Systems of Linear Equations & Matrices

Figure 1.4: Geometric Configurations of 3 Planes in 3D Space

1. Matrix Representation of Linear Systems

A system of three linear equations in three variables $(x, y, z)$:

$$\begin{cases} a_1 x + b_1 y + c_1 z = d_1 \\ a_2 x + b_2 y + c_2 z = d_2 \\ a_3 x + b_3 y + c_3 z = d_3 \end{cases}$$

can be expressed compactly in matrix equation form $A\mathbf{x} = \mathbf{b}$:

$$\begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} d_1 \\ d_2 \\ d_3 \end{pmatrix}$$

To analyze solvability analytically without technology (Paper 1), we form the augmented matrix $[A | \mathbf{b}]$ and perform elementary row operations (EROs):

$$[A | \mathbf{b}] = \left(\begin{array}{ccc|c} a_1 & b_1 & c_1 & d_1 \\ a_2 & b_2 & c_2 & d_2 \\ a_3 & b_3 & c_3 & d_3 \end{array}\right)$$

2. Elementary Row Operations (EROs) & Row Echelon Form (REF)

The three permissible EROs that preserve the complete solution space are:

Row swap: $R_i \leftrightarrow R_j$ (interchange two rows to position a convenient pivot).
Scalar multiplication: $R_i \to k R_i$ where $k \neq 0$.
Row replacement: $R_i \to R_i + k R_j$ (add a scalar multiple of row $j$ to row $i$).

We systematically apply Gaussian elimination to reduce $[A|\mathbf{b}]$ to upper-triangular row echelon form (REF):

$$\left(\begin{array}{ccc|c} 1 & u_{12} & u_{13} & w_1 \\ 0 & 1 & u_{23} & w_2 \\ 0 & 0 & p & q \end{array}\right)$$

where the bottom row $(0\; 0\; p \mid q)$ dictates the rank and nature of solutions.

3. Rouché–Capelli Theorem & Rank Analysis

The rank of a matrix $\text{rank}(M)$ is the number of non-zero rows in its row echelon form. The fundamental criterion for consistency is governed by the Rouché–Capelli Theorem:

Consistent Unique Solution: $\text{rank}(A) = \text{rank}([A \mid \mathbf{b}]) = 3 \iff \det(A) \neq 0$ ($p \neq 0$). The system has a unique solution $(x_0, y_0, z_0)$.
Consistent Infinitely Many Solutions: $\text{rank}(A) = \text{rank}([A \mid \mathbf{b}]) = r < 3$ ($p = 0, q = 0$).
- If $r = 2$: 1 free parameter (the solution set is a straight line $\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}$).
- If $r = 1$: 2 free parameters (the solution set is a plane, meaning all three equations represent the same plane).
Inconsistent (No Solution): $\text{rank}(A) < \text{rank}([A \mid \mathbf{b}])$ ($p = 0, q \neq 0$, producing $0 = q$ where $q \neq 0$).

4. Complete 3D Geometric Classification of 3 Planes

Each equation $a_i x + b_i y + c_i z = d_i$ represents a flat plane $\Pi_i$ in $\mathbb{R}^3$ with normal vector $\mathbf{n}_i = (a_i, b_i, c_i)$. The algebraic solution directly maps to 3D geometry:

Bottom Row	Rank Relationship	Normal Vectors Condition	Exact Geometric Classification
$p \neq 0$	$\text{rank}(A) = \text{rank}([A\|\mathbf{b}]) = 3$	$\mathbf{n}_1, \mathbf{n}_2, \mathbf{n}_3$ linearly independent	Single Point of Intersection: The 3 planes intersect at a single unique point $(x_0, y_0, z_0)$.
$p = 0, q = 0$	$\text{rank}(A) = \text{rank}([A\|\mathbf{b}]) = 2$	No two normal vectors are parallel	Line of Intersection (Sheaf of Planes): The 3 planes intersect along a single common line $\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}$.
$p = 0, q = 0$	$\text{rank}(A) = \text{rank}([A\|\mathbf{b}]) = 2$	Two planes have parallel normals & identical RHS	Two Coincident Planes intersecting the third plane along a line.
$p = 0, q = 0$	$\text{rank}(A) = \text{rank}([A\|\mathbf{b}]) = 1$	All three normals are parallel & identical RHS	Three Coincident Planes: All three equations define the exact same plane.
$p = 0, q \neq 0$	$\text{rank}(A) = 2 < \text{rank}([A\|\mathbf{b}]) = 3$	No two normal vectors are parallel: $\mathbf{n}_i \times \mathbf{n}_j \neq \mathbf{0}$	Triangular Prism: The three planes intersect pairwise in 3 distinct, mutually parallel lines. They enclose an open prism with triangular cross-section.
$p = 0, q \neq 0$	$\text{rank}(A) = 2 < \text{rank}([A\|\mathbf{b}]) = 3$	Two normals parallel ($\mathbf{n}_1 \parallel \mathbf{n}_2$), but $d_1/d_2 \neq \text{ratio}$	Two Parallel Planes cut by a third intersecting plane (forming two parallel lines of intersection).
$p = 0, q \neq 0$	$\text{rank}(A) = 1 < \text{rank}([A\|\mathbf{b}]) = 2$	All three normals parallel: $\mathbf{n}_1 \parallel \mathbf{n}_2 \parallel \mathbf{n}_3$	Three Mutually Parallel Planes with distinct constants (no intersections whatsoever).

