# 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=$$\frac{1}{8} \begin{bmatrix}2&-3&-1\\4&-2&-6\\-6&9&11\end{bmatrix}$$

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

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) $$\begin{bmatrix}3&5&-1\\-1&-9&1\\-2&-4&-5\end{bmatrix}$$

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.

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) $$\begin{bmatrix}42&-7&70\\21&+21&-21\\105&119&203\end{bmatrix}$$

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) $$\begin{bmatrix}-100&0&-12\\12&-16&-12\\-36&48&-28\end{bmatrix}$$

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

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) $$\begin{bmatrix}40&-126&-504\\126&-172&-63\\252&-316&-46\end{bmatrix}$$

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

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)=$$\begin{bmatrix}80&-126&-504\\126&-172&-63\\252&-315&-46\end{bmatrix}$$.

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