Cayley–Hamilton Theorem, Matrix Inverse, and a Linear System: 04-BS-5, May 2013, Q7

Question 7: Cayley–Hamilton Theorem, Matrix Inverse, and a Linear System (a) 6, (b) 6, (c) 8 marks

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $A=\begin{pmatrix}5&-1&-2\\-1&3&4\\-2&4&6\end{pmatrix}$, $U=I_3$, $O$ = the $3\times3$ zero matrix; a linear system with coefficient matrix $A$ and right-hand side $b=(9,-6,-11)^T$.

Find. (a) Verify $A^3-14A^2+42A-8U=O$. (b) $A^{-1}$. (c) $x_1,x_2,x_3$.

Approach. (a) Compute $A$'s characteristic polynomial (trace, sum of principal $2\times2$ minors, determinant) and confirm it matches the stated identity — this IS the Cayley–Hamilton theorem for $A$. (b) Rearrange the identity to isolate $A^{-1}$. (c) Multiply $A^{-1}b$.

(a) Compute the characteristic polynomial's coefficients. $$\operatorname{tr}(A)=5+3+6=14$$ $$M_{11}+M_{22}+M_{33}=(5\cdot3-(-1)(-1))+(5\cdot6-(-2)(-2))+(3\cdot6-4\cdot4)=(15-1)+(30-4)+(18-16)=14+26+2=42$$ $$\det(A)=5(3\cdot6-4\cdot4)-(-1)\big((-1)(6)-4(-2)\big)+(-2)\big((-1)(4)-3(-2)\big)=5(2)-(-1)(2)+(-2)(2)=10+2-4=8$$ so the characteristic equation is $\lambda^3-14\lambda^2+42\lambda-8=0$.
(a) Invoke Cayley–Hamilton and verify directly. By the Cayley–Hamilton theorem, every square matrix satisfies its own characteristic equation, so $A^3-14A^2+42A-8U=O$ automatically. Direct computation confirms it: $$A^2=\begin{pmatrix}30&-16&-26\\-16&26&38\\-26&38&56\end{pmatrix},\qquad A^3=\begin{pmatrix}218&-182&-280\\-182&246&364\\-280&364&540\end{pmatrix}$$ $$\boxed{A^3-14A^2+42A-8U=\begin{pmatrix}0&0&0\\0&0&0\\0&0&0\end{pmatrix}=O}\qquad\blacksquare$$ (e.g. entry $(1,1)$: $218-14(30)+42(5)-8=218-420+210-8=0$ ✓.)
(b) Solve the identity for $A^{-1}$. $$A^3-14A^2+42A=8U\ \Longrightarrow\ A(A^2-14A+42U)=8U\ \Longrightarrow\ A^{-1}=\tfrac18(A^2-14A+42U)$$ (valid since $\det A=8\ne0$). Computing $A^2-14A+42U$ entrywise, $$A^2-14A+42U=\begin{pmatrix}2&-2&2\\-2&26&-18\\2&-18&14\end{pmatrix}$$ $$\boxed{A^{-1}=\dfrac18\begin{pmatrix}2&-2&2\\-2&26&-18\\2&-18&14\end{pmatrix}=\begin{pmatrix}\tfrac14&-\tfrac14&\tfrac14\\[2pt]-\tfrac14&\tfrac{13}4&-\tfrac94\\[2pt]\tfrac14&-\tfrac94&\tfrac74\end{pmatrix}}$$
(c) Solve $Ax=b$ via $x=A^{-1}b$, with $b=(9,-6,-11)^T$. $$x_1=\tfrac14(9)-\tfrac14(-6)+\tfrac14(-11)=\tfrac14(9+6-11)=\boxed{1}$$ $$x_2=-\tfrac14(9)+\tfrac{13}4(-6)-\tfrac94(-11)=\tfrac14(-9-78+99)=\boxed{3}$$ $$x_3=\tfrac14(9)-\tfrac94(-6)+\tfrac74(-11)=\tfrac14(9+54-77)=\boxed{-\tfrac72}$$ (Check: substituting $(1,3,-3.5)$ back into all three original equations reproduces $9,-6,-11$ exactly.)

Quantity	Result
Characteristic polynomial	$\lambda^3-14\lambda^2+42\lambda-8=0$
$A^{-1}$	$\tfrac18\begin{pmatrix}2&-2&2\\-2&26&-18\\2&-18&14\end{pmatrix}$
$x_1,x_2,x_3$	$1,\ 3,\ -\tfrac72$

Concept & Theory

Topic. The Cayley–Hamilton theorem and matrix-inverse-via-characteristic-polynomial.

Key relations. Every square matrix satisfies its own characteristic equation; for $3\times3$ $A$, $\lambda^3-\operatorname{tr}(A)\lambda^2+(\text{sum of principal }2\times2\text{ minors})\lambda-\det(A)=0$, giving $A^{-1}=\big[A^2-\operatorname{tr}(A)A+(\text{minors sum})I\big]/\det(A)$ whenever $\det A\ne0$.

Why this works. Cayley–Hamilton expresses any higher power of $A$ (and, rearranged, $A^{-1}$) as a LINEAR combination of $I,A,A^2$ — turning matrix inversion into ordinary matrix arithmetic once the characteristic polynomial is known, with no Gaussian elimination required.

Common pitfall. Sign errors in the alternating characteristic-polynomial coefficients, or forgetting the rearrangement to $A^{-1}$ is only valid when the constant term ($\det A$) is nonzero — a singular matrix has no such formula.

Source. Strang, Introduction to Linear Algebra, Ch. 6 (Eigenvalues, Cayley–Hamilton); Kreyszig, Advanced Engineering Mathematics, Ch. 8 (Linear Systems, Matrix Inversion).

Question 7 of 7: Cayley–Hamilton Theorem, Matrix Inverse, and a Linear System

Question 7: Cayley–Hamilton Theorem, Matrix Inverse, and a Linear System (a) 6, (b) 6, (c) 8 marks