5. Matrix Inverses & Determinants ($3 \times 3$)

For square matrix $A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$:

$$\det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)$$

If $\det(A) \neq 0$, the inverse exists and is given by the adjugate formula:

$$A^{-1} = \frac{1}{\det(A)} \text{adj}(A) = \frac{1}{\det(A)} C^T$$

where $C_{ij} = (-1)^{i+j} M_{ij}$ is the matrix of cofactors. The unique solution to $A\mathbf{x} = \mathbf{b}$ is $\mathbf{x} = A^{-1}\mathbf{b}$.

Worked Examples

Worked Example 1: Gaussian Elimination & Back-Substitution

Question: Solve the following system algebraically using augmented matrix elementary row operations:

$$\begin{cases} x + 2y - z = 2 \\ 2x + 5y + z = 11 \\ 3x + 7y + 2z = 15 \end{cases}$$

Solution:

Set up the augmented matrix $[A \mid \mathbf{b}]$: [M1]

$$\left(\begin{array}{ccc|c} 1 & 2 & -1 & 2 \\ 2 & 5 & 1 & 11 \\ 3 & 7 & 2 & 15 \end{array}\right)$$

Eliminate $x$ from rows 2 and 3 using $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$: [M1]

$$\left(\begin{array}{ccc|c} 1 & 2 & -1 & 2 \\ 0 & 1 & 3 & 7 \\ 0 & 1 & 5 & 9 \end{array}\right) \quad [\text{A1}]$$

Eliminate $y$ from row 3 using $R_3 \to R_3 - R_2$: [M1]

$$\left(\begin{array}{ccc|c} 1 & 2 & -1 & 2 \\ 0 & 1 & 3 & 7 \\ 0 & 0 & 2 & 2 \end{array}\right) \quad [\text{A1}]$$

Back-substitution:

From Row 3: $2z = 2 \implies z = 1$. [A1]
From Row 2: $y + 3(1) = 7 \implies y = 4$. [A1]
From Row 1: $x + 2(4) - (1) = 2 \implies x + 7 = 2 \implies x = -5$. [A1]

Conclusion: The unique solution is $(x, y, z) = (-5, 4, 1)$. Geometrically, the 3 planes intersect at the single point $(-5, 4, 1)$.

Examiner Advice: Always substitute the final coordinates back into the original second or third equation to verify arithmetic before moving on: $2(-5) + 5(4) + 1 = -10 + 20 + 1 = 11$ (verified).

Worked Example 2: Quadratic Parameter in Bottom Row & Line Equation

Question: Consider the system of linear equations in $x, y, z$ with real parameter $k$:

$$\begin{cases} x + y + z = 4 \\ x + 2y + 3z = 7 \\ 2x + 3y + (k^2 - 5)z = k + 7 \end{cases}$$

(a) Find the values of $k$ for which the system has: (i) a unique solution, (ii) infinitely many solutions, (iii) no solution.
(b) For the value of $k$ leading to infinitely many solutions, find the vector equation of the line of intersection of the three planes.

Solution:

(a) Row Reduction: Form the augmented matrix: [M1]

$$\left(\begin{array}{ccc|c} 1 & 1 & 1 & 4 \\ 1 & 2 & 3 & 7 \\ 2 & 3 & k^2-5 & k+7 \end{array}\right)$$

Perform $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - 2R_1$: [M1]

$$\left(\begin{array}{ccc|c} 1 & 1 & 1 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 1 & k^2-7 & k-1 \end{array}\right) \quad [\text{A1}]$$

Perform $R_3 \to R_3 - R_2$: [M1]

$$\left(\begin{array}{ccc|c} 1 & 1 & 1 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & k^2-9 & k-3 \end{array}\right) \quad [\text{A1}]$$

Thus the reduced third equation is $(k^2 - 9)z = k - 3$.

