Home / A-Level Further Maths 9231 / Matrices: systems and eigenvalues / Systems of three linear equations
Cambridge International AS & A Level · Topic 2.2
A-Level Further Maths 9231 / Topic 2.2 / AS & A Level

Three planes, one matrix.

Write three equations in three unknowns as \(\mathbf{A}\mathbf{x} = \mathbf{b}\). If \(\det\mathbf{A} \ne 0\) there is exactly one solution and the three planes meet at a point. If \(\det\mathbf{A} = 0\), elimination decides: a whole line of solutions (a sheaf) or none at all (a prism or parallel planes).

Topic 2.2All matrices: systems and eigenvalues lessons →
The key idea

Three equations in \(x, y, z\) are one matrix equation \(\mathbf{A}\mathbf{x} = \mathbf{b}\), and each equation is a plane. If \(\det\mathbf{A} \ne 0\), \(\mathbf{x} = \mathbf{A}^{-1}\mathbf{b}\) is the unique solution: the planes meet at one point. If \(\det\mathbf{A} = 0\) there is no inverse, and you must eliminate. A final row \(0 = 0\) means the system is consistent with infinitely many solutions on a common line (a sheaf); a final row \(0 = c\) with \(c \ne 0\) means it is inconsistent: a triangular prism, or two or more parallel planes.

Sheaf Triangular prism Parallel planes all three planes share one line each pair meets in a line; the three lines are parallel two planes never meet; the third cuts both consistent: a line inconsistent inconsistent
Fig. 1: when det A = 0 the three normals lie in one plane, so there is a direction parallel to all three planes. Looking along that direction, each plane is seen edge-on as a line. Left: the planes share a common line (a sheaf), so there are infinitely many solutions. Middle: they form a triangular prism. Right: two are parallel. In the last two cases no point lies on all three planes
Section 01

The matrix equation and the unique solution.

Line up the unknowns in the same order in every equation, writing a \(0\) for any missing term. The coefficients form \(\mathbf{A}\), the unknowns form \(\mathbf{x}\) and the right-hand sides form \(\mathbf{b}\). Each row of \(\mathbf{A}\) is the normal vector of one plane. When \(\det\mathbf{A} \ne 0\), multiply on the left by \(\mathbf{A}^{-1}\) to get \(\mathbf{x} = \mathbf{A}^{-1}\mathbf{b}\): one solution, and the three planes meet at a single point.

For example, the system \(x + y + z = 6\), \(2x - y + z = 3\), \(x + 2y - z = 2\) becomes

\[\begin{pmatrix} 1 & 1 & 1 \\ 2 & -1 & 1 \\ 1 & 2 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 6 \\ 3 \\ 2 \end{pmatrix}.\]

Reading it backwards turns a matrix equation into three equations: row times column gives the left side of each one. The inverse is quickest when you need it anyway (or the question gives it); for a single system, elimination is often less work. Either way the condition is the same: \(\det\mathbf{A} \ne 0\) if and only if there is exactly one solution, whatever \(\mathbf{b}\) is.

