Question

If \( I \) is an identity matrix, the polynomial equation that satisfies the Cayley-Hamilton theorem for the matrix \[ P = \begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix} \] is

• A. \( P^2 - 5P + 6I = 0 \)
• B. \( P^2 - 5P + 5I = 0 \)
• C. \( P^2 - 6P + 5I = 0 \)
• D. \( P^2 - 4P + 3I = 0 \)

Hint:

Find the characteristic equation of P from \( \det(P-\lambda I)=0 \), then replace \( \lambda \) with P itself.

Answer 1

The Correct Option is A

Another way to find the equation a matrix satisfies under the Cayley-Hamilton theorem is to compute $P^2$ directly and then test which combination of $P$, $P^2$ and $I$ gives the zero matrix.

The matrix is $P = \begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix}$.

First find $P^2$ by multiplying $P$ with itself:

\[ P^2 = \begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 4 & 5 \\ 0 & 9 \end{bmatrix} \]

Now test option (A), $P^2 - 5P + 6I = 0$, by substituting the matrices:

\[ \begin{bmatrix} 4 & 5 \\ 0 & 9 \end{bmatrix} - 5\begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix} + 6\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]

Work it entry by entry. Top left: $4 - 10 + 6 = 0$. Top right: $5 - 5 + 0 = 0$. Bottom left: $0 - 0 + 0 = 0$. Bottom right: $9 - 15 + 6 = 0$.

Every entry comes out zero, so $P^2 - 5P + 6I = 0$ holds for this matrix. That confirms option (A) directly, without using the trace and determinant shortcut.

1. $P^2-5P+5I=0$: plugging in the same matrices gives a bottom right entry of $9-15+5=-1$, not zero, so this fails.
2. $P^2-6P+5I=0$: the top left entry becomes $4-12+5=-3$, not zero, so this fails too.
3. $P^2-4P+3I=0$: the top left entry becomes $4-8+3=-1$, not zero, so this also fails.

Only option (A) makes every entry of the matrix expression vanish, so $P^2-5P+6I=0$ is the equation this matrix satisfies under the Cayley-Hamilton theorem.