(i) Unique Solution: Coefficient of $z$ is non-zero $\iff k^2 - 9 \neq 0 \implies k \neq \pm 3$. [A1]
(ii) Infinitely Many Solutions: Both sides vanish $\iff k^2 - 9 = 0$ and $k - 3 = 0 \implies k = 3$. [A1]
(iii) No Solution: Inconsistent equation $(0z = c \neq 0) \iff k^2 - 9 = 0$ and $k - 3 \neq 0 \implies k = -3$. [A1]

(b) Line of Intersection for $k = 3$:
For $k = 3$, Row 3 is $0z = 0$. From Row 2: $y + 2z = 3 \implies y = 3 - 2z$.
From Row 1: $x + (3 - 2z) + z = 4 \implies x = 1 + z$.
Setting parameter $z = \lambda \in \mathbb{R}$, the vector equation is: [M1][A1]

$$\mathbf{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 + \lambda \\ 3 - 2\lambda \\ \lambda \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}, \quad \lambda \in \mathbb{R}$$

Worked Example 3: Proving 3 Planes Form a Triangular Prism

Question: Show that the following system of linear equations is inconsistent, and prove that the geometric configuration of the three planes is a triangular prism:

$$\begin{cases} \Pi_1: x - y + 2z = 4 \\ \Pi_2: 2x + y - z = 1 \\ \Pi_3: 4x - y + 3z = 7 \end{cases}$$

Solution:

Part 1: Proof of Inconsistency: Set up augmented matrix: [M1]

$$\left(\begin{array}{ccc|c} 1 & -1 & 2 & 4 \\ 2 & 1 & -1 & 1 \\ 4 & -1 & 3 & 7 \end{array}\right)$$

Apply $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 4R_1$: [M1]

$$\left(\begin{array}{ccc|c} 1 & -1 & 2 & 4 \\ 0 & 3 & -5 & -7 \\ 0 & 3 & -5 & -9 \end{array}\right) \quad [\text{A1}]$$

Apply $R_3 \to R_3 - R_2$: [M1]

$$\left(\begin{array}{ccc|c} 1 & -1 & 2 & 4 \\ 0 & 3 & -5 & -7 \\ 0 & 0 & 0 & -2 \end{array}\right) \quad [\text{A1}]$$

The third row translates to $0x + 0y + 0z = -2 \iff 0 = -2$, which is an algebraic impossibility. Hence the system has no solution (inconsistent). [R1]

Part 2: Establishing Triangular Prism Geometry:
Extract normal vectors of the three planes: [M1]

$$\mathbf{n}_1 = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}, \quad \mathbf{n}_2 = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}, \quad \mathbf{n}_3 = \begin{pmatrix} 4 \\ -1 \\ 3 \end{pmatrix}$$

Examine pairwise cross-products / collinearity: [R1]

$\frac{2}{1} \neq \frac{1}{-1} \implies \mathbf{n}_1 \nparallel \mathbf{n}_2$.
$\frac{4}{1} \neq \frac{-1}{-1} \implies \mathbf{n}_1 \nparallel \mathbf{n}_3$.
$\frac{4}{2} \neq \frac{-1}{1} \implies \mathbf{n}_2 \nparallel \mathbf{n}_3$.

Because no two normal vectors are parallel, no two planes are parallel. [R1]
Since the system is inconsistent and no two planes are parallel, the three planes must intersect pairwise in three parallel lines, forming a triangular prism. [AG] [A1]

Worked Example 4: Homogeneous System & Non-Trivial Kernel

Question: Consider the homogeneous system with parameter $\lambda \in \mathbb{R}$:

$$\begin{cases} x + 2y - z = 0 \\ 3x + \lambda y + z = 0 \\ 2x + y + 2z = 0 \end{cases}$$

(a) State why the system is guaranteed to have at least one solution for all $\lambda$.
(b) Find the value of $\lambda$ for which non-trivial solutions exist.
(c) For this value of $\lambda$, find the general non-trivial solution.

