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

Step 1: Recall the Cayley-Hamilton theorem.
The Cayley-Hamilton theorem says every square matrix satisfies its own characteristic equation.
For a 2x2 matrix \( P \), we get the characteristic equation from \( \det(P - \lambda I) = 0 \), then replace \( \lambda \) with \( P \) itself.

Step 2: Write the characteristic equation.
The given matrix is \( P = \begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix} \).
\[ P - \lambda I = \begin{bmatrix} 2-\lambda & 1 \\ 0 & 3-\lambda \end{bmatrix} \]
This matrix is upper triangular, so its determinant is just the product of the diagonal terms minus the product of the off diagonal terms.
\[ \det(P - \lambda I) = (2-\lambda)(3-\lambda) - (1)(0) \]

Step 3: Expand the determinant.
\[ (2-\lambda)(3-\lambda) = 6 - 2\lambda - 3\lambda + \lambda^2 = \lambda^2 - 5\lambda + 6 \]
So the characteristic equation is \( \lambda^2 - 5\lambda + 6 = 0 \).
We can check this a faster way too: for a 2x2 matrix the characteristic equation is always \( \lambda^2 - (\text{trace})\lambda + \det = 0 \). Here trace of P is \( 2+3=5 \) and determinant of P is \( 2\times3 - 1\times0 = 6 \), which gives the same \( \lambda^2 - 5\lambda + 6 = 0 \).

Step 4: Apply the Cayley-Hamilton theorem.
The theorem says P satisfies its own characteristic equation, so we replace \( \lambda \) with \( P \) and the constant term with \( 6I \).
\[ P^2 - 5P + 6I = 0 \]

Step 5: Check the wrong options.
Option B changes the constant to 5I, but the determinant of P is 6, not 5, so this is wrong.
Option C changes the middle term to 6P, but the trace of P is 5, not 6, so this is wrong.
Option D, \( P^2-4P+3I=0 \), belongs to a matrix with trace 4 and determinant 3, not this matrix.

Final Answer:
The matrix P satisfies \( P^2 - 5P + 6I = 0 \).
\[ \boxed{P^2 - 5P + 6I = 0} \]