Engineering Mathematics II • Digital Workbook • Page 24

Cayley-Hamilton Theorem

Characteristic Equation, Verification and Matrix Inverse

The Cayley-Hamilton theorem is used to obtain a matrix inverse. Form the characteristic equation by the S1, S2, S3 method, replace the scalar by the matrix and simplify.

UnitMatrices
TopicCayley-Hamilton Theorem
Question Type4 Mark Problems

Theory Required for This Problem

Cayley-Hamilton Theorem

Every square matrix satisfies its own characteristic equation. If

\[\lambda^3-S_1\lambda^2+S_2\lambda-S_3=0,\]
\[\lambda^3-S_1\lambda^2\]\[{}+S_2\lambda-S_3=0,\]

then

\[A^3-S_1A^2+S_2A-S_3I=0.\]
\[A^3-S_1A^2\]\[{}+S_2A-S_3I=0.\]

The S1, S2, S3 Method

\[S_1=\mathrm{trace}(A).\]

\[S_2=\mathrm{sum\ of\ principal\ minors}.\]

\[S_3=|A|.\]

\[\lambda^3-S_1\lambda^2+S_2\lambda-S_3=0.\]
\[\lambda^3-S_1\lambda^2\]\[{}+S_2\lambda-S_3=0.\]

Calculator check: Use matrix mode to verify \(S_3=|A|\) and the eigenvalues.

Finding the Inverse

When \(S_3\ne0\), multiply the matrix equation by \(A^{-1}\). Then isolate \(A^{-1}\).

Calculator check: Use matrix mode to verify the determinant and the final inverse.

Verification

After finding the inverse, check

\[AA^{-1}=A^{-1}A=I.\]

Method used: Find \(S_1,S_2,S_3\), write the characteristic equation, replace \(\lambda\) by \(A\), multiply by \(A^{-1}\), and simplify.
Continue through the IOE Engineering Mathematics II Digital Workbook. This cayley-hamilton theorem page follows the Engineering Mathematics II course sequence in the Tribhuvan University curriculum.
55

Inverse Using the Cayley-Hamilton Theorem

4 Marks
PAST-PAPER REFERENCE: TU IOE • 2082 Bhadra • Regular (New Course) • ENSH 151 • Q. 13 • 4 Marks

Use the Cayley-Hamilton theorem to find the inverse of

\[A=\left[\begin{array}{ccc}1&0&1\\1&1&0\\1&0&2\end{array}\right].\]

Solution

The S1, S2, S3 method

Given

\[A=\left[\begin{array}{ccc}1&0&1\\1&1&0\\1&0&2\end{array}\right].\]

First,

\[S_1=\mathrm{trace}(A)=1+1+2=4.\]

The principal minors give

\[S_2=1+1+2=4.\]

Also,

\[S_3=|A|=1.\]

Therefore, the characteristic equation is

\[\lambda^3-4\lambda^2+4\lambda-1=0.\]

By the Cayley-Hamilton theorem,

\[A^3-4A^2+4A-I=0.\]

Multiply by \(A^{-1}\):

\[A^2-4A+4I-A^{-1}=0.\]

Hence,

\[A^{-1}=A^2-4A+4I.\]

Now,

\[A^2=\left[\begin{array}{ccc}2&0&3\\2&1&1\\3&0&5\end{array}\right].\]

Substituting,

\[A^{-1}=\left[\begin{array}{ccc}2&0&3\\2&1&1\\3&0&5\end{array}\right]-4\left[\begin{array}{ccc}1&0&1\\1&1&0\\1&0&2\end{array}\right]+4I.\]

Therefore,

\[A^{-1}=\left[\begin{array}{ccc}2&0&-1\\-2&1&1\\-1&0&1\end{array}\right].\]

Verification:

\[AA^{-1}=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right].\]
\[A^{-1}=\left[\begin{array}{ccc}2&0&-1\\-2&1&1\\-1&0&1\end{array}\right]\]

A Few Important Questions

Verification and Inverse

Verify the Cayley-Hamilton theorem and find the inverse of

\[A=\left[\begin{array}{ccc}1&0&3\\2&1&-1\\1&-1&1\end{array}\right].\]
Show Solution

Characteristic equation:

\[\lambda^3-3\lambda^2-\lambda+9=0.\]
\[A^{-1}=\frac19\left[\begin{array}{ccc}0&3&3\\3&2&-7\\3&-1&-1\end{array}\right].\]

Characteristic Equation by S1, S2, S3

Find the characteristic equation of

\[\left[\begin{array}{ccc}1&0&2\\0&2&1\\2&0&3\end{array}\right].\]
Show Solution
\[\lambda^3-6\lambda^2+7\lambda+2=0.\]
Formulae

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