Solution:

(a) Trivial Solution: The system is homogeneous ($A\mathbf{x} = \mathbf{0}$). The trivial solution $(x, y, z) = (0, 0, 0)$ satisfies all equations for all $\lambda \in \mathbb{R}$. [A1]

(b) Condition for Non-Trivial Solutions: A non-trivial solution exists if and only if $\det(A) = 0$ (matrix $A$ is singular): [M1]

$$\det\begin{pmatrix} 1 & 2 & -1 \\ 3 & \lambda & 1 \\ 2 & 1 & 2 \end{pmatrix} = 0$$

Expand along Row 1: [M1]

$$\det(A) = 1(2\lambda - 1) - 2(6 - 2) + (-1)(3 - 2\lambda) = (2\lambda - 1) - 8 - (3 - 2\lambda) = 4\lambda - 12$$

Set $4\lambda - 12 = 0 \implies \lambda = 3$. [A1]

(c) General Solution for $\lambda = 3$: Form augmented matrix: [M1]

$$\left(\begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ 3 & 3 & 1 & 0 \\ 2 & 1 & 2 & 0 \end{array}\right) \xrightarrow{R_2-3R_1,\; R_3-2R_1} \left(\begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ 0 & -3 & 4 & 0 \\ 0 & -3 & 4 & 0 \end{array}\right) \xrightarrow{R_3-R_2} \left(\begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ 0 & -3 & 4 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right)$$

Let $z = 3t$ (choosing $3t$ to avoid fractions). [M1]
From Row 2: $-3y + 4(3t) = 0 \implies 3y = 12t \implies y = 4t$. [A1]
From Row 1: $x + 2(4t) - 3t = 0 \implies x + 5t = 0 \implies x = -5t$. [A1]
General solution: $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = t \begin{pmatrix} -5 \\ 4 \\ 3 \end{pmatrix}$ for any $t \in \mathbb{R}$ ($t \neq 0$ for non-trivial). [A1]

Worked Example 5: Paper 2 Applied Model with GDC Matrix Inverse

Question: An agricultural researcher blends three fertilizers: Grade A ($x$), Grade B ($y$), and Grade C ($z$) to supply nitrogen (N), phosphorus (P), and potassium (K). The percentages by weight are:
• Nitrogen: $10\% x + 20\% y + 40\% z = 180 \text{ kg}$
• Phosphorus: $20\% x + 30\% y + 50\% z = 290 \text{ kg}$
• Potassium: $30\% x + 10\% y + 20\% z = 210 \text{ kg}$
(a) Write this system as a matrix equation $A\mathbf{x} = \mathbf{b}$.
(b) Determine $\det(A)$ and find $A^{-1}$.
(c) Hence calculate the exact mass of each fertilizer required.

Solution:

(a) Matrix Equation: [M1] [A1]

$$\begin{pmatrix} 0.1 & 0.2 & 0.4 \\ 0.2 & 0.3 & 0.5 \\ 0.3 & 0.1 & 0.2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 180 \\ 290 \\ 210 \end{pmatrix}$$

Multiply each row by 10 for convenient integer arithmetic:

$$\begin{pmatrix} 1 & 2 & 4 \\ 2 & 3 & 5 \\ 3 & 1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1800 \\ 2900 \\ 2100 \end{pmatrix}$$

(b) Determinant & Inverse: [M1]

$$\det(M) = 1(6 - 5) - 2(4 - 15) + 4(2 - 9) = 1(1) - 2(-11) + 4(-7) = 1 + 22 - 28 = -5 \quad [\text{A1}]$$

Since $\det(M) = -5 \neq 0$, the inverse exists. The adjugate matrix is: [M1]

$$\text{adj}(M) = \begin{pmatrix} 1 & 0 & -2 \\ 11 & -10 & 3 \\ -7 & 5 & -1 \end{pmatrix} \implies M^{-1} = -\frac{1}{5} \begin{pmatrix} 1 & 0 & -2 \\ 11 & -10 & 3 \\ -7 & 5 & -1 \end{pmatrix} \quad [\text{A1}]$$

(c) Solving for $(x, y, z)$: [M1]

$$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = -\frac{1}{5} \begin{pmatrix} 1 & 0 & -2 \\ 11 & -10 & 3 \\ -7 & 5 & -1 \end{pmatrix} \begin{pmatrix} 1800 \\ 2900 \\ 2100 \end{pmatrix}$$

$x = -\frac{1}{5} [1(1800) + 0 - 2(2100)] = -\frac{1}{5} [1800 - 4200] = -\frac{1}{5}(-2400) = 480 \text{ kg}$. [A1]
$y = -\frac{1}{5} [11(1800) - 10(2900) + 3(2100)] = -\frac{1}{5} [19800 - 29000 + 6300] = -\frac{1}{5}(-2900) = 580 \text{ kg}$. [A1]
$z = -\frac{1}{5} [-7(1800) + 5(2900) - 1(2100)] = -\frac{1}{5} [-12600 + 14500 - 2100] = -\frac{1}{5}(-200) = 40 \text{ kg}$. [A1]

Requirements: 480 kg of Grade A, 580 kg of Grade B, and 40 kg of Grade C.

