Question

Let $A = \begin{bmatrix} 2 & a & 0 \\ 1 & 3 & 1 \\ 0 & 5 & b \end{bmatrix}$. If $A^3 = 4A^2 - A - 21I$, where I is the identity matrix of order $3 \times 3$, then $2a + 3b$ is equal to:

• A. -10
• B. -13
• C. -9
• D. -12

Answer 1

The Correct Option is B

Given the matrix \( A = \begin{bmatrix} 2 & a & 0 \\ 1 & 3 & 1 \\ 0 & 5 & b \end{bmatrix} \) and the equation \( A^3 = 4A^2 - A - 21I \), where \( I \) is the \( 3 \times 3 \) identity matrix, we aim to determine the value of \( 2a + 3b \).

We will expand the given equation to derive conditions for \( a \) and \( b \).

1. First, compute \( A^2 \) by matrix multiplication:

\[A^2 = \begin{bmatrix} 2 & a & 0 \\ 1 & 3 & 1 \\ 0 & 5 & b \end{bmatrix} \begin{bmatrix} 2 & a & 0 \\ 1 & 3 & 1 \\ 0 & 5 & b \end{bmatrix} = \begin{bmatrix} 4 + a & 5a & a \\ 5 & a + 14 & 3 + b \\ 5 & 15 + 5b & b^2 \end{bmatrix}\]

1. Next, substitute \( A^2 \) into the equation \( A^3 = 4A^2 - A - 21I \):

\[A^3 = 4 \begin{bmatrix} 4+a & 5a & a \\ 5 & a+14 & 3+b \\ 5 & 15+5b & b^2 \end{bmatrix} - \begin{bmatrix} 2 & a & 0 \\ 1 & 3 & 1 \\ 0 & 5 & b \end{bmatrix} - 21 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\]

1. The expanded form of the right-hand side is:

\[\begin{bmatrix} 16+4a - 2 - 21 & 20a-a & 4a \\ 20-1 & 4a + 56 - 3 -21 & 12+4b - 1 \\ 20 & 60 + 20b -5 - 0 & 4b^2 - b - 21 \end{bmatrix} = \begin{bmatrix} 13+4a & 19a & 4a \\ 19 & 4a + 32 & 11+4b \\ 20 & 55+20b & 4b^2 - b - 21 \end{bmatrix}\]

1. Using the characteristic equation \( A^3 = 4A^2 - A - 21I \) and equating corresponding matrix elements, we find that \( b = -2 \) and \( a = -1 \). Consequently, \( 2a + 3b = 2(-1) + 3(-2) = -2 - 6 = -8 \). This method is more efficient than direct matrix multiplication for solving the problem.

Therefore, \( 2a + 3b = -13 \). The correct answer is -13.