Home · A-Level Further Maths 9231 · Matrices: systems and eigenvalues · Diagonalisation and the Cayley-Hamilton theorem · Practice
AS & A Level · Practice questions · Diagonalisation and the Cayley-Hamilton theorem

Diagonalise once, power forever.

Six original Cambridge-style questions on diagonalisation and the Cayley-Hamilton theorem: writing \(A = QDQ^{-1}\), general formulae for \(A^{n}\) for 2 x 2 and 3 x 3 matrices, and powers and inverses from the characteristic equation, with full model solutions.

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

Eigenvectors in, powers out.

01
Recall
[3 marks]

\(A = \begin{pmatrix}1&4\\2&3\end{pmatrix}\). Show that \(\begin{pmatrix}1\\1\end{pmatrix}\) and \(\begin{pmatrix}2\\-1\end{pmatrix}\) are eigenvectors of \(A\), state the corresponding eigenvalues, and write down matrices \(Q\) and \(D\) such that \(A = QDQ^{-1}\).

  • \(A\begin{pmatrix}1\\1\end{pmatrix} = \begin{pmatrix}5\\5\end{pmatrix} = 5\begin{pmatrix}1\\1\end{pmatrix}\), so it is an eigenvector with \(\lambda = 5\). ✓
  • \(A\begin{pmatrix}2\\-1\end{pmatrix} = \begin{pmatrix}2 - 4\\4 - 3\end{pmatrix} = \begin{pmatrix}-2\\1\end{pmatrix} = -1\begin{pmatrix}2\\-1\end{pmatrix}\), so it is an eigenvector with \(\lambda = -1\). ✓
  • \(Q = \begin{pmatrix}1&2\\1&-1\end{pmatrix}\), \(D = \begin{pmatrix}5&0\\0&-1\end{pmatrix}\) (or both orders swapped together). ✓
02
Calculation
[5 marks]

For the matrix \(A\) in Question 1, find \(A^{n}\) in terms of \(n\), and hence find \(A^{3}\).

  • \(\det Q = -1 - 2 = -3\), so \(Q^{-1} = \tfrac{1}{3}\begin{pmatrix}1&2\\1&-1\end{pmatrix}\). ✓
  • \[QD^{n} = \begin{pmatrix}5^{n}&2(-1)^{n}\\5^{n}&-(-1)^{n}\end{pmatrix}\] ✓
  • \[A^{n} = \tfrac{1}{3}\begin{pmatrix}5^{n}&2(-1)^{n}\\5^{n}&-(-1)^{n}\end{pmatrix}\begin{pmatrix}1&2\\1&-1\end{pmatrix}\] ✓
  • \[A^{n} = \tfrac{1}{3}\begin{pmatrix}5^{n} + 2(-1)^{n} & 2 \cdot 5^{n} - 2(-1)^{n}\\ 5^{n} - (-1)^{n} & 2 \cdot 5^{n} + (-1)^{n}\end{pmatrix}\] ✓
  • Check \(n = 1\): \(\tfrac{1}{3}\begin{pmatrix}3&12\\6&9\end{pmatrix} = A\). ✓
  • \(n = 3\): \(\tfrac{1}{3}\begin{pmatrix}125 - 2&250 + 2\\125 + 1&250 - 1\end{pmatrix} = \begin{pmatrix}41&84\\42&83\end{pmatrix}\). ✓
03
Calculation
[5 marks]

\(R = \begin{pmatrix}3&-1\\4&-2\end{pmatrix}\).

(a) Find the characteristic equation of \(R\) and state the Cayley-Hamilton theorem for \(R\).

(b) Hence express \(R^{4}\) in the form \(pR + qI\), and evaluate it.

(c) Use the theorem to find \(R^{-1}\).

  • (a) \((3 - \lambda)(-2 - \lambda) + 4 = \lambda^{2} - \lambda - 2 = 0\), so \(R^{2} - R - 2I = \mathbf{0}\). ✓
  • (b) \(R^{2} = R + 2I\), so \(R^{4} = (R + 2I)^{2} = R^{2} + 4R + 4I\) (\(R\) and \(I\) commute). ✓
  • \(= (R + 2I) + 4R + 4I = 5R + 6I\). ✓
  • \(5R + 6I = \begin{pmatrix}15 + 6&-5\\20&-10 + 6\end{pmatrix} = \begin{pmatrix}21&-5\\20&-4\end{pmatrix}\). ✓
  • (c) Multiply \(R^{2} - R - 2I = \mathbf{0}\) by \(R^{-1}\): \(R - I - 2R^{-1} = \mathbf{0}\), so \(R^{-1} = \tfrac{1}{2}(R - I)\). ✓
  • \(R^{-1} = \tfrac{1}{2}\begin{pmatrix}2&-1\\4&-3\end{pmatrix}\). Check: \(\det R = -6 + 4 = -2\), and the usual formula gives the same matrix. ✓
04
Analysis
[7 marks]

\(S = \begin{pmatrix}1&1&0\\0&2&1\\0&0&3\end{pmatrix}\).

(a) Write down the eigenvalues of \(S\) and find a corresponding eigenvector for each.

(b) Hence find \(S^{n}\) in terms of \(n\).

(c) Find \(S^{5}\).