Practice Exercises (10 Tiered Questions)

Try these 10 exam-style questions, from core skills to harder problems. Open each question to view its solution.

Exercise 1 · Row Operations & Back-Substitution

Given the augmented matrix in upper triangular row echelon form: $$\left(\begin{array}{ccc|c} 2 & 1 & -1 & 5 \\ 0 & 3 & 2 & 11 \\ 0 & 0 & 4 & 4 \end{array}\right)$$ Find the values of $x$, $y$, and $z$ using systematic back-substitution.

Show Full Solution

Step 1: Solve for $z$ from Row 3:
$4z = 4 \implies z = 1$. [A1]

Step 2: Solve for $y$ from Row 2:
$3y + 2(1) = 11 \implies 3y = 9 \implies y = 3$. [A1]

Step 3: Solve for $x$ from Row 1:
$2x + 1(3) - 1(1) = 5 \implies 2x + 2 = 5 \implies 2x = 3 \implies x = \frac{3}{2}$. [A1]

Solution: $(x, y, z) = \left(\frac{3}{2}, 3, 1\right)$. [A1]

Exercise 2 · Row Echelon Transformation

Perform the row operations $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$ on the augmented matrix: $$\left(\begin{array}{ccc|c} 1 & -1 & 2 & 3 \\ 2 & 1 & 1 & 7 \\ 3 & 2 & -1 & 8 \end{array}\right)$$ and state the resulting intermediate matrix.

Show Full Solution

For $R_2$: $2 - 2(1) = 0$, $1 - 2(-1) = 3$, $1 - 2(2) = -3$, $7 - 2(3) = 1$. [M1]
For $R_3$: $3 - 3(1) = 0$, $2 - 3(-1) = 5$, $-1 - 3(2) = -7$, $8 - 3(3) = -1$. [M1]

Resulting matrix: $$\left(\begin{array}{ccc|c} 1 & -1 & 2 & 3 \\ 0 & 3 & -3 & 1 \\ 0 & 5 & -7 & -1 \end{array}\right) \quad [\text{A1}]$$

Exercise 3 · Identifying Inconsistency

After Gaussian elimination, a system yields the augmented matrix: $$\left(\begin{array}{ccc|c} 1 & 4 & -2 & 5 \\ 0 & 2 & 3 & 1 \\ 0 & 0 & 0 & 7 \end{array}\right)$$ State the number of solutions and justify your answer algebraically.

Show Full Solution

The third row translates to the algebraic equation $0x + 0y + 0z = 7 \iff 0 = 7$. [M1]
Because $0 = 7$ is false for all real values of $x, y, z$, this is a contradiction. [R1]
Therefore, the system is inconsistent and has no solution ($\text{rank}(A) = 2 < \text{rank}([A|\mathbf{b}]) = 3$). [A1]

Exercise 4 · Complete 3-Variable Reduction

Solve the following system by Gaussian elimination: $$\begin{cases} x + y + 2z = 8 \\ -x - 2y + 3z = 1 \\ 3x - 7y + 4z = 10 \end{cases}$$

Show Full Solution

Augmented matrix: $\left(\begin{array}{ccc|c} 1 & 1 & 2 & 8 \\ -1 & -2 & 3 & 1 \\ 3 & -7 & 4 & 10 \end{array}\right)$ [M1]
$R_2 \to R_2 + R_1$, $R_3 \to R_3 - 3R_1$:
$\left(\begin{array}{ccc|c} 1 & 1 & 2 & 8 \\ 0 & -1 & 5 & 9 \\ 0 & -10 & -2 & -14 \end{array}\right)$ [A1]
$R_3 \to R_3 - 10R_2$:
Row 3: $0 - 10(0) = 0$; $-10 - 10(-1) = 0$; $-2 - 10(5) = -52$; $-14 - 10(9) = -104$.
$\left(\begin{array}{ccc|c} 1 & 1 & 2 & 8 \\ 0 & -1 & 5 & 9 \\ 0 & 0 & -52 & -104 \end{array}\right)$ [A1]
From $R_3$: $-52z = -104 \implies z = 2$. [A1]
From $R_2$: $-y + 5(2) = 9 \implies -y + 10 = 9 \implies y = 1$. [A1]
From $R_1$: $x + 1 + 2(2) = 8 \implies x + 5 = 8 \implies x = 3$. [A1]
Unique solution: $(x, y, z) = (3, 1, 2)$.

