Linear Algebra Questions and Answers – Cayley Hamilton Theorem

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This set of Linear Algebra Multiple Choice Questions & Answers (MCQs) focuses on “Cayley Hamilton Theorem”.

1. Find the inverse of the given Matrix, using Cayley Hamilton’s Theorem.
A=\(\begin{bmatrix}1&2&3\\2&3&4\\3&4&5\end{bmatrix}\)
a) A-1=\(\frac{1}{16} \begin{bmatrix}2&-3&-1\\4&-2&-6\\-6&9&11\end{bmatrix}\)
b) A-1=\(\frac{1}{8} \begin{bmatrix}2&-3&-1\\4&-2&-3\\-6&9&11\end{bmatrix}\)
c) A-1=\(\frac{1}{16} \begin{bmatrix}2&-1&-1\\4&-2&-6\\-6&9&11\end{bmatrix}\)
d) A-1=\(\frac{1}{8} \begin{bmatrix}2&-3&-1\\4&-2&-6\\-6&9&11\end{bmatrix}\)
View Answer

Answer: d
Explanation: For the given Matrix,
A=\(\begin{bmatrix}1&2&3\\2&3&4\\3&4&5\end{bmatrix}\)
The characteristic polynomial is given by –
α3-(Sum of diagonal elements) α2+(Sum of minor of diagonal element)α-|A|=0
α3-7α2+11α-8=0
The Cayley Hamilton’s Theorem states that every matrix satisfies its Characteristic Polynomial.
Thus,
A3-7A2+11A-8I=0
To find A-1, multiply both the sides of the equation by A-1
A2 A A-1-7AA A-1+11A A-1-8I A-1=0
We know that A A-1=I
A2I-7AI+11I-8IA-1=0
A2-7A+11-8 A-1=0
A2-7A+11=8 A-1

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8A-1=\(\begin{bmatrix}19&18&13\\-3&1&1\\15&9&7\end{bmatrix}-7\begin{bmatrix}4&3&2\\-1&2&1\\3&0&1\end{bmatrix}+11\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)

8A-1=\(\begin{bmatrix}19-28+11&18-21&13-14\\-3+7&1-14+11&1-7\\15-21&9&7-7+11\end{bmatrix}\)

8A-1=\(\begin{bmatrix}2&-3&-1\\4&-2&-6\\-6&9&11\end{bmatrix}\)

A-1=\(\frac{1}{8} \begin{bmatrix}2&-3&-1\\4&-2&-6\\-6&9&11\end{bmatrix}\).

2. Find the value of A3 where A=\(\begin{bmatrix}-1&-1&2\\0&1&-1\\2&2&1\end{bmatrix}\).
a) \(\begin{bmatrix}3&5&-1\\-2&-9&2\\-2&-4&-5\end{bmatrix}\)
b) \(\begin{bmatrix}3&5&-1\\1&-9&1\\-2&-4&-5\end{bmatrix}\)
c) \(\begin{bmatrix}3&5&-1\\-2&-9&1\\-2&-4&-5\end{bmatrix}\)
d) \(\begin{bmatrix}3&5&-1\\-1&-9&1\\-2&-4&-5\end{bmatrix}\)
View Answer

Answer: c
Explanation: For the given Matrix,
A=\(\begin{bmatrix}-1&-1&2\\0&1&-1\\2&2&1\end{bmatrix}\)
The characteristic polynomial is given by-
α3-(Sum of diagonal elements) α2+(Sum of minor of diagonal element)α-|A|=0

α32+3α+5=0
The Cayley Hamilton’s Theorem states that every matrix satisfies its Characteristic Polynomial.

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Thus,
A3-A2+3A+5I=0
A3=A2-3A-5I
A3=\(\begin{bmatrix}5&2&5\\-2&-1&-2\\4&2&3\end{bmatrix}-3\begin{bmatrix}-1&-1&2\\0&1&-1\\2&2&1\end{bmatrix}-5\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)

A3=\(\begin{bmatrix}5+3-5&2+3&5-6\\-2+0&-1-3-5&-2+3\\4-6&2-6&3-3-5\end{bmatrix}\)

A3=\(\begin{bmatrix}3&5&-1\\-2&-9&1\\-2&-4&-5\end{bmatrix}\).

3. Find the value of A3+19A, A=\(\begin{bmatrix}2&-3&1\\2&0&-1\\1&4&5\end{bmatrix}\).
a) \(\begin{bmatrix}42&-14&70\\21&+21&-21\\105&119&203\end{bmatrix}\)
b) \(\begin{bmatrix}42&-7&70\\21&-21&-21\\105&119&203\end{bmatrix}\)
c) \(\begin{bmatrix}42&-14&70\\21&-21&-21\\105&119&203\end{bmatrix}\)
d) \(\begin{bmatrix}42&-7&70\\21&+21&-21\\105&119&203\end{bmatrix}\)
View Answer

Answer: c
Explanation: Explanation: For the given Matrix,
A=\(\begin{bmatrix}2&-3&1\\2&0&-1\\1&4&5\end{bmatrix}\)
The characteristic polynomial is given by –
α3-(Sum of diagonal elements) α2+(Sum of minor of diagonal element)α-|A|=0

α3-7α2+19α-49=0
The Cayley Hamilton’s Theorem states that every matrix satisfies its Characteristic Polynomial.

