Question

Find the value of A^3+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) \(\begin{bmatrix}42&-7&70\\21&+21&-21\\105&119&203\end{bmatrix}\)

I had been asked this question in a job interview.

My question comes from Cayley Hamilton Theorem in division Eigen Values and Eigen Vectors of Engineering Mathematics

Answer 1

Right answer is (c) \(\begin{bmatrix}42&-14&70\\21&-21&-21\\105&119&203\end{bmatrix}\)

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 –

α^3-(Sum of diagonal elements) α^2+(Sum of minor of diagonal element)α-|A|=0

α^3-7α^2+19α-49=0

The Cayley Hamilton’s Theorem states that every matrix satisfies its Characteristic Polynomial.

Thus,

A^3-7A^2+19A-49I=0

A^3+19A=7A^2+49I

A^3+19A=7\(\begin{bmatrix}-1&-2&10\\3&-10&-3\\15&17&22\end{bmatrix}+49\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)

A^3+19A=\(\begin{bmatrix}-7+49&-17&70\\21&-70+49&-21\\105&119&154+49\end{bmatrix}\)

A^3+19A=\(\begin{bmatrix}42&-14&70\\21&-21&-21\\105&119&203\end{bmatrix}\).