Exercise 5 · Parametric Solution Line

Solve the dependent system and express the solution in vector form using real parameter $\lambda$: $$\begin{cases} x + 2y - z = 3 \\ 2x + 5y - 3z = 7 \\ 3x + 7y - 4z = 10 \end{cases}$$

Show Full Solution

Augmented matrix: $\left(\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 2 & 5 & -3 & 7 \\ 3 & 7 & -4 & 10 \end{array}\right)$ [M1]
$R_2 \to R_2 - 2R_1$, $R_3 \to R_3 - 3R_1$:
$\left(\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 0 & 1 & -1 & 1 \\ 0 & 1 & -1 & 1 \end{array}\right)$ [A1]
$R_3 \to R_3 - R_2$:
$\left(\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 0 & 1 & -1 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right)$ (Consistent with infinitely many solutions). [M1]
Let $z = \lambda \in \mathbb{R}$.
From Row 2: $y - \lambda = 1 \implies y = 1 + \lambda$. [A1]
From Row 1: $x + 2(1 + \lambda) - \lambda = 3 \implies x + 2 + \lambda = 3 \implies x = 1 - \lambda$. [A1]
Vector equation of the line: $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}, \; \lambda \in \mathbb{R}$. [A1]

Exercise 6 · Dual Parameter Analysis

A system is given by: $$\begin{cases} x + y + 2z = 3 \\ x + 2y + 3z = 4 \\ 2x + 3y + kz = m \end{cases}$$ Determine the values of $k$ and $m$ such that the system has: (a) a unique solution, (b) infinitely many solutions, (c) no solution.

Show Full Solution

Augmented matrix: $\left(\begin{array}{ccc|c} 1 & 1 & 2 & 3 \\ 1 & 2 & 3 & 4 \\ 2 & 3 & k & m \end{array}\right)$ [M1]
$R_2 \to R_2 - R_1$, $R_3 \to R_3 - 2R_1$:
$\left(\begin{array}{ccc|c} 1 & 1 & 2 & 3 \\ 0 & 1 & 1 & 1 \\ 0 & 1 & k-4 & m-6 \end{array}\right)$
$R_3 \to R_3 - R_2$:
$\left(\begin{array}{ccc|c} 1 & 1 & 2 & 3 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & k-5 & m-7 \end{array}\right)$ [A1]
(a) Unique solution: $k - 5 \neq 0 \implies k \neq 5$, with $m \in \mathbb{R}$. [A1]
(b) Infinitely many solutions: $k - 5 = 0$ AND $m - 7 = 0 \implies k = 5, m = 7$. [A1]
(c) No solution: $k - 5 = 0$ AND $m - 7 \neq 0 \implies k = 5, m \neq 7$. [A1]

Exercise 7 · Geometric Sheaf vs Prism

Given three planes $\Pi_1: x - 2y + z = 1$, $\Pi_2: 2x - y + 3z = 4$, $\Pi_3: 3x - 3y + 4z = 6$:
(a) Show that the system is inconsistent.
(b) By examining normal vectors, deduce whether the planes form a triangular prism or contain parallel planes.

Show Full Solution

(a) Inconsistency test: [M1]
Augmented matrix: $\left(\begin{array}{ccc|c} 1 & -2 & 1 & 1 \\ 2 & -1 & 3 & 4 \\ 3 & -3 & 4 & 6 \end{array}\right)$
$R_2 \to R_2 - 2R_1$, $R_3 \to R_3 - 3R_1$: $\left(\begin{array}{ccc|c} 1 & -2 & 1 & 1 \\ 0 & 3 & 1 & 2 \\ 0 & 3 & 1 & 3 \end{array}\right)$ [A1]
$R_3 \to R_3 - R_2$: $\left(\begin{array}{ccc|c} 1 & -2 & 1 & 1 \\ 0 & 3 & 1 & 2 \\ 0 & 0 & 0 & 1 \end{array}\right)$
Row 3 gives $0 = 1$, contradiction. System is inconsistent. [A1]

(b) Geometric configuration:
Normal vectors: $\mathbf{n}_1 = (1, -2, 1)$, $\mathbf{n}_2 = (2, -1, 3)$, $\mathbf{n}_3 = (3, -3, 4)$. [M1]
No two normal vectors are scalar multiples (i.e. $\mathbf{n}_1 \nparallel \mathbf{n}_2$, $\mathbf{n}_1 \nparallel \mathbf{n}_3$, $\mathbf{n}_2 \nparallel \mathbf{n}_3$). [R1]
Therefore, the three planes pairwise intersect along three non-coincident parallel lines, forming an open triangular prism. [A1]

