CAIE A-Level Further Math A2 2.2 Matrices Questions

Practise analysing linear systems, characteristic equations, eigenvectors and diagonal forms, linking matrix calculations to plane geometry, powers and inverses.

Exam points

evaluate a determinant to classify a three-equation system and interpret the corresponding planes
find eigenvalues and matched eigenvectors before constructing the diagonal and modal matrices
substitute a matrix into its characteristic equation to reduce powers or calculate an inverse

Covered syllabus points

2.2.1—Matrix equations 2.2.2—Determinants and inverses 2.2.3—Eigen terminology 2.2.4—Eigenvalues and eigenvectors 2.2.5—Eigenvalues and eigenvectors 2.2.6—Eigen terminology

[Maximum number: 4]

Show that the system of equations

\begin{aligned} 14 x-4 y+6 z & =5 \\ x+y+k z & =3 \\ -21 x+6 y-9 z & =14 \end{aligned}

where k is a constant, does not have a unique solution and interpret this situation geometrically.

$\begin{aligned} \left|\begin{array}{ccc} 14 -4 1 1 k -21 6 -9 \end{array}\right|=14\left|\begin{array}{cc} 1 k 6 -9 \end{array}\right|+4\left|\begin{array}{cc} 1 k -21 -9 \end{array}\right|+6\left|\begin{array}{cc} 1 -21 6 \end{array}\right| =14(-9-6 k)+4(-9+21 k)+6(6+21)=0 \end{aligned}$
Evaluates determinant. Can expand along any row e.g. -(36-36)+(-126+126)+k(84-84).

If using row operations, they must show an inconsistent system for M1. All their row operations must be correct for A1.

Two parallel planes, not identical.
Other plane not parallel.
4

[Maximum number: 14]

The matrix A is given by

\mathbf{A}=\left(\begin{array}{ccc} a & -6 a & 2 a+2 \\ 0 & 1-a & 0 \\ 0 & 2-a & -1 \end{array}\right)

where a is a constant with $a \neq 0$ and $a \neq 1$ .

(a)

Show that the equation $\mathbf{A}\left(\begin{array}{c}x \\ y \\ z\end{array}\right)=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)$ has a unique solution and interpret this situation geometrically.

[ 3 ]

$\left|\begin{array}{ccc}a -6 a 2 a+2 0 1-a 0 2-a -1\end{array}\right|=a(a-1) \neq 0$ or
$\begin{gathered} a x=1+\frac{12 a}{1-a}-2 \frac{(a+1)^{2}}{1-a}=\frac{-2 a^{2}+7 a-1}{1-a} y=\frac{2}{1-a} z=\frac{2(2-a)}{1-a}-3=\frac{a+1}{1-a} \end{gathered}$
Shows that determinant is non-zero or solves
equations.

Three planes intersect at a single point.
Must be clear that there is just one point where all 3
planes intersect.

3

(b)

Show that the eigenvalues of A are a, 1-a and -1.

[ 2 ]

$\left|\begin{array}{ccc}a-\lambda -6 a 2 a+2 0 1-a-\lambda 0 2-a -1-\lambda\end{array}\right|=0$
Equates determinant to zero or working to find
characteristic equation.

$(a-\lambda)(1-a-\lambda)(-1-\lambda)=0 \Rightarrow \lambda=a, 1-a,-1$
AG. Factorisation must be clear.

2

(c)

Find a matrix P and a diagonal matrix D such that $\mathbf{A}^{4}=\mathbf{P D P}^{-1}$ .

[ 6 ]

$\lambda=a:\left|\begin{array}{ccc}\mathbf{i} \mathbf{j} \mathbf{k} 0 1-2 a 0 2-a -1-a\end{array}\right|=\left(\begin{array}{c}2 a^{2}+a-1 0 0\end{array}\right) \sim\left(\begin{array}{l}1 0 0\end{array}\right)$
Uses vector product (or equations) to find
corresponding eigenvectors.