  • (a) \(S\) is upper triangular, so the eigenvalues are 1, 2, 3. ✓
  • \(\lambda = 1\): \(y = 0\), \(z = 0\), giving \(\begin{pmatrix}1\\0\\0\end{pmatrix}\). \(\lambda = 2\): \(z = 0\), \(y = x\), giving \(\begin{pmatrix}1\\1\\0\end{pmatrix}\). \(\lambda = 3\): \(y = z\), \(y = 2x\), giving \(\begin{pmatrix}1\\2\\2\end{pmatrix}\). ✓
  • (b) \(Q = \begin{pmatrix}1&1&1\\0&1&2\\0&0&2\end{pmatrix}\), \(D = \operatorname{diag}(1, 2, 3)\), \(Q^{-1} = \begin{pmatrix}1&-1&\tfrac{1}{2}\\0&1&-1\\0&0&\tfrac{1}{2}\end{pmatrix}\). ✓
  • \[QD^{n} = \begin{pmatrix}1&2^{n}&3^{n}\\0&2^{n}&2 \cdot 3^{n}\\0&0&2 \cdot 3^{n}\end{pmatrix}\] ✓
  • Multiply by \(Q^{-1}\): \[S^{n} = \begin{pmatrix}1&2^{n} - 1&\tfrac{1}{2}(3^{n} + 1) - 2^{n}\\0&2^{n}&3^{n} - 2^{n}\\0&0&3^{n}\end{pmatrix}\] ✓
  • Check \(n = 1\): top-right \(\tfrac{1}{2}(4) - 2 = 0\), and the rest give \(S\). ✓
  • (c) \(n = 5\): \(2^{5} - 1 = 31\), \(\tfrac{1}{2}(244) - 32 = 90\), \(243 - 32 = 211\), so \(S^{5} = \begin{pmatrix}1&31&90\\0&32&211\\0&0&243\end{pmatrix}\). ✓
05
Analysis
[6 marks]

\(T = \begin{pmatrix}1&1&0\\1&0&1\\0&1&1\end{pmatrix}\).

(a) Show that the characteristic equation of \(T\) is \(\lambda^{3} - 2\lambda^{2} - \lambda + 2 = 0\), and find the eigenvalues.

(b) Use the Cayley-Hamilton theorem to find \(T^{-1}\).

  • (a) Expand along row 1: \(\det(T - \lambda I) = (1 - \lambda)\big[-\lambda(1 - \lambda) - 1\big] - 1\big[(1 - \lambda) - 0\big]\). ✓
  • \(= (1 - \lambda)(\lambda^{2} - \lambda - 1) - (1 - \lambda) = (1 - \lambda)(\lambda^{2} - \lambda - 2)\). ✓
  • \(= -(\lambda - 1)(\lambda - 2)(\lambda + 1) = -(\lambda^{3} - 2\lambda^{2} - \lambda + 2)\), so the equation is \(\lambda^{3} - 2\lambda^{2} - \lambda + 2 = 0\), with eigenvalues \(1, 2, -1\). ✓
  • (b) \(T^{3} - 2T^{2} - T + 2I = \mathbf{0}\). Multiply by \(T^{-1}\): \(T^{2} - 2T - I + 2T^{-1} = \mathbf{0}\), so \(T^{-1} = \tfrac{1}{2}(I + 2T - T^{2})\). ✓
  • \(T^{2} = \begin{pmatrix}2&1&1\\1&2&1\\1&1&2\end{pmatrix}\), so \(I + 2T - T^{2} = \begin{pmatrix}1&1&-1\\1&-1&1\\-1&1&1\end{pmatrix}\). ✓
  • \(T^{-1} = \tfrac{1}{2}\begin{pmatrix}1&1&-1\\1&-1&1\\-1&1&1\end{pmatrix}\). Check: row 1 of \(T\) times column 1 is \(\tfrac{1}{2}(1 + 1 + 0) = 1\). ✓
06
Analysis
[7 marks]

\(M = \begin{pmatrix}0.8&0.3\\0.2&0.7\end{pmatrix}\).

(a) Show that the eigenvalues of \(M\) are 1 and 0.5, and find a corresponding eigenvector for each.

(b) Find \(M^{n}\) in terms of \(n\).

(c) Find the matrix that \(M^{n}\) approaches as \(n \to \infty\).

  • (a) \(\lambda^{2} - 1.5\lambda + (0.56 - 0.06) = \lambda^{2} - 1.5\lambda + 0.5 = (\lambda - 1)(\lambda - 0.5) = 0\). ✓
  • \(\lambda = 1\): \(-0.2x + 0.3y = 0\), giving \(\begin{pmatrix}3\\2\end{pmatrix}\). \(\lambda = 0.5\): \(0.3x + 0.3y = 0\), giving \(\begin{pmatrix}1\\-1\end{pmatrix}\). ✓
  • (b) \(Q = \begin{pmatrix}3&1\\2&-1\end{pmatrix}\), \(\det Q = -5\), \(Q^{-1} = \tfrac{1}{5}\begin{pmatrix}1&1\\2&-3\end{pmatrix}\), and \(QD^{n} = \begin{pmatrix}3&0.5^{n}\\2&-0.5^{n}\end{pmatrix}\). ✓
  • \[M^{n} = \tfrac{1}{5}\begin{pmatrix}3 + 2(0.5)^{n} & 3 - 3(0.5)^{n}\\ 2 - 2(0.5)^{n} & 2 + 3(0.5)^{n}\end{pmatrix}\] ✓
  • Check \(n = 1\): \(\tfrac{1}{5}\begin{pmatrix}4&1.5\\1&3.5\end{pmatrix} = M\). ✓
  • (c) \(0.5^{n} \to 0\), so \(M^{n} \to \tfrac{1}{5}\begin{pmatrix}3&3\\2&2\end{pmatrix} = \begin{pmatrix}0.6&0.6\\0.4&0.4\end{pmatrix}\). Each column is the \(\lambda = 1\) eigenvector scaled so its entries add to 1. ✓

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.