Matrix Inverse by the Cayley–Hamilton Theorem: 04-BS-5, Undated paper, Q7

Question 7: Matrix Inverse by the Cayley–Hamilton Theorem (20 marks: (a) 10, (b) 10)

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 symmetric $3\times3$ matrix $A$, its characteristic polynomial, the identity $U$ and zero matrix $O$, the Cayley–Hamilton rearrangement (3), and a three-equation linear system whose coefficient matrix is $A$.

Given data
Quantity	Value
Matrix $A$ (rows)	$(13,\,3,\,5)$; $(3,\,7,\,2)$; $(5,\,2,\,6)$
Characteristic equation	$\lambda^{3}-26\lambda^{2}+173\lambda-325=0$
Invariants implied	$\operatorname{tr}A = 26$; sum of principal $2\times2$ minors $=173$; $\det A = 325$
Working relation (3)	$A^{-1} = \tfrac{1}{325}\big(A^{2}-26A+173U\big)$
Right-hand side of the system	$b=(17,\,-7,\,13)^{\mathsf T}$

Find. $A^{-1}$ from relation (3), a verification that $A^{-1}A=U$, and the solution $(x,y,z)$ of the linear system.

Check: recovery of the entries of $A$. The matrix used here is recovered from the printed characteristic equation itself, whose coefficients fix $\operatorname{tr}A=26$, the principal-minor sum $=173$ and $\det A=325$; the legible rows $(3,7,2)$ and $(5,2,6)$ then force the first row to be $(13,3,5)$, and the coefficients of the part-(b) system printed on the same page are exactly those three rows. Every invariant checks: $13+7+6=26$, $38+53+82=173$ and $\det A=325$.

Approach. Form $A^{2}$, substitute into relation (3) to get $A^{-1}$ as one over the determinant times an integer matrix, confirm $A^{-1}A=U$ by direct multiplication, then obtain the solution as the single matrix–vector product $A^{-1}b$.

Read the invariants off the characteristic equation. For a $3\times3$ matrix the characteristic polynomial is $$\lambda^{3} - (\operatorname{tr}A)\lambda^{2} + (M_{11}+M_{22}+M_{33})\lambda - \det A = 0 ,$$ so comparing with the printed equation gives $\operatorname{tr}A=26$, principal-minor sum $=173$ and $\det A=325$. Because $\det A = 325 \ne 0$, $A$ is invertible and relation (3) is legitimate.
Square the matrix. Multiplying $A$ by itself row into column, $$A^{2} = \begin{pmatrix} 13 & 3 & 5 \\ 3 & 7 & 2 \\ 5 & 2 & 6 \end{pmatrix}\begin{pmatrix} 13 & 3 & 5 \\ 3 & 7 & 2 \\ 5 & 2 & 6 \end{pmatrix} = \begin{pmatrix} 203 & 70 & 101 \\ 70 & 62 & 41 \\ 101 & 41 & 65 \end{pmatrix} ,$$ where for instance the leading entry is $13^{2}+3^{2}+5^{2}=203$. The result is symmetric because $A$ is.
Substitute into relation (3). Subtracting $26A$ and adding $173U$ entry by entry, $$A^{2}-26A+173U = \begin{pmatrix} 203-338+173 & 70-78 & 101-130 \\ 70-78 & 62-182+173 & 41-52 \\ 101-130 & 41-52 & 65-156+173 \end{pmatrix} = \begin{pmatrix} 38 & -8 & -29 \\ -8 & 53 & -11 \\ -29 & -11 & 82 \end{pmatrix} ,$$ so that $$\boxed{\;A^{-1} = \frac{1}{325}\begin{pmatrix} 38 & -8 & -29 \\ -8 & 53 & -11 \\ -29 & -11 & 82 \end{pmatrix}\;}$$ The integer matrix is the adjugate of $A$, exactly as the identity $A^{-1}=\operatorname{adj}A/\det A$ predicts.
Check that $A^{-1}A=U$, as the question requires. Taking the first row of the integer matrix against the three columns of $A$, $$38(13)-8(3)-29(5) = 325,\qquad 38(3)-8(7)-29(2) = 0,\qquad 38(5)-8(2)-29(6) = 0 ,$$ and the remaining rows behave the same way, so after division by $325$ the product is $U$. This also re-confirms $\det A = 325$, since the diagonal entries of the unscaled product all come out equal to the determinant.
Write the system in matrix form (part b). The three equations are $Ax=b$ with $x=(x,y,z)^{\mathsf T}$ and $b=(17,-7,13)^{\mathsf T}$; the coefficient matrix is precisely the $A$ of part (a), which is why part (b) says to use the result already obtained. Hence $x = A^{-1}b$.
Multiply out. Component by component, $$x = \frac{38(17)+(-8)(-7)+(-29)(13)}{325} = \frac{646+56-377}{325} = \frac{325}{325} ,$$ and likewise $$y = \frac{-8(17)+53(-7)+(-11)(13)}{325} = \frac{-650}{325},\qquad z = \frac{-29(17)+(-11)(-7)+82(13)}{325} = \frac{650}{325} ,$$ giving $$\boxed{\;x = 1,\qquad y = -2,\qquad z = 2\;}$$
Substitute back into the original equations. $13(1)+3(-2)+5(2) = 17$; $3(1)+7(-2)+2(2) = -7$; $5(1)+2(-2)+6(2) = 13$. All three are satisfied exactly, so the solution is confirmed and, since $\det A \ne 0$, it is the only one.

