Question

Suppose \(A=\begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{bmatrix}\) is an adjoint of the matrix \(\begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{bmatrix}\). Then the value of \(\frac{a_!+b_2+c_3}{b_1a_2}\) is

• A. 0
• B. 3
• C. 1
• D. 2
• E. 4

Answer 1

The Correct Option is B

Given: The adjoint of matrix \( M = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \\ \end{bmatrix} \) is \( A = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\ \end{bmatrix} \).

We know the relationship between a matrix and its adjugate: \[ \text{adj}(M) = C^T \] where \( C \) is the cofactor matrix of \( M \).

First compute the cofactor matrix \( C \): \[ C_{11} = +\begin{vmatrix} 4 & 3 \\ 3 & 4 \end{vmatrix} = 7 \] \[ C_{12} = -\begin{vmatrix} 1 & 3 \\ 1 & 4 \end{vmatrix} = -1 \] \[ C_{13} = +\begin{vmatrix} 1 & 4 \\ 1 & 3 \end{vmatrix} = -1 \] \[ C_{21} = -\begin{vmatrix} 3 & 3 \\ 3 & 4 \end{vmatrix} = -3 \] \[ C_{22} = +\begin{vmatrix} 1 & 3 \\ 1 & 4 \end{vmatrix} = 1 \] \[ C_{23} = -\begin{vmatrix} 1 & 3 \\ 1 & 3 \end{vmatrix} = 0 \] \[ C_{31} = +\begin{vmatrix} 3 & 3 \\ 4 & 3 \end{vmatrix} = -3 \] \[ C_{32} = -\begin{vmatrix} 1 & 3 \\ 1 & 3 \end{vmatrix} = 0 \] \[ C_{33} = +\begin{vmatrix} 1 & 3 \\ 1 & 4 \end{vmatrix} = 1 \]

Thus the adjugate matrix is: \[ A = \text{adj}(M) = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \\ \end{bmatrix} \]

Now evaluate the expression: \[ \frac{a_1 + b_2 + c_3}{b_1 a_2} = \frac{7 + 1 + 1}{(-3)(-1)} = \frac{9}{3} = 3 \]

Therefore, the correct answer is 3.

Answer 2

Let \( B = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix} \). Then the adjoint of \( B \), denoted by \( \text{adj}(B) \), is given by:

\[ \text{adj}(B) = \begin{bmatrix} M_{11} & -M_{12} & M_{13} \\ -M_{21} & M_{22} & -M_{23} \\ M_{31} & -M_{32} & M_{33} \end{bmatrix} \]

where \( M_{ij} \) are the minors of \( B \).

We have:

\[ M_{11} = \begin{vmatrix} 4 & 3 \\ 3 & 4 \end{vmatrix} = 16 - 9 = 7 \] \[ M_{12} = \begin{vmatrix} 1 & 3 \\ 1 & 4 \end{vmatrix} = 4 - 3 = 1 \] \[ M_{13} = \begin{vmatrix} 1 & 4 \\ 1 & 3 \end{vmatrix} = 3 - 4 = -1 \] \[ M_{21} = \begin{vmatrix} 3 & 3 \\ 3 & 4 \end{vmatrix} = 12 - 9 = 3 \] \[ M_{22} = \begin{vmatrix} 1 & 3 \\ 1 & 4 \end{vmatrix} = 4 - 3 = 1 \] \[ M_{23} = \begin{vmatrix} 1 & 3 \\ 1 & 3 \end{vmatrix} = 3 - 3 = 0 \] \[ M_{31} = \begin{vmatrix} 3 & 3 \\ 4 & 3 \end{vmatrix} = 9 - 12 = -3 \] \[ M_{32} = \begin{vmatrix} 1 & 3 \\ 1 & 3 \end{vmatrix} = 3 - 3 = 0 \] \[ M_{33} = \begin{vmatrix} 1 & 3 \\ 1 & 4 \end{vmatrix} = 4 - 3 = 1 \]

Thus,

\[ \text{adj}(B) = \begin{bmatrix} 7 & -1 & -1 \\ -3 & 1 & 0 \\ -3 & 0 & 1 \end{bmatrix} \]

So we have \( a_1 = 7 \), \( b_2 = 1 \), \( c_3 = 1 \), \( b_1 = -1 \), \( a_2 = -3 \).

Then

\[ \frac{a_1 + b_2 + c_3}{b_1 a_2} = \frac{7 + 1 + 1}{(-1)(-3)} = \frac{9}{3} = 3 \]