumn replacement
Strategic Toolkit
$$x_i = \frac{|A_i|}{|A|}$$
Step 1
Compute |A|
Expand the 3×3 determinant along row 1.
$$|A| = 3(-1+4) - 1(-1-2) + (-1)(2+1) = 9 + 3 - 3 = 9$$
Step 2
Find x₁ via |A₁|
Replace column 1 of $A$ with $B = [-4, -4, 1]^T$ and compute the determinant.
$$|A_1| = -4(3) - 1(6) + (-1)(-9) = -12 - 6 + 9 = -9 \quad \Rightarrow x_1 = \frac{-9}{9} = -1$$
Step 3
Find x₂ via |A₂|
Replace column 2 of $A$ with $B$ and compute.
$$|A_2| = 3(6) - 12 + 3 = 9 \quad \Rightarrow x_2 = \frac{9}{9} = 1$$
Step 4
Find x₃ via |A₃|
Replace column 3 of $A$ with $B$ and compute.
$$|A_3| = 3(9) + 3 - 12 = 18 \quad \Rightarrow x_3 = \frac{18}{9} = 2$$
Step 5
Final Answer
The solution is the triple $(x_1, x_2, x_3)$.
$$x_1 = -1, \quad x_2 = 1, \quad x_3 = 2$$
Problem Statement
Determine if the system has a non-trivial solution: $x_1 + x_2 + x_3 = 0$, $x_1 - x_2 + 3x_3 = 0$, $x_1 + 3x_2 - x_3 = 0$.
Homogeneous systemsDeterminant testParameterised solutions
Strategic Toolkit
$$|A| = 0 \iff \text{non-trivial solutions exist}$$
Step 1
Compute |A|
Expand along row 1.
$$|A| = 1(1-9) - 1(-1-3) + 1(3+1) = -8 + 4 + 4 = 0$$
Step 2
Conclude Non-Trivial Solutions Exist
Since $|A| = 0$, the system has infinitely many solutions beyond $(0,0,0)$.
$|A| = 0$ — non-trivial solutions guaranteed
Step 3
Find x₁ in Terms of x₃
Add equations 1 and 2 to eliminate $x_2$:
$$2x_1 + 4x_3 = 0 \quad \Rightarrow \quad x_1 = -2x_3$$
Step 4
Find x₂ in Terms of x₃
Subtract equation 2 from equation 1 to eliminate $x_1$:
$$2x_2 - 2x_3 = 0 \quad \Rightarrow \quad x_2 = x_3$$
Step 5
Parameterise and Verify
Let $x_3 = t$. Check in equation 3: $(-2t) + 3(t) - t = 0$ ✓
$$x_1 = -2t, \quad x_2 = t, \quad x_3 = t \quad (t \in \mathbb{R})$$
Problem Statement
Solve using $X = A^{-1}B$: $2x + 5y - z = 5$, $3x + 4y + 2z = 11$, $x + 2y - 2z = -3$.
Matrix inverseAdjoint methodMatrix multiplication
Strategic Toolkit
$$X = A^{-1}B = \frac{1}{|A|}(\text{adj}\ A) \cdot B$$
Step 1
Compute |A|
The coefficient matrix has determinant $|A| = 2(-8-4) - 5(-6-2) - 1(6-4) = -24 + 40 - 2 = 14$.
$$|A| = 14 \neq 0$$
Step 2
Find Inverse via Row Operations
Augment $[A|I]$ and row-reduce to $[I|A^{-1}]$. This avoids cofactor computation.
Row-reduce $[A|I_{3}]$ to obtain $A^{-1}$
Step 3
Compute X = A⁻¹B
Multiply the inverse matrix by the constants vector to get the solution.
$$X = A^{-1} \begin{bmatrix} 5 \\\\ 11 \\\\ -3 \end{bmatrix} = \begin{bmatrix} -1 \\\\ 2 \\\\ 3 \end{bmatrix}$$
Step 4
Verify
Check: $2(-1) + 5(2) - 3 = -2 + 10 - 3 = 5$ ✓, $3(-1) + 4(2) + 2(3) = -3 + 8 + 6 = 11$ ✓
$$x = -1, \quad y = 2, \quad z = 3$$
Memorize formulas
Cramer's Rule
Each variable is the ratio of two determinants: modified matrix over original.
Formula
$x_i = \frac{|A_i|}{|A|}$
Matrix Inverse Solution
Direct solution when the coefficient matrix is invertible.