Final results — Question 7
Quantity	Result
$A^{2}$	rows $(203,\,70,\,101)$; $(70,\,62,\,41)$; $(101,\,41,\,65)$
$A^{2}-26A+173U$	rows $(38,\,-8,\,-29)$; $(-8,\,53,\,-11)$; $(-29,\,-11,\,82)$
$A^{-1}$	$\tfrac{1}{325}$ times the matrix above
Verification	$A^{-1}A = U$ (off-diagonal products vanish; diagonal products all $325$)
Solution of the system	$x=1$, $y=-2$, $z=2$
Uniqueness	guaranteed by $\det A = 325 \ne 0$

Topic. Inverting a matrix through the Cayley–Hamilton theorem, and solving a linear system with the inverse.

Key relations. The characteristic polynomial $\lambda^{3}-(\operatorname{tr}A)\lambda^{2}+(\sum M_{ii})\lambda-\det A$; the Cayley–Hamilton theorem $p(A)=O$; the rearrangement $A^{-1}=\big(A^{2}-(\operatorname{tr}A)A+(\sum M_{ii})U\big)/\det A$; and $Ax=b \Rightarrow x=A^{-1}b$ when $\det A \ne 0$.

Why this works. Cayley–Hamilton says a matrix annihilates its own characteristic polynomial, so $A^{3}-26A^{2}+173A = 325U$. Because the constant term $325 = \det A$ is non-zero, the equation can be divided by it and one factor of $A$ pulled out, which expresses $A^{-1}$ as a polynomial in $A$ — no cofactor expansion, no row reduction, just one matrix multiplication and two scalar multiples. For a $3\times3$ matrix the resulting integer matrix must be the adjugate, which is a useful cross-check. The same identity explains why a singular matrix has no inverse in this framework: the constant term vanishes and there is nothing to divide by. Once $A^{-1}$ is in hand, any number of right-hand sides can be solved for the cost of one matrix–vector product each, which is the reason part (b) is worth as much as part (a) yet takes three lines.

Common pitfall. Mis-stating relation (3) as $\tfrac{1}{325}(A^{2}-26A-173U)$ which destroys the check $A^{-1}A=U$; and forgetting that $U$ here denotes the identity, not an arbitrary matrix, so that $173U$ contributes only to the diagonal.

Source. Strang, Introduction to Linear Algebra, 6th ed., Ch. 6 (eigenvalues, characteristic polynomial, Cayley–Hamilton); Kreyszig, Advanced Engineering Mathematics, 10th ed., §8.1 and §7.8 (inverse of a matrix).

Question 7 of 7: Matrix Inverse by the Cayley–Hamilton Theorem

Question 7: Matrix Inverse by the Cayley–Hamilton Theorem (20 marks: (a) 10, (b) 10)