AS & A Level · Practice questions · Systems of three linear equations

Point, line or nothing?

Six original Cambridge-style questions on systems of three linear equations: the inverse method, Gaussian elimination, consistency of singular systems with lines of solutions, the classic find-\(k\)-then-\(m\) question and geometric interpretation as a point, sheaf, prism or parallel planes.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Determinant first, then eliminate.

Set up \(\mathbf{A}\mathbf{x} = \mathbf{b}\): one row per equation, columns \(x, y, z\), zeros for missing terms.
Test \(\det\mathbf{A}\): non-zero means one solution, \(\mathbf{x} = \mathbf{A}^{-1}\mathbf{b}\), a single point.
Eliminate \((\mathbf{A} \mid \mathbf{b})\) when \(\det\mathbf{A} = 0\): \(0 = 0\) is consistent, \(0 = d \ne 0\) is inconsistent.
Parametrise a consistent singular system: one unknown \(= t\), answer as \(\mathbf{r} = \mathbf{a} + t\mathbf{d}\), checked in all three equations.
Interpret: sheaf if consistent; if not, parallel planes when two normals are parallel, otherwise a prism.

Recall

[3 marks]

Given that \(\mathbf{A} = \begin{pmatrix} 2 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix}\) has inverse \(\mathbf{A}^{-1} = \dfrac{1}{3}\begin{pmatrix} 1 & 1 & -1 \\ 1 & -2 & 2 \\ -1 & 2 & 1 \end{pmatrix}\), solve the equations \(2x + y = 4\), \(x + z = 4\), \(y + z = 5\).

The system is \(\mathbf{A}\mathbf{x} = \mathbf{b}\) with \(\mathbf{b} = \begin{pmatrix} 4 \\ 4 \\ 5 \end{pmatrix}\), so \(\mathbf{x} = \mathbf{A}^{-1}\mathbf{b}\). ✓
\[\mathbf{x} = \frac{1}{3}\begin{pmatrix} 4 + 4 - 5 \\ 4 - 8 + 10 \\ -4 + 8 + 5 \end{pmatrix} = \frac{1}{3}\begin{pmatrix} 3 \\ 6 \\ 9 \end{pmatrix}.\] ✓
\(x = 1\), \(y = 2\), \(z = 3\). Check: \(2 + 2 = 4\), \(1 + 3 = 4\), \(2 + 3 = 5\). ✓

Calculation

[4 marks]

Use Gaussian elimination to solve \(x + y + z = 2\), \(2x + 3y - z = -1\), \(x - y + 2z = 7\).

\[\left(\begin{array}{ccc|c} 1 & 1 & 1 & 2 \\ 2 & 3 & -1 & -1 \\ 1 & -1 & 2 & 7 \end{array}\right)\] ✓
\(R_{2} - 2R_{1}\): \((0 \;\; 1 \;\; -3 \mid -5)\); \(R_{3} - R_{1}\): \((0 \;\; -2 \;\; 1 \mid 5)\). ✓
\(R_{3} + 2R_{2}\): \((0 \;\; 0 \;\; -5 \mid -5)\), so \(z = 1\). ✓
Back-substitute: \(y = -5 + 3z = -2\), then \(x = 2 - y - z = 3\). ✓
Check in the third equation: \(3 + 2 + 2 = 7\). Solution \(x = 3\), \(y = -2\), \(z = 1\): the planes meet at one point. ✓

Calculation

[5 marks]

The equations are \(x - y + 2z = 2\), \(2x + y - z = 1\), \(x + 2y - 3z = -1\).

(a) Show that the matrix of coefficients is singular.

(b) Show that the equations are consistent, and find the line of solutions in vector form.

(a) \[\det\mathbf{A} = 1(-3 + 2) + 1(-6 + 1) + 2(4 - 1) = -1 - 5 + 6 = 0.\] ✓
(b) \(R_{2} - 2R_{1}\): \((0 \;\; 3 \;\; -5 \mid -3)\); \(R_{3} - R_{1}\): \((0 \;\; 3 \;\; -5 \mid -3)\); \(R_{3} - R_{2}\) is \((0 \;\; 0 \;\; 0 \mid 0)\), so the system is consistent. ✓
Adding the first two equations gives \(3x + z = 3\). Let \(x = t\): \(z = 3 - 3t\), and from the first equation \(y = x + 2z - 2 = 4 - 5t\). ✓
\(\mathbf{r} = \begin{pmatrix} 0 \\ 4 \\ 3 \end{pmatrix} + t\begin{pmatrix} 1 \\ -5 \\ -3 \end{pmatrix}\). ✓
Check in the third equation: \(t + 2(4 - 5t) - 3(3 - 3t) = -1\) for every \(t\). The planes form a sheaf. ✓

Analysis

[7 marks]

The equations are \(x + 2y + kz = 1\), \(2x - y + z = 2\), \(x + y + 2z = m\).

