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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).