Linear Algebra Questions and Answers – Cayley Hamilton Theorem

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

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

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

α³-α²+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=\(\begin{bmatrix}3&5&-1\\-2&-9&1\\-2&-4&-5\end{bmatrix}\).

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=\(\begin{bmatrix}42&-14&70\\21&-21&-21\\105&119&203\end{bmatrix}\).

Answer: a
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=\(\begin{bmatrix}-200&0&-24\\24&-32&-24\\-72&96&-56\end{bmatrix}\).

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

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)
(α³-3α²-28α) = -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)=\(\begin{bmatrix}80&-126&-504\\126&-172&-63\\252&-315&-46\end{bmatrix}\).