Worked example Solve with the inverse
Write the system \(x + y + z = 6\), \(2x - y + z = 3\), \(x + 2y - z = 2\) as \(\mathbf{A}\mathbf{x} = \mathbf{b}\) and solve it using \(\mathbf{A}^{-1}\).
  1. \(\mathbf{A} = \begin{pmatrix} 1 & 1 & 1 \\ 2 & -1 & 1 \\ 1 & 2 & -1 \end{pmatrix}\), \(\mathbf{b} = \begin{pmatrix} 6 \\ 3 \\ 2 \end{pmatrix}\).
  2. Expand along row 1: \[\det\mathbf{A} = 1(1 - 2) - 1(-2 - 1) + 1(4 + 1) = -1 + 3 + 5 = 7 \ne 0.\]
  3. Cofactors, transposed, give the adjugate: \[\mathbf{A}^{-1} = \frac{1}{7}\begin{pmatrix} -1 & 3 & 2 \\ 3 & -2 & 1 \\ 5 & -1 & -3 \end{pmatrix}.\]
  4. \[\mathbf{x} = \mathbf{A}^{-1}\mathbf{b} = \frac{1}{7}\begin{pmatrix} -6 + 9 + 4 \\ 18 - 6 + 2 \\ 30 - 3 - 6 \end{pmatrix} = \frac{1}{7}\begin{pmatrix} 7 \\ 14 \\ 21 \end{pmatrix}.\]
  5. Check in the third equation: \(1 + 4 - 3 = 2\). The three planes meet at the point \((1, 2, 3)\).
Answer: \(x = 1\), \(y = 2\), \(z = 3\): the planes meet at the single point \((1, 2, 3)\)
Worked example When is the solution unique?
The equations \(x - y + 2z = 6\), \(3x + ky + z = 4\), \(2x + y + z = 3\) are given. (a) Find the value of \(k\) for which there is NOT a unique solution. (b) Solve the system when \(k = 1\).
  1. (a) \[\det\mathbf{A} = \begin{vmatrix} 1 & -1 & 2 \\ 3 & k & 1 \\ 2 & 1 & 1 \end{vmatrix} = 1(k - 1) + 1(3 - 2) + 2(3 - 2k) = 6 - 3k.\]
  2. There is a unique solution exactly when \(6 - 3k \ne 0\), so the exception is \(k = 2\).
  3. (b) With \(k = 1\), \(\det\mathbf{A} = 3 \ne 0\). The second equation is \(3x + y + z = 4\); subtract the third, \(2x + y + z = 3\), to get \(x = 1\).
  4. Put \(x = 1\) into the third equation: \(y + z = 1\). Into the first: \(-y + 2z = 5\).
  5. Add these: \(3z = 6\), so \(z = 2\) and \(y = -1\). Check in the second: \(3 - 1 + 2 = 4\).
Answer: (a) \(k = 2\)   (b) \(x = 1\), \(y = -1\), \(z = 2\), a single point of intersection
Your turn
Write \(x - 2y + z = 0\), \(2x + y - z = 1\), \(x + y + 2z = 9\) as a matrix equation, show that it has a unique solution, and find it.
Examiner trap

Order matters. \(\mathbf{A}\mathbf{x} = \mathbf{b}\) gives \(\mathbf{x} = \mathbf{A}^{-1}\mathbf{b}\), not \(\mathbf{b}\mathbf{A}^{-1}\) (which is not even defined for a column \(\mathbf{b}\)). And a missing variable is a zero coefficient in its own column: shuffling the other numbers left to fill the gap gives the wrong matrix.

Section 02

Gaussian elimination and consistency.

Write the augmented matrix \((\mathbf{A} \mid \mathbf{b})\) and use row operations (swap two rows, multiply a row by a non-zero number, add a multiple of one row to another) to clear the entries below the leading diagonal. Each operation replaces the system with an equivalent one, so the solutions do not change. Then read upwards from the last row.

Last row after eliminationMeaningSolutions
\((0 \;\; 0 \;\; c \mid d)\) with \(c \ne 0\)\(\det\mathbf{A} \ne 0\)exactly one: back-substitute
\((0 \;\; 0 \;\; 0 \mid 0)\)\(\det\mathbf{A} = 0\), consistentinfinitely many: set one unknown \(= t\)
\((0 \;\; 0 \;\; 0 \mid d)\) with \(d \ne 0\)\(\det\mathbf{A} = 0\), inconsistentnone: the row says \(0 = d\)

In the consistent case two independent equations remain. Let one unknown be a parameter \(t\), express the other two in terms of \(t\), and write the answer as a line \(\mathbf{r} = \mathbf{a} + t\mathbf{d}\). The direction \(\mathbf{d}\) is parallel to \(\mathbf{n}_{1} \times \mathbf{n}_{2}\), the cross product of two of the normals: a useful check.

