Let’s break down the solution step by step. We are given the matrix

$$ X = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix}, $$

and its adjugate (or classical adjoint) is given by

$$ \text{adj}(X) = A = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}. $$

Remember that the adjugate matrix is the transpose of the cofactor matrix. Therefore:

• The entry in the first row and first column, $$a_1,$$ equals the cofactor of $$X$$ at position $$ (1,1) $$; thus, $$a_1 = C_{11}.$$

• The entry $$b_2$$ (at position $$ (2,2) $$) equals $$C_{22}.$$

• The entry $$c_3$$ (at position $$ (3,3) $$) equals $$C_{33}.$$

• Similarly, $$b_1$$ (position $$ (1,2) $$) equals $$C_{21},$$ and $$a_2$$ (position $$ (2,1) $$) equals $$C_{12}.$$

We need to find the value of

$$ \frac{a_1 + b_2 + c_3}{b_1 a_2}. $$

Let’s compute each required cofactor.

Compute $$a_1 = C_{11}:$$

Remove the 1st row and 1st column of $$X$$ to get

$$ \begin{bmatrix} 4 & 3 \\ 3 & 4 \end{bmatrix}. $$

The determinant is

$$ 4 \times 4 - 3 \times 3 = 16 - 9 = 7. $$

Since $$(-1)^{1+1} = 1,$$ we have

$$ a_1 = 7. $$

Compute $$b_2 = C_{22}:$$

Remove the 2nd row and 2nd column to obtain

$$ \begin{bmatrix} 1 & 3 \\ 1 & 4 \end{bmatrix}. $$

Its determinant is

$$ 1 \times 4 - 3 \times 1 = 4 - 3 = 1, $$

and since $$(-1)^{2+2} = 1,$$ we get

$$ b_2 = 1. $$

Compute $$c_3 = C_{33}:$$

Remove the 3rd row and 3rd column, which gives

$$ \begin{bmatrix} 1 & 3 \\ 1 & 4 \end{bmatrix}. $$

Its determinant is also

$$ 1 \times 4 - 3 \times 1 = 1. $$

With $$(-1)^{3+3} = 1,$$ we have

$$ c_3 = 1. $$

Compute $$b_1 = C_{21}:$$

The cofactor of the element at position $$ (2,1) $$ is computed by removing the 2nd row and 1st column:

$$ \begin{bmatrix} 3 & 3 \\ 3 & 4 \end{bmatrix}. $$

The determinant is

$$ 3 \times 4 - 3 \times 3 = 12 - 9 = 3. $$

The sign factor is $$(-1)^{2+1} = -1,$$ so

$$ b_1 = -3. $$

Compute $$a_2 = C_{12}:$$

For position $$ (1,2),$$ remove the 1st row and 2nd column:

$$ \begin{bmatrix} 1 & 3 \\ 1 & 4 \end{bmatrix}. $$

The determinant is

$$ 1 \times 4 - 3 \times 1 = 4 - 3 = 1. $$

Here, the sign factor is $$(-1)^{1+2} = -1,$$ so

$$ a_2 = -1. $$

Now, substitute these values into the expression:

• Numerator:

$$ a_1 + b_2 + c_3 = 7 + 1 + 1 = 9. $$

• Denominator:

$$ b_1 \cdot a_2 = (-3) \times (-1) = 3. $$

Thus, the ratio is

$$ \frac{9}{3} = 3. $$

So, the value of $$\frac{a_1+b_2+c_3}{b_1 a_2}$$ is 3, which corresponds to Option A.

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