Formula
$X = A^{-1}B = \frac{1}{|A|}(\text{adj}\ A) \cdot B$
Homogeneous Non-Trivial Condition
A homogeneous system has solutions beyond zero iff the determinant vanishes.
Formula
$|A| = 0 \iff \text{non-trivial solutions exist}$
Consistency by Rank
Compare ranks of coefficient and augmented matrices to classify the system.
Formula
$\text{rank}(A) = \text{rank}([A|B]) \implies \text{consistent}$
No Solution Condition (2×2)
Parallel lines with different intercepts have no intersection.
Formula
$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$
Review
1
The augmented matrix of a system reduces to $\begin{bmatrix} 1 & 0 & 3 & | & 4 \\\\ 0 & 1 & -1 & | & -3 \\\\ 0 & 0 & 0 & | & 0 \end{bmatrix}$. What is the general solution?
Ans
$x = 4 - 3t, \ y = t - 3, \ z = t$
Review
Tap to reveal
Take the free variable from the zero row
The last row gives a tautology, so it adds no restriction. Let the free variable be $z = t$.
Use the second row to express y
Substitute $z = t$ into the reduced equation $y - z = -3$ and solve for $y$.
$$y = t - 3$$
Back-substitute into the first row
Now use $x + 3z = 4$ with $z = t$ to get the final parametric form.
$$x = 4 - 3t$$
2
Using Cramer's Rule, solve for $x_2$ in: $3x_1 + x_2 - x_3 = -4$, $x_1 + x_2 - 2x_3 = -4$, $-x_1 + 2x_2 - x_3 = 1$, given that $|A| = 9$.
Ans
$x_2 = 1$
Review
Tap to reveal
Form the replacement determinant
For $x_2$, replace the second column of $A$ with the constants column $B$ and evaluate the new determinant.
$$|A_2| = \begin{vmatrix} 3 & -4 & -1 \\\\ 1 & -4 & -2 \\\\ -1 & 1 & -1 \end{vmatrix} = 9$$
Divide by the original determinant
Cramer's Rule gives $x_2 = |A_2| / |A|$. Since both determinants are 9, the ratio is 1.
$$x_2 = \frac{9}{9} = 1$$
3
For what value of $\lambda$ does the system $x_1 + 4x_2 + \lambda x_3 = 2$, $2x_1 + x_2 - 2x_3 = 11$, $3x_1 + 2x_2 - 2x_3 = 16$ fail to have a unique solution?
Ans
$\lambda = 6$
Review
Tap to reveal
Set the determinant condition
A 3×3 system loses uniqueness exactly when the coefficient determinant is zero.
$$|A| = \lambda - 6$$
Solve for the singular value
Set the determinant equal to zero and isolate the parameter.
$$\lambda - 6 = 0 \Rightarrow \lambda = 6$$
4
A 3×3 homogeneous system $AX = O$ has $|A| = 0$. Which statement is TRUE?
Ans
The system has infinitely many solutions including $X \neq O$
Review
Tap to reveal
Use the trivial solution first
Because the right-hand side is zero, $X = O$ always satisfies the system, so inconsistency is impossible.
$$AX = O \text{ has at least } X = O$$
Use singularity to infer free variables
When $|A| = 0$, the matrix is singular, so the solution set is not just the trivial vector; there must be non-trivial solutions.
$$|A| = 0 \Rightarrow \text{infinitely many solutions}$$
5
For what value of $k$ will the system $2x + 3y = 7$ and $4x + ky = 15$ have no solution?
Ans
$k = 6$
Review
Tap to reveal
Match the coefficient ratios
For no solution, the two lines must be parallel, so the $x$ and $y$ coefficients must be proportional.
$$\frac{2}{4} = \frac{3}{k} \Rightarrow k = 6$$
Check that the constants do not match
With $k = 6$, the left-hand sides are proportional but the constants are not, so the lines are distinct and never intersect.
$$\frac{7}{15} \ne \frac{1}{2}$$
6
Find the rank of $\begin{bmatrix} 2 & 1 & 3 \\\\ 4 & 2 & 6 \\\\ 1 & 3 & 5 \end{bmatrix}$.