Worked example A consistent singular system
Solve \(x + y + z = 3\), \(x - y + 2z = 2\), \(2x + 3z = 5\).
  1. \[\left(\begin{array}{ccc|c} 1 & 1 & 1 & 3 \\ 1 & -1 & 2 & 2 \\ 2 & 0 & 3 & 5 \end{array}\right) \to \left(\begin{array}{ccc|c} 1 & 1 & 1 & 3 \\ 0 & -2 & 1 & -1 \\ 0 & -2 & 1 & -1 \end{array}\right)\] using \(R_{2} - R_{1}\) and \(R_{3} - 2R_{1}\).
  2. \(R_{3} - R_{2}\) gives \((0 \;\; 0 \;\; 0 \mid 0)\): the system is consistent and \(\det\mathbf{A} = 0\), so there are infinitely many solutions.
  3. Let \(y = t\). Row 2: \(-2t + z = -1\), so \(z = 2t - 1\). Row 1: \(x = 3 - t - (2t - 1) = 4 - 3t\).
  4. Check with the second equation: \((4 - 3t) - t + 2(2t - 1) = 2\) for every \(t\).
  5. Direction check: \((1, 1, 1) \times (1, -1, 2) = (3, -1, -2)\), parallel to \((-3, 1, 2)\).
Answer: \(\mathbf{r} = \begin{pmatrix} 4 \\ 0 \\ -1 \end{pmatrix} + t\begin{pmatrix} -3 \\ 1 \\ 2 \end{pmatrix}\): the three planes form a sheaf
Worked example An inconsistent singular system
Show that \(x + y + z = 3\), \(x - y + 2z = 2\), \(2x + 3z = 7\) has no solution.
  1. The same row operations as before: \(R_{2} - R_{1}\) gives \((0 \;\; -2 \;\; 1 \mid -1)\) and \(R_{3} - 2R_{1}\) gives \((0 \;\; -2 \;\; 1 \mid 1)\).
  2. \(R_{3} - R_{2}\) gives \((0 \;\; 0 \;\; 0 \mid 2)\), that is \(0x + 0y + 0z = 2\), which is impossible.
  3. So the system is inconsistent. The normals \((1, 1, 1)\), \((1, -1, 2)\), \((2, 0, 3)\) are not multiples of each other, so no two planes are parallel.
Answer: No solution: the three planes form a triangular prism
Your turn
Solve \(x + 2y - z = 1\), \(2x + y + z = 5\), \(4x + 5y - z = 7\).
Examiner trap

\(\det\mathbf{A} = 0\) does not mean 'no solution'. It means 'not a unique solution': there may be none or infinitely many, and only the right-hand sides decide which. So never stop at the determinant; finish the elimination including the last column. And when you give a line of solutions, check the point and direction in ALL THREE equations, not just the two you used.

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Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

Quick check+10 XP

The system \(2x - y = 3\), \(x + y + z = 0\), \(3y - z = 4\) is written as \(\mathbf{A}\mathbf{x} = \mathbf{b}\). What is \(\mathbf{A}\)?

Quick check+10 XP

For a system \(\mathbf{A}\mathbf{x} = \mathbf{b}\) of three equations, \(\det\mathbf{A} = 3\). Which statement is true?

Quick check+10 XP

Elimination on \((\mathbf{A} \mid \mathbf{b})\) leaves two independent rows and a last row \((0 \;\; 0 \;\; 0 \mid 0)\). What does this tell you?

Quick check+10 XP

Elimination ends with the row \((0 \;\; 0 \;\; 0 \mid 5)\), and no two of the normal vectors are parallel. How do the three planes meet?

Section 03

Singular matrices and the geometry of planes.

A classic exam question has two constants. First find the value of \(k\) that makes \(\det\mathbf{A} = 0\) (for every other \(k\) the planes meet at a point). Then, with that \(k\), eliminate to find the value of the second constant that makes the system consistent. Finally describe each case geometrically, which needs one more check: are any two normals parallel?

AlgebraGeometry (Fig. 1)
\(\det\mathbf{A} \ne 0\): one solutionthree planes meet at a single point
\(\det\mathbf{A} = 0\), consistent: infinitely many solutionsthe planes share a common line (a sheaf); if two equations are multiples of each other those planes coincide
\(\det\mathbf{A} = 0\), inconsistent, no two normals parallela triangular prism: each pair of planes meets in a line, and the three lines are parallel
\(\det\mathbf{A} = 0\), inconsistent, two normals paralleltwo parallel planes, both cut by the third; if all three normals are parallel, three parallel planes

Why \(\det\mathbf{A} = 0\) matters geometrically: then one row of \(\mathbf{A}\) is a combination of the other two, so the three normals lie in one plane and some direction is parallel to all three planes. That is why the planes can line up as a sheaf or a prism. Finding the combination (for example \(R_{3} = R_{2} - R_{1}\)) also tells you the consistency condition: the right-hand sides must combine the same way.