Thus,
A3-7A2+19A-49I=0
A3+19A=7A2+49I
A3+19A=7\(\begin{bmatrix}-1&-2&10\\3&-10&-3\\15&17&22\end{bmatrix}+49\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)
A3+19A=\(\begin{bmatrix}-7+49&-17&70\\21&-70+49&-21\\105&119&154+49\end{bmatrix}\)
A3+19A=\(\begin{bmatrix}42&-14&70\\21&-21&-21\\105&119&203\end{bmatrix}\).

4. Find the value of 2A3+4A2, where = \(\begin{bmatrix}5&0&-1\\1&2&-1\\-3&4&1\end{bmatrix}\).
a) \(\begin{bmatrix}-200&0&-24\\24&-32&-24\\-72&96&-56\end{bmatrix}\)
b) \(\begin{bmatrix}-200&0&-24\\24&-32&-12\\-72&96&-56\end{bmatrix}\)
c) \(\begin{bmatrix}-200&0&-24\\12&-32&-24\\-72&96&-56\end{bmatrix}\)
d) \(\begin{bmatrix}-100&0&-12\\12&-16&-12\\-36&48&-28\end{bmatrix}\)
View Answer

Answer: a
Explanation: Explanation: For the given Matrix,
A=\(\begin{bmatrix}5&0&-1\\1&2&-1\\-3&4&1\end{bmatrix}\)

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The characteristic polynomial is given by-
α3-(Sum of diagonal elements) α2+(Sum of minor of diagonal element)α-|A|=0

α3+2α2-12α-40=0
The Cayley Hamilton’s Theorem states that every matrix satisfies its Characteristic Polynomial.

Thus,
A3+2A2-12A+40I=0
A3+2A2=12A-40I
A3+2A2=\(12\begin{bmatrix}5&0&-1\\1&2&-1\\-3&4&1\end{bmatrix}-40\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)

A3+2A2=\(\begin{bmatrix}-60-40&0&-12\\12&24-40&-12\\-36&48&12-40\end{bmatrix}\)

A3+2A2=\(\begin{bmatrix}-100&0&-12\\12&-16&-12\\-36&48&-28\end{bmatrix}\)

2A3+4A2=\(\begin{bmatrix}-200&0&-24\\24&-32&-24\\-72&96&-56\end{bmatrix}\).

5. Find the value of A3-3A2-28A, A = \(\begin{bmatrix}-1&2&8\\-2&3&0\\-4&5&1\end{bmatrix}\).
a) \(\begin{bmatrix}80&-126&-504\\126&-172&-63\\252&-316&-46\end{bmatrix}\)
b) \(\begin{bmatrix}80&-126&-504\\126&-172&-63\\252&-315&-46\end{bmatrix}\)
c) \(\begin{bmatrix}40&-126&-504\\126&-172&-63\\252&-315&-46\end{bmatrix}\)
d) \(\begin{bmatrix}40&-126&-504\\126&-172&-63\\252&-316&-46\end{bmatrix}\)
View Answer

Answer: b
Explanation: For the given Matrix,
A=\(\begin{bmatrix}-1&2&8\\-2&3&0\\-4&5&1\end{bmatrix}\)

The characteristic polynomial is given by-
α3-(Sum of diagonal elements) α2+(Sum of minor of diagonal element)α-|A|=0

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α3-3α2+35α-17=0
The Cayley Hamilton’s Theorem states that every matrix satisfies its Characteristic Polynomial.

Thus,
A3-3A2+35A-17I=0
On performing long division (α3-3α2+35α-17)/(α2-7α)
Q=α+4 and R=63α-17
Using division properties,
α3-3α2+35α-17=(α2-7α)×(α+4)+(63α-17)
α3-3α2+35α-17=(α3-3α2-28α)+( 63α-17)
0=(α3-3α2-28α)+(63α-17) ————— (From Characteristic Polynomial)
3-3α2-28α) = -63α+17
(A3-3A2-28A) = -63A+17I
(A3-3A2-28A) = \(-63\begin{bmatrix}-1&2&8\\-2&3&0\\-4&5&1\end{bmatrix}+17\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)
(A3-3A2-28A)=\(\begin{bmatrix}63+17&-126&-504\\126&17-189&-63\\252&-315&17-63\end{bmatrix}\)
(A3-3A2-28A)=\(\begin{bmatrix}80&-126&-504\\126&-172&-63\\252&-315&-46\end{bmatrix}\).

Sanfoundry Global Education & Learning Series – Linear Algebra.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn