If $$P = \begin{bmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{bmatrix}$$ is the adjoint of a $$3 \times 3$$ matrix $$A$$ and $$|A| = 4$$, then $$\alpha$$ is equal to

We are told that the matrix

$$P=\begin{bmatrix}1&\alpha&3\\1&3&3\\2&4&4\end{bmatrix}$$

is the adjoint (classical adjugate) of a square matrix $$A$$ of order $$3$$ and that the determinant of $$A$$ is

$$|A|=4.$$

The standard relation between a non-singular square matrix and its adjoint is

$$A\;\operatorname{adj}(A)=\operatorname{adj}(A)\,A=|A|\,I.$$

Because $$P=\operatorname{adj}(A)$$ and $$|A|=4,$$ we have simultaneously

$$A\,P=4I \qquad\text{and}\qquad P\,A=4I.$$

Either product is sufficient; we shall work with the first one,

$$A\,P=4I,$$

and write the entries of $$A$$ as unknowns:

$$A=\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}.$$

Multiplying $$A$$ by $$P$$ gives

$$A\,P=\begin{bmatrix} a+b+2c & a\alpha+3b+4c & 3a+3b+4c\\ d+e+2f & d\alpha+3e+4f & 3d+3e+4f\\ g+h+2i & g\alpha+3h+4i & 3g+3h+4i \end{bmatrix}.$$

This product must equal $$4I,$$ that is,

$$A\,P=\begin{bmatrix}4&0&0\\0&4&0\\0&0&4\end{bmatrix}.$$

Equating corresponding entries produces nine linear equations.

From the first row we obtain

$$\begin{aligned} a+b+2c&=4,\\ a\alpha+3b+4c&=0,\\ 3a+3b+4c&=0. \end{aligned}$$

Solving this triple:

1. From the third equation $$3a+3b+4c=0$$ we isolate $$b$$:

$$b=-\dfrac{3a+4c}{3}.$$

2. Substitute this in $$a+b+2c=4$$:

$$a-\dfrac{3a+4c}{3}+2c=4 \;\Longrightarrow\;2c=12 \;\Longrightarrow\;c=6.$$

3. Then $$b=-a-8.$$

4. Insert $$b$$ and $$c$$ into $$a\alpha+3b+4c=0$$:

$$a\alpha+3(-a-8)+24=0 \;\Longrightarrow\;a(\alpha-3)=0.$$

Thus either $$\alpha=3$$ or $$a=0.$$ The choice $$\alpha=3$$ is quickly discarded because it would make two different diagonal requirements for the second row contradict each other; hence

$$a=0\qquad\text{and}\qquad\alpha\neq3.$$ With $$a=0$$ we have

$$b=-8,\qquad c=6.$$

Proceeding to the second row of equations,

$$\begin{aligned} d+e+2f&=0,\\ d\alpha+3e+4f&=4,\\ 3d+3e+4f&=0, \end{aligned}$$

subtracting the first multiplied by $$3$$ from the third gives $$2f=0,$$ so

$$f=0\quad\text{and}\quad e=-d.$$

Substituting into the middle equation:

$$d\alpha+3(-d)=4\;\Longrightarrow\;d(\alpha-3)=4 \;\Longrightarrow\;d=\dfrac{4}{\alpha-3}.$$

The third row supplies

$$\begin{aligned} g+h+2i&=0,\\ g\alpha+3h+4i&=0,\\ 3g+3h+4i&=4. \end{aligned}$$

Eliminating as before, the first and third give $$i=-2,$$ then $$h=4-g,$$ and the second yields

$$g\alpha-3g+4=0\;\Longrightarrow\;g(\alpha-3)=-4 \;\Longrightarrow\;g=-\dfrac{4}{\alpha-3},$$

so

$$h=4+\dfrac{4}{\alpha-3}.$$

Collecting all results, the matrix $$A$$ is

$$A=\begin{bmatrix} 0 & -8 & 6\\[2mm] \dfrac{4}{\alpha-3} & -\dfrac{4}{\alpha-3} & 0\\[3mm] -\dfrac{4}{\alpha-3} & 4+\dfrac{4}{\alpha-3} & -2 \end{bmatrix}.$$

To exploit the given determinant, we compute $$|A|.$$ Expanding along the first row (cofactor rule):

$$|A|=0-\!(-8)\,\bigl(d\,i-f\,g\bigr)+6\bigl(d\,h-e\,g\bigr).$$

Because $$f=0$$ and $$g=-d,$$ this simplifies to

$$|A|=-16d+6d(h+e).$$

But $$e=-d$$ and $$h=4+d,$$ so $$h+e=4.$$ Therefore

$$|A|=-16d+24d=8d.$$

The problem states $$|A|=4,$$ hence

$$8d=4\;\Longrightarrow\;d=\dfrac12.$$

Finally substituting $$d=\dfrac12$$ into $$d(\alpha-3)=4$$ gives

$$\dfrac12(\alpha-3)=4 \;\Longrightarrow\;\alpha-3=8 \;\Longrightarrow\;\boxed{\alpha=11}.$$

Among the alternatives, this corresponds to Option D.

Hence, the correct answer is Option D.