Question

Let \( M = \begin{bmatrix} \tfrac{1}{4} & \tfrac{3}{4} \\ \\ \tfrac{3}{5} & \tfrac{2}{5} \end{bmatrix}. \) If \( I \) is the \( 2 \times 2 \) identity matrix and \( 0 \) is the \( 2 \times 2 \) zero matrix, then

• A. \( 20 M^2 - 13 M + 7 I = 0 \)
• B. \( 20 M^2 - 13 M - 7 I = 0 \)
• C. \( 20 M^2 + 13 M + 7 I = 0 \)
• D. \( 20 M^2 + 13 M - 7 I = 0 \)

Hint:

When solving matrix equations, always ensure to perform matrix multiplications and additions/subtractions step by step. Verify the result to avoid common calculation errors.

Answer 1

The Correct Option is B

Step 1: Find \(\operatorname{tr}(M)\) and \(\det(M)\)

\[\operatorname{tr}(M)=\frac14+\frac25=\frac{13}{20}, \qquad \det(M)=\frac14\cdot\frac25-\frac34\cdot\frac35 =\frac{2}{20}-\frac{9}{20}=-\frac{7}{20}.\]

Step 2: Use Cayley–Hamilton theorem

The characteristic polynomial of (M) is
\[\lambda^2-(\operatorname{tr}M)\lambda+\det(M)=0,\]
that is,
\[\lambda^2-\frac{13}{20}\lambda-\frac{7}{20}=0.\]

By the Cayley–Hamilton theorem,
\[M^2-\frac{13}{20}M-\frac{7}{20}I=0.\]

Multiplying throughout by \(20\),
\[20M^2-13M-7I=0.\]

\[\boxed{20M^2-13M-7I=0}\]

Thus, the correct answer is option (B)