Exercise 8 · Quadratic Factorization in Pivot

A linear system has the augmented matrix: $$\left(\begin{array}{ccc|c} 1 & 2 & 1 & 4 \\ 0 & 1 & 3 & 5 \\ 0 & 0 & a^2 - 9 & a - 3 \end{array}\right)$$ Find all values of $a \in \mathbb{R}$ for which the system has (a) no solution, (b) infinitely many solutions, (c) a unique solution with $z = 1$.

Show Full Solution

The third equation is $(a^2 - 9)z = a - 3$, or $(a - 3)(a + 3)z = a - 3$. [M1]

(a) No solution: Requires $(a^2 - 9) = 0$ while $a - 3 \neq 0$.
$(a-3)(a+3)=0 \implies a = 3$ or $a = -3$.
When $a = -3$: coefficient $= 0$, RHS $= -6 \neq 0$.
Thus $a = -3$ yields no solution. [A1]

(b) Infinitely many solutions: Requires coefficient $= 0$ AND RHS $= 0$.
When $a = 3$: $0z = 0$, which holds for all $z \in \mathbb{R}$. System has infinitely many solutions. [A1]

(c) Unique solution with $z = 1$:
For $a \neq \pm 3$, $z = \frac{a-3}{(a-3)(a+3)} = \frac{1}{a+3}$. [M1]
Setting $z = 1 \implies \frac{1}{a+3} = 1 \implies a+3 = 1 \implies a = -2$.
Since $-2 \neq \pm 3$, $a = -2$ is valid! [A1]

Exercise 9 · Homogeneous System & Non-Trivial Kernel

Consider the homogeneous linear system: $$\begin{cases} x + 2y - z = 0 \\ 3x + ky + z = 0 \\ 2x + y + 2z = 0 \end{cases}$$ (a) State the trivial solution.
(b) Find the value of $k$ for which the system possesses non-trivial solutions.
(c) For this value of $k$, find the general solution.

Show Full Solution

(a) Trivial solution: $(x, y, z) = (0, 0, 0)$. [A1]

(b) Non-trivial solutions condition:
A homogeneous system has non-trivial solutions if and only if $\det(A) = 0$. [M1]
$\det(A) = 1(2k - 1) - 2(6 - 2) - 1(3 - 2k) = (2k - 1) - 8 - (3 - 2k) = 4k - 12$.
Setting $\det(A) = 0 \implies 4k - 12 = 0 \implies k = 3$. [A1]

(c) General solution for $k = 3$:
$\left(\begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ 3 & 3 & 1 & 0 \\ 2 & 1 & 2 & 0 \end{array}\right) \xrightarrow{R_2-3R_1,\; R_3-2R_1} \left(\begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ 0 & -3 & 4 & 0 \\ 0 & -3 & 4 & 0 \end{array}\right) \xrightarrow{R_3-R_2} \left(\begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ 0 & -3 & 4 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right)$ [M1]
Let $z = 3\lambda$. Then $-3y + 4(3\lambda) = 0 \implies 3y = 12\lambda \implies y = 4\lambda$.
$x + 2(4\lambda) - 3\lambda = 0 \implies x + 5\lambda = 0 \implies x = -5\lambda$.
General solution: $(x, y, z) = (-5\lambda, 4\lambda, 3\lambda) = \lambda(-5, 4, 3)$ for $\lambda \in \mathbb{R}$. [A1]

Exercise 10 · 4 Planes Concurrency Problem

A fourth plane $\Pi_4: 2x - y + z = d$ is added to the system in Exemplar 1 whose intersection point was $(-5, 4, 1)$.
(a) Find the value of $d$ such that all four planes pass concurrently through the same point.
(b) If $d = 10$, describe the geometric relationship between $\Pi_4$ and the common point of $\Pi_1, \Pi_2, \Pi_3$.

Show Full Solution

(a) All 4 planes concurrent at $(-5, 4, 1)$: [M1]
Substitute $(-5, 4, 1)$ into $\Pi_4$:
$2(-5) - (4) + 1 = d \implies -10 - 4 + 1 = d \implies d = -13$. [A1]
Thus, when $d = -13$, all four planes pass concurrently through $(-5, 4, 1)$.

