Question

Which one of the following matrices has \(p(x) = x^3 - 8x^2 + 5x + 7\) as the minimal polynomial?

• A. \[\begin{pmatrix}0&0&7\\1&0&5\\0&1&8\end{pmatrix}\]
• B. \[\begin{pmatrix}1&0&8\\0&0&5\\0&1&7\end{pmatrix}\]
• C. \[\begin{pmatrix}0&0&-7\\1&0&-5\\0&1&8\end{pmatrix}\]
• D. \[\begin{pmatrix}0&0&7\\1&0&5\\0&1&-8\end{pmatrix}\]

Hint:

Use the standard companion matrix form for a monic cubic \(x^3+a_2x^2+a_1x+a_0\).

Answer 1

The Correct Option is C

Step 1: For a monic cubic polynomial \(p(x) = x^3 + a_2x^2 + a_1x + a_0\), the standard companion matrix is \[C = \begin{pmatrix}0&0&-a_0\\1&0&-a_1\\0&1&-a_2\end{pmatrix}\] and this companion matrix always has \(p(x)\) as both its characteristic polynomial and its minimal polynomial.

Step 2: Here \(p(x) = x^3 - 8x^2 + 5x + 7\), so comparing with \(x^3+a_2x^2+a_1x+a_0\) gives \(a_2 = -8\), \(a_1 = 5\), \(a_0 = 7\). Hence \(-a_0 = -7\), \(-a_1 = -5\), \(-a_2 = 8\).

Step 3: Substituting these values into the companion matrix form gives \[C = \begin{pmatrix}0&0&-7\\1&0&-5\\0&1&8\end{pmatrix}\] which is exactly option (C).

Step 4: As a check, compute \(xI - C = \begin{pmatrix}x&0&7\\-1&x&5\\0&-1&x-8\end{pmatrix}\). Expanding along the first row, \[\det(xI-C) = x\big(x(x-8)-5(-1)\big) + 7\big((-1)(-1) - x\cdot 0\big) = x(x^2-8x+5) + 7 = x^3 - 8x^2 + 5x + 7 = p(x).\] Since this companion matrix is non-derogatory, its minimal polynomial equals its characteristic polynomial \(p(x)\), confirming option (C).

\[\boxed{\begin{pmatrix}0&0&-7\\1&0&-5\\0&1&8\end{pmatrix}}\]