(a) Find the value of \(k\) for which the equations do not have a unique solution.

(b) For this value of \(k\), find the value of \(m\) for which the equations are consistent, and solve them.

(a) \[\det\mathbf{A} = 1(-2 - 1) - 2(4 - 1) + k(2 + 1) = 3k - 9,\] so \(k = 3\). ✓
(b) With \(k = 3\): \(R_{2} - 2R_{1}\) gives \((0 \;\; -5 \;\; -5 \mid 0)\) and \(R_{3} - R_{1}\) gives \((0 \;\; -1 \;\; -1 \mid m - 1)\). ✓
\(5R_{3} - R_{2}\) gives \((0 \;\; 0 \;\; 0 \mid 5m - 5)\), so the equations are consistent if and only if \(m = 1\). ✓
Then \(y = -z\). Let \(z = t\): \(y = -t\) and \(x = 1 - 2y - 3z = 1 - t\), so \(\mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} + t\begin{pmatrix} -1 \\ -1 \\ 1 \end{pmatrix}\). ✓
Check in the second equation: \(2(1 - t) + t + t = 2\). ✓
(c) For \(m \ne 1\) the system is inconsistent. The normals \((1, 2, 3)\), \((2, -1, 1)\), \((1, 1, 2)\) are pairwise non-parallel, so the planes form a triangular prism. ✓

Analysis

[5 marks]

The equations are \(x - 2y + z = 3\), \(2x + ky + 2z = 1\), \(x + y - z = 4\).

(a) Find the value of \(k\) for which the matrix of coefficients is singular, and explain geometrically why there is then no solution.

(b) Solve the equations when \(k = 1\).

(a) \[\det\mathbf{A} = 1(-k - 2) + 2(-2 - 2) + 1(2 - k) = -2k - 8,\] so \(k = -4\). ✓
With \(k = -4\) the second plane is \(2x - 4y + 2z = 1\), with normal \(2(1, -2, 1)\), parallel to the first plane. But \(x - 2y + z = \tfrac{1}{2}\) and \(x - 2y + z = 3\) cannot both hold: the planes are parallel and distinct, so no point lies on both. ✓
(b) With \(k = 1\): second minus twice the first gives \(5y = -5\), so \(y = -1\). ✓
Then \(x + z = 3 + 2y = 1\) and \(x - z = 4 - y = 5\), so \(x = 3\), \(z = -2\). ✓
Check in the second equation: \(6 - 1 - 4 = 1\). The planes meet at the point \((3, -1, -2)\). ✓

Modelling

[6 marks]

A shop sells pens, pencils and rulers at fixed prices of \(x\), \(y\) and \(z\) dollars. Order P (3 pens, 2 pencils, 1 ruler) costs 10 dollars, order Q (1 pen, 4 pencils, 2 rulers) costs 10 dollars and order R (2 pens, 1 pencil, 3 rulers) costs 11 dollars.

(a) Write this information as a matrix equation and use the inverse matrix to find the three prices.

(b) Explain why order S (4 pens, 6 pencils, 3 rulers) together with orders P and Q could never be enough to find the prices.

(a) \[\begin{pmatrix} 3 & 2 & 1 \\ 1 & 4 & 2 \\ 2 & 1 & 3 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 10 \\ 10 \\ 11 \end{pmatrix}\] ✓
\(\det\mathbf{A} = 3(12 - 2) - 2(3 - 4) + 1(1 - 8) = 25\), and \[\mathbf{A}^{-1} = \frac{1}{25}\begin{pmatrix} 10 & -5 & 0 \\ 1 & 7 & -5 \\ -7 & 1 & 10 \end{pmatrix}.\] ✓
\[\mathbf{x} = \frac{1}{25}\begin{pmatrix} 100 - 50 + 0 \\ 10 + 70 - 55 \\ -70 + 10 + 110 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix}.\] ✓
A pen costs 2 dollars, a pencil 1 dollar and a ruler 2 dollars. ✓
(b) Order S is order P plus order Q: \((4, 6, 3) = (3, 2, 1) + (1, 4, 2)\). So its row is the sum of the other two and \(\det\begin{pmatrix} 3 & 2 & 1 \\ 1 & 4 & 2 \\ 4 & 6 & 3 \end{pmatrix} = 0\). ✓
The matrix is singular, so there is never a unique solution. S adds no new information: it must cost 20 dollars, the total of P and Q, and then infinitely many sets of prices fit (if it cost anything else, the data would be inconsistent). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.