(b) For $d = 10$:
Substituting $(-5, 4, 1)$ yields $2(-5) - 4 + 1 = -13 \neq 10$.
The common intersection point of the first three planes does not lie on $\Pi_4$. Geometrically, the point of intersection of $\Pi_1, \Pi_2, \Pi_3$ is at a perpendicular distance from $\Pi_4$ given by $D = \frac{|2(-5) - 4 + 1 - 10|}{\sqrt{2^2 + (-1)^2 + 1^2}} = \frac{23}{\sqrt{6}}$. [A1]

Formula Booklet References

Syllabus AHL 1.15: Systems of Linear Equations

The IB DP Mathematics AA Formula Booklet contains standard matrix multiplication and inverse formulas for $2 \times 2$ matrices, but specifies that solving $3 \times 3$ systems and row reductions relies on elementary row operations and GDC matrix commands (Paper 2).

Augmented Matrix: $[A \mid \mathbf{b}]$. Elementary row operations preserve the solution space.

Key Formulas & Identities

1. Conditions for Solvability via Echelon Form $(0\; 0\; p \mid q)$

$p \neq 0$: Unique solution $\iff \text{rank}(A) = \text{rank}([A|\mathbf{b}]) = 3 \iff \det(A) \neq 0$.
$p = 0, q = 0$: Infinitely many solutions $\iff \text{rank}(A) = \text{rank}([A|\mathbf{b}]) < 3$. Planes meet in a common line (sheaf) or are coincident.
$p = 0, q \neq 0$: Inconsistent (no solution) $\iff \text{rank}(A) < \text{rank}([A|\mathbf{b}])$. Planes meet in a triangular prism (if no two normals parallel) or parallel planes.

2. Homogeneous Systems $A\mathbf{x} = \mathbf{0}$

Always consistent ($x=y=z=0$ is the trivial solution). Has non-trivial solutions if and only if $\det(A) = 0$.

3. $3 \times 3$ Determinant & Adjugate Inverse

$$\det(A) = a_1(b_2 c_3 - b_3 c_2) - b_1(a_2 c_3 - a_3 c_2) + c_1(a_2 b_3 - a_3 b_2)$$ $$A^{-1} = \frac{1}{\det(A)} C^T, \quad C_{ij} = (-1)^{i+j} M_{ij}$$

Common Exam Pitfalls

Common Exam Mistake: Forgetting to Swap Rows before Dividing

Avoid introducing unnecessary fractions in row operations. If the leading entry in row 1 is not 1, look for a row with leading coefficient 1 and perform a row swap ($R_1 \leftrightarrow R_2$) before reducing.

Prism vs Parallel Planes Confusion

When a system has no solution, IB examiners frequently award 2 marks for naming the geometry. If no two normal vectors $\mathbf{n}_i$ are parallel, write "The three planes form a triangular prism". Do NOT simply say "the planes do not intersect".

Free Variable in Parametric Solutions

When a system has infinitely many solutions, remember to explicitly declare your parameter (e.g., "Let $z = \lambda, \lambda \in \mathbb{R}$") and express your final answer as a vector line equation $\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}$. Leaving it in terms of unresolved variables loses the final answer mark [A1].

Figure 1.4: Geometric Configurations of 3 Planes in 3D Space

1. Matrix Representation of Linear Systems

2. Elementary Row Operations (EROs) & Row Echelon Form (REF)

3. Rouché–Capelli Theorem & Rank Analysis

4. Complete 3D Geometric Classification of 3 Planes

5. Matrix Inverses & Determinants ($3 \times 3$)

Worked Examples

Worked Example 1: Gaussian Elimination & Back-Substitution

Worked Example 2: Quadratic Parameter in Bottom Row & Line Equation

Worked Example 3: Proving 3 Planes Form a Triangular Prism

Worked Example 4: Homogeneous System & Non-Trivial Kernel

Worked Example 5: Paper 2 Applied Model with GDC Matrix Inverse

Practice Exercises (10 Tiered Questions)

Formula Booklet References

Syllabus AHL 1.15: Systems of Linear Equations

Key Formulas & Identities

1. Conditions for Solvability via Echelon Form $(0\; 0\; p \mid q)$

2. Homogeneous Systems $A\mathbf{x} = \mathbf{0}$

3. $3 \times 3$ Determinant & Adjugate Inverse

Common Exam Pitfalls

Common Exam Mistake: Forgetting to Swap Rows before Dividing

Prism vs Parallel Planes Confusion

Free Variable in Parametric Solutions

Topic Diagnostic Test

Section A: Diagnostic Multiple Choice Questions

Section B: Examination-Style Free Response Questions