250+ TOP MCQs on Cayley Hamilton Theorem and Answers

Linear Algebra Multiple Choice Questions 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&5end{bmatrix})
a) A-1=(frac{1}{16} begin{bmatrix}2&-3&-1\4&-2&-6\-6&9&11end{bmatrix})
b) A-1=(frac{1}{8} begin{bmatrix}2&-3&-1\4&-2&-3\-6&9&11end{bmatrix})
c) A-1=(frac{1}{16} begin{bmatrix}2&-1&-1\4&-2&-6\-6&9&11end{bmatrix})
d) A-1=(frac{1}{8} begin{bmatrix}2&-3&-1\4&-2&-6\-6&9&11end{bmatrix})
Answer: d
Explanation: For the given Matrix,
A=(begin{bmatrix}1&2&3\2&3&4\3&4&5end{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&7end{bmatrix}-7begin{bmatrix}4&3&2\-1&2&1\3&0&1end{bmatrix}+11begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})

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

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

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

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

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

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

3. Find the value of A3+19A, A=(begin{bmatrix}2&-3&1\2&0&-1\1&4&5end{bmatrix}).
a) (begin{bmatrix}42&-14&70\21&+21&-21\105&119&203end{bmatrix})
b) (begin{bmatrix}42&-7&70\21&-21&-21\105&119&203end{bmatrix})
c) (begin{bmatrix}42&-14&70\21&-21&-21\105&119&203end{bmatrix})
d) (begin{bmatrix}42&-7&70\21&+21&-21\105&119&203end{bmatrix})
Answer: c
Explanation: Explanation: For the given Matrix,
A=(begin{bmatrix}2&-3&1\2&0&-1\1&4&5end{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&22end{bmatrix}+49begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})
A3+19A=(begin{bmatrix}-7+49&-17&70\21&-70+49&-21\105&119&154+49end{bmatrix})
A3+19A=(begin{bmatrix}42&-14&70\21&-21&-21\105&119&203end{bmatrix}).

4. Find the value of 2A3+4A2, where = (begin{bmatrix}5&0&-1\1&2&-1\-3&4&1end{bmatrix}).
a) (begin{bmatrix}-200&0&-24\24&-32&-24\-72&96&-56end{bmatrix})
b) (begin{bmatrix}-200&0&-24\24&-32&-12\-72&96&-56end{bmatrix})
c) (begin{bmatrix}-200&0&-24\12&-32&-24\-72&96&-56end{bmatrix})
d) (begin{bmatrix}-100&0&-12\12&-16&-12\-36&48&-28end{bmatrix})
Answer: a
Explanation: Explanation: For the given Matrix,
A=(begin{bmatrix}5&0&-1\1&2&-1\-3&4&1end{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=(12begin{bmatrix}5&0&-1\1&2&-1\-3&4&1end{bmatrix}-40begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})

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

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

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

5. Find the value of A3-3A2-28A, A = (begin{bmatrix}-1&2&8\-2&3&0\-4&5&1end{bmatrix}).
a) (begin{bmatrix}80&-126&-504\126&-172&-63\252&-316&-46end{bmatrix})
b) (begin{bmatrix}80&-126&-504\126&-172&-63\252&-315&-46end{bmatrix})
c) (begin{bmatrix}40&-126&-504\126&-172&-63\252&-315&-46end{bmatrix})
d) (begin{bmatrix}40&-126&-504\126&-172&-63\252&-316&-46end{bmatrix})
Answer: b
Explanation: For the given Matrix,
A=(begin{bmatrix}-1&2&8\-2&3&0\-4&5&1end{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) = (-63begin{bmatrix}-1&2&8\-2&3&0\-4&5&1end{bmatrix}+17begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})
(A3-3A2-28A)=(begin{bmatrix}63+17&-126&-504\126&17-189&-63\252&-315&17-63end{bmatrix})
(A3-3A2-28A)=(begin{bmatrix}80&-126&-504\126&-172&-63\252&-315&-46end{bmatrix}).

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