Cayley–Hamilton, Matrix Inversion and a Linear System: 04-BS-5, May 2017, Q7

Question 7: Cayley–Hamilton, Matrix Inversion and a Linear System (20 marks: a 7, b 6, c 7)

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. The matrix $A=\begin{pmatrix}2&-2&-2\\4&-4&-2\\-2&1&-1\end{pmatrix}$, the identity $U$, the zero matrix $O$, and the right-hand side $\mathbf{b}=(3,\ 12,\ -10)^{T}$ for part (c).

Find. (a) a proof that $A^3+3A^2-4U=O$; (b) $A^{-1}$ from equation (3); (c) the solution $\mathbf{x}=A^{-1}\mathbf{b}$.

Approach. Form $A^2$ and $A^3$ by direct multiplication and add; then read $A^{-1}$ straight off equation (3); finally apply that inverse to the right-hand side. The identity in (a) is no coincidence — it is the Cayley–Hamilton theorem for this matrix, which is worth confirming as an independent check.

Part (a) — compute $A^2$. Multiplying $A$ by itself row-into-column, for instance the $(1,1)$ entry is $2(2)+(-2)(4)+(-2)(-2)=4-8+4=0$: $$A^2=\begin{pmatrix}0&2&2\\-4&6&2\\2&-1&3\end{pmatrix}$$
Compute $A^3=A^2A$. Again row-into-column, the $(1,1)$ entry being $0(2)+2(4)+2(-2)=4$: $$A^3=\begin{pmatrix}4&-6&-6\\12&-14&-6\\-6&3&-5\end{pmatrix}$$
Form the combination and conclude. Adding three times $A^2$ to $A^3$, $$A^3+3A^2=\begin{pmatrix}4&-6&-6\\12&-14&-6\\-6&3&-5\end{pmatrix}+\begin{pmatrix}0&6&6\\-12&18&6\\6&-3&9\end{pmatrix}=\begin{pmatrix}4&0&0\\0&4&0\\0&0&4\end{pmatrix}=4U$$ Therefore $$\boxed{A^3+3A^2-4U=O}$$ as required. Independent check: this is exactly the Cayley–Hamilton statement for $A$. Its characteristic polynomial is $\lambda^3-(\operatorname{tr}A)\lambda^2+c_2\lambda-\det A$, and here $\operatorname{tr}A=2-4-1=-3$; the sum of the three principal $2\times2$ minors is $c_2=6+(-6)+0=0$; and $\det A=4$. So the characteristic polynomial is $\lambda^3+3\lambda^2-4$, and substituting $A$ for $\lambda$ reproduces equation (1) term for term.
Part (b) — apply equation (3). Since $\det A=4\ne0$ the inverse exists, and equation (3) requires only the two matrices already computed: $$A^{-1}=\tfrac14\big(A^2+3A\big)=\tfrac14\left[\begin{pmatrix}0&2&2\\-4&6&2\\2&-1&3\end{pmatrix}+\begin{pmatrix}6&-6&-6\\12&-12&-6\\-6&3&-3\end{pmatrix}\right]=\tfrac14\begin{pmatrix}6&-4&-4\\8&-6&-4\\-4&2&0\end{pmatrix}$$ $$\boxed{A^{-1}=\begin{pmatrix}3/2&-1&-1\\2&-3/2&-1\\-1&1/2&0\end{pmatrix}}$$
Verify the inverse. Multiplying out, the first row of $AA^{-1}$ is $$\big[2(\tfrac32)+(-2)(2)+(-2)(-1),\ \ 2(-1)+(-2)(-\tfrac32)+(-2)(\tfrac12),\ \ 2(-1)+(-2)(-1)+(-2)(0)\big]=[1,\,0,\,0]$$ and the remaining rows give $[0,1,0]$ and $[0,0,1]$, so $AA^{-1}=U$ exactly.
Part (c) — solve the system. The three equations are precisely $A\mathbf{x}=\mathbf{b}$ with $\mathbf{b}=(3,12,-10)^T$, since the coefficient matrix reproduces $A$ row for row. Hence $\mathbf{x}=A^{-1}\mathbf{b}$: $$x_1=\tfrac32(3)+(-1)(12)+(-1)(-10)=4.5-12+10=2.5$$ $$x_2=2(3)+(-\tfrac32)(12)+(-1)(-10)=6-18+10=-2$$ $$x_3=(-1)(3)+\tfrac12(12)+0(-10)=-3+6=3$$ $$\boxed{x_1=\tfrac52,\qquad x_2=-2,\qquad x_3=3}$$
Check the solution in the original equations. Substituting back: $$2(2.5)-2(-2)-2(3)=5+4-6=3\ \checkmark$$ $$4(2.5)-4(-2)-2(3)=10+8-6=12\ \checkmark$$ $$-2(2.5)+(-2)-(3)=-5-2-3=-10\ \checkmark$$ All three equations are satisfied exactly, confirming both the inverse and the solution.