Worked example Find k, then the value for consistency
The system is \(x + y + 2z = 3\), \(2x - y + kz = 1\), \(x - 2y + z = m\). (a) Find the value of \(k\) for which the matrix of coefficients is singular. (b) For this \(k\), find the value of \(m\) for which the system is consistent, and solve it. (c) Interpret the other values of \(m\) geometrically.
  1. (a) \[\begin{vmatrix} 1 & 1 & 2 \\ 2 & -1 & k \\ 1 & -2 & 1 \end{vmatrix} = (-1 + 2k) - (2 - k) + 2(-4 + 1) = 3k - 9,\] so \(k = 3\).
  2. (b) \(R_{2} - 2R_{1}\): \((0 \;\; -3 \;\; -1 \mid -5)\); \(R_{3} - R_{1}\): \((0 \;\; -3 \;\; -1 \mid m - 3)\); then \(R_{3} - R_{2}\): \((0 \;\; 0 \;\; 0 \mid m + 2)\).
  3. Consistent if and only if \(m + 2 = 0\), so \(m = -2\).
  4. Let \(y = 2 - s\). Row 2: \(z = 5 - 3y = 3s - 1\). Row 1: \(x = 3 - y - 2z = 3 - 5s\).
  5. (c) For \(m \ne -2\) there is no solution. The normals \((1, 1, 2)\), \((2, -1, 3)\), \((1, -2, 1)\) are pairwise non-parallel, so the planes form a triangular prism.
Answer: (a) \(k = 3\)   (b) \(m = -2\), \(\mathbf{r} = \begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix} + s\begin{pmatrix} -5 \\ -1 \\ 3 \end{pmatrix}\), a sheaf   (c) \(m \ne -2\): a triangular prism
Worked example A singular value that gives parallel planes
The system is \(x + 2y - z = 4\), \(kx + 4y - 2z = 5\), \(x - y + z = 1\). Find the value of \(k\) for which there is no unique solution, and describe the planes for this \(k\).
  1. \[\det\mathbf{A} = 1(4 - 2) - 2(k + 2) - 1(-k - 4) = 2 - k,\] so \(k = 2\).
  2. With \(k = 2\) the second plane is \(2x + 4y - 2z = 5\), that is \(x + 2y - z = 2.5\).
  3. Its normal \((2, 4, -2)\) is \(2(1, 2, -1)\), parallel to the first plane's normal, but \(2.5 \ne 4\): the planes are parallel and distinct, so the system is inconsistent.
  4. The third normal \((1, -1, 1)\) is not parallel to \((1, 2, -1)\), so the third plane cuts both of them.
Answer: \(k = 2\): no solution; two parallel planes, each cut by the third plane in a line (two parallel lines)
Your turn
Show that the matrix of \(x + y + z = 1\), \(x + 2y + 3z = 4\), \(2x + 3y + 4z = c\) is singular, find \(c\) for consistency, and describe the planes when \(c\) takes any other value.
Examiner trap

Do not call every inconsistent system a prism. Check the normals: if two are multiples of each other, the answer is parallel planes. And the value of \(k\) that makes \(\mathbf{A}\) singular never depends on \(\mathbf{b}\): find \(k\) from the determinant first, and only then use the right-hand sides to find the second constant.

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Find the value of \(k\) for which \(\begin{pmatrix} 1 & 2 & 3 \\ 2 & k & 1 \\ 1 & 1 & 1 \end{pmatrix}\) is singular.

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The equations \(x + 2y - z = 2\), \(3x - y + 2z = 1\), \(5x + 3y = m\) have a singular matrix of coefficients. Find the value of \(m\) for which they are consistent.

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Describe the planes \(2x - 4y + 6z = 1\), \(x - 2y + 3z = 5\) and \(x + y + z = 0\).

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The planes \(x + y - z = 1\), \(2x - y + z = 5\), \(3x = 6\) form a sheaf. Find their common line.

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Six original Cambridge-style questions
Solve with a given inverse and by Gaussian elimination, find the common line of a sheaf, handle the two-constant question (singular \(k\), then consistent \(m\)), spot parallel planes, and set up a pricing problem as \(\mathbf{A}\mathbf{x} = \mathbf{b}\).