$\lambda=-1:\left|\begin{array}{ccc}\mathbf{i} \mathbf{j} \mathbf{k} a+1 -6 a 2 a+2 0 2-a 0\end{array}\right|=\left(\begin{array}{c}-(2-a)(2 a+2) 0 (a+1)(2-a)\end{array}\right) \sim\left(\begin{array}{c}-2 0 1\end{array}\right)$ $\lambda=1-a:\left|\begin{array}{ccc}\mathbf{i} \mathbf{j} \mathbf{k} 2 a-1 -6 a 2 a+2 0 2-a -2+a\end{array}\right|=\left(\begin{array}{l}-4 a^{2}+10 a-4 -2 a^{2}+5 a-2 -2 a^{2}+5 a-2\end{array}\right) \sim\left(\begin{array}{l}2 1 1\end{array}\right)$ $\mathbf{P}=\left(\begin{array}{ccc}1 -2 0 0 0 1 1\end{array}\right) \quad$ and $\quad \mathbf{D}=\left(\begin{array}{ccc}a^{4} 0 0 1 0 0 (1-a)^{4}\end{array}\right)$
Correctly matched permutations of columns. Their
eigenvectors must be non-zero for M1.

6

(d)

Use the characteristic equation of A to find $\mathbf{A}^{4}$ in terms of A and a.

[ 3 ]

$(a-\lambda)(1-a-\lambda)(-1-\lambda)=\lambda^{3}-\left(a^{2}-a+1\right) \lambda-\left(a^{2}-a\right)=0$
Finds characteristic equation multiplied out.

$\mathbf{A}^{4}=\left(a^{2}-a+1\right) \mathbf{A}^{2}+\left(a^{2}-a\right) \mathbf{A}$
Multiplies through by A.

3

[Maximum number: 11]

The matrix A is given by

\mathbf{A}=\left(\begin{array}{rrr} -6 & 2 & 13 \\ 0 & -2 & 5 \\ 0 & 0 & 8 \end{array}\right) .

(a)

Find a matrix P and a diagonal matrix D such that $\mathbf{A}^{-1}=\mathbf{P D P}^{-1}$ .

[ 7 ]

$\lambda=-6, \quad \lambda=-2, \quad \lambda=8$ $\lambda=-6:\left|\begin{array}{ccc}\mathbf{i} \mathbf{j} \mathbf{k} 0 4 0 0 2\end{array}\right|=\left(\begin{array}{l}8 0 0\end{array}\right) \sim\left(\begin{array}{l}1 0 0\end{array}\right)$
Uses vector product (or equations) to find corresponding
eigenvectors.

$\lambda=-2:\left|\begin{array}{ccc}\mathbf{i} \mathbf{j} \mathbf{k} -4 2 0 0 5\end{array}\right|=\left(\begin{array}{c}10 20 0\end{array}\right) \sim\left(\begin{array}{c}1 2 0\end{array}\right)$ $\lambda=8:\left|\begin{array}{ccc}\mathbf{i} \mathbf{j} \mathbf{k} -14 2 0 -10 5\end{array}\right|=\left(\begin{array}{c}140 70 140\end{array}\right) \sim\left(\begin{array}{l}2 1 2\end{array}\right)$
Thus $\mathbf{P}=\left(\begin{array}{lll}1 1 0 2 0 0 2\end{array}\right)$ and $\mathbf{D}=\left(\begin{array}{ccc}-\frac{1}{6} 0 0 -\frac{1}{2} 0 0 \frac{1}{8}\end{array}\right)$
Or correctly matched permutations of columns.
M1 for their (non-zero) eigenvectors matched to their
eigenvalues.

7

(b)

Use the characteristic equation of A to find $\mathbf{A}^{-1}$ .

[ 4 ]

$\boldsymbol{A}^{3}-52 \boldsymbol{A}-96 \mathbf{I}=0$
Substitutes A into characteristic equation.

$96 \mathbf{A}^{-1}=\mathbf{A}^{2}-52 \mathbf{I}$
Multiples through by $\mathbf{A}^{-1}$ .

$\mathbf{A}^{2}=\left(\begin{array}{ccc}36 -16 0 4 0 0 64\end{array}\right)$
7(b)

$\mathbf{A}^{-1}=\left(\begin{array}{ccc}-\frac{1}{6} -\frac{1}{6} \frac{3}{8} 0 -\frac{1}{2} \frac{5}{16} 0 0 \frac{1}{8}\end{array}\right)$
4

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Covered syllabus points

Question 1

Question 2

Question (a)

Question (b)

Question (c)

Question (d)

Question 3

Question (a)

Question (b)