Ans
2
Review
Tap to reveal
Identify the dependent row
Row 2 is exactly twice row 1, so it does not add a new pivot. Row 3 is not a multiple of row 1, so it contributes a second independent row.
$$R_2 - 2R_1 = [0,0,0]$$
Count the pivots
Two non-zero rows remain in echelon form, so the rank is 2. The non-zero $2 \times 2$ minor confirms the rank cannot fall below 2.
$$\operatorname{rank}(A) = 2$$
7
A system of 3 equations in 3 variables has rank($A$) = 2 and rank($[A|B]$) = 2. The system has:
Ans
Infinitely many solutions with 1 free variable
Review
Tap to reveal
Apply the consistency test
Equal ranks for the coefficient and augmented matrices mean the system is consistent, so at least one solution exists.
$$\operatorname{rank}(A) = \operatorname{rank}([A|B]) = 2$$
Count the free variables
With 3 variables and rank 2, there is exactly one free variable, so the solution set is infinite and one-parameter.
$$3 - 2 = 1$$
8
Which of the following matrices is NOT in echelon form?
Ans
$\begin{bmatrix} 0 & 0 & 1 & 2 \\\\ 0 & 1 & 3 & -1 \\\\ 0 & 0 & 0 & 0 \end{bmatrix}$
Review
Tap to reveal
Check the staircase rule
In echelon form, the leading entry in each lower non-zero row must appear strictly to the right of the leading entry in the row above it.
$$\text{leading column must move rightward}$$
9
Given the homogeneous system $x_1 + x_2 + x_3 = 0$, $x_1 - x_2 + 3x_3 = 0$, $x_1 + 3x_2 - x_3 = 0$ with $|A| = 0$, the solution set is $x_1 = -2t, \ x_2 = t, \ x_3 = t$. Which of the following is a valid non-trivial solution?
Ans
$(-4, 2, 2)$
Review
Tap to reveal
Use the parameterized form
The solution set has the pattern $x_1 = -2t$, $x_2 = t$, $x_3 = t$, so any valid vector must keep the ratio $-2:1:1$.
$$x_1 : x_2 : x_3 = -2 : 1 : 1$$
Choose a matching parameter value
Take $t = 2$ to produce a non-trivial solution that fits the pattern exactly.
$$(-4, 2, 2)$$
10
If $AX = B$ where $A$ is a 3×3 matrix with $|A| = -5$ and $\text{adj}(A) \cdot B = \begin{bmatrix} 10 \\\\ -15 \\\\ 5 \end{bmatrix}$, what is the solution vector $X$?
Ans
$(-2, 3, -1)$
Review
Tap to reveal
Apply the adjugate formula
For an invertible 3×3 matrix, the solution is $X = A^{-1}B = \frac{1}{|A|}\,\text{adj}(A)B$.
$$X = \frac{1}{-5}\begin{bmatrix} 10 \\\\ -15 \\\\ 5 \end{bmatrix}$$
Scale the vector by -1/5
Divide each entry by $-5$, which flips the signs and gives the final vector.
$$X = \begin{bmatrix} -2 \\\\ 3 \\\\ -1 \end{bmatrix}$$
11
The system $x - y + 2z = 1$, $2x - 6y + 5z = 7$, $3x + 5y + 4z = -3$ reduces to an echelon form with the last row $[0 \ 0 \ 0 \ | \ 4]$. This means:
Ans
The system is inconsistent — rank($A$) < rank($[A|B]$)
Review
Tap to reveal
Read the contradictory row
The final row becomes $0 = 4$, which is impossible for any values of the variables.
$$0 = 4$$
Use the rank mismatch
Because the augmented matrix has one more non-zero row than the coefficient matrix, the ranks are different and the system has no solution.
$$\operatorname{rank}(A) < \operatorname{rank}([A|B])$$
12
To solve $AX = B$ where $A = \begin{bmatrix} 2 & 5 & -1 \\\\ 3 & 4 & 2 \\\\ 1 & 2 & -2 \end{bmatrix}$ using row operations, you form $[A|I]$ and reduce to $[I|A^{-1}]$. The first row operation should be:
Ans
$R_1 \leftrightarrow R_3$ (to get a leading 1 in position (1,1))
Review
Tap to reveal
Bring the natural pivot to the top
The first column already contains a 1 in row 3, so swapping row 1 with row 3 is the cleanest first move toward reduced row echelon form.
$$R_1 \leftrightarrow R_3$$
Prefer integer arithmetic early
This swap avoids fractions at the start, which makes the elimination steps simpler and less error-prone than scaling row 1 immediately.
$$\text{keep integer pivots as long as possible}$$
Question 1 of 12
0:00
1
The augmented matrix of a system reduces to $\begin{bmatrix} 1 & 0 & 3 & | & 4 \\\\ 0 & 1 & -1 & | & -3 \\\\ 0 & 0 & 0 & | & 0 \end{bmatrix}$. What is the general solution?