Quantity	Result
$A^2$	$\begin{pmatrix}0&2&2\\-4&6&2\\2&-1&3\end{pmatrix}$
$A^3$	$\begin{pmatrix}4&-6&-6\\12&-14&-6\\-6&3&-5\end{pmatrix}$
Part (a)	$A^3+3A^2=4U$, hence $A^3+3A^2-4U=O$
Part (b) $A^{-1}$	$\begin{pmatrix}3/2&-1&-1\\2&-3/2&-1\\-1&1/2&0\end{pmatrix}$
$\det A$, $\operatorname{tr}A$	$4$, $-3$
Part (c) solution	$x_1=2.5,\ x_2=-2,\ x_3=3$

Topic: The Cayley–Hamilton theorem and its use for matrix inversion and linear-system solution.

Key relations: characteristic polynomial $\det(A-\lambda U)=-\big(\lambda^3-(\operatorname{tr}A)\lambda^2+c_2\lambda-\det A\big)$ for $3\times3$; Cayley–Hamilton (every square matrix satisfies its own characteristic equation); $A^{-1}=\frac14(A^2+3A)$ here; $\mathbf{x}=A^{-1}\mathbf{b}$.

Why this works: Cayley–Hamilton says that substituting $A$ into its own characteristic polynomial returns the zero matrix. For this $A$ that polynomial is $\lambda^3+3\lambda^2-4$, whose constant term $-4$ is (up to sign) the determinant. Because that constant is non-zero, the relation can be divided through by $A$ once: multiplying $4U=A^3+3A^2$ by $A^{-1}$ lowers every power by one and expresses $A^{-1}$ as a polynomial in $A$ itself. This is far more than a curiosity — it means an inverse can be obtained with two matrix multiplications and no elimination, no cofactors and no division except by the single scalar $\det A$. Part (c) then reduces to one matrix–vector product, and the same inverse would serve any number of further right-hand sides at negligible extra cost.

Common pitfall: mis-stating the middle coefficient $c_2$ as the sum of the diagonal $2\times2$ minors of only the top-left block rather than the sum of all three principal minors, which produces a characteristic polynomial that $A$ does not satisfy. A second is dropping the factor $\frac14$ when reading $A^{-1}$ off equation (3), and a third is forgetting that the reduction from equation (2) to (3) is legitimate only because $\det A\ne0$ — for a singular matrix the constant term vanishes, the relation cannot be divided by $A$, and no inverse exists.

Source: Kreyszig, Advanced Engineering Mathematics, 10th ed., Ch. 8 §8.1 (Eigenvalues, Eigenvectors) and Ch. 20 §20.1–20.3 (Numeric Linear Algebra); Strang, Introduction to Linear Algebra, 6th ed., Ch. 6.

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

Question 7: Cayley–Hamilton, Matrix Inversion and a Linear System (20 marks: a 7, b 6, c 7)