Consider the system of linear equations
$$-x + y + 2z = 0$$
$$3x - ay + 5z = 1$$
$$2x - 2y - az = 7$$
Let $$S_1$$ be the set of all $$a \in R$$ for which the system is inconsistent and $$S_2$$ be the set of all $$a \in R$$ for which the system has infinitely many solutions. If $$nS_1$$ and $$nS_2$$ denote the number of elements in $$S_1$$ and $$S_2$$ respectively, then

We are given the linear system

$$-x + y + 2z = 0$$ $$3x - ay + 5z = 1$$ $$2x - 2y - az = 7$$

and we have to find those real numbers $$a$$ for which the system is inconsistent as well as those for which it has infinitely many solutions.

First we collect the coefficients in the matrix form $$A\mathbf x=\mathbf b$$, where

$$A=\begin{bmatrix}-1 & 1 & 2\\ 3 & -a & 5\\ 2 & -2 & -a\end{bmatrix},\qquad \mathbf b=\begin{bmatrix}0\\1\\7\end{bmatrix},\qquad \mathbf x=\begin{bmatrix}x\\y\\z\end{bmatrix}.$$

The standard criterion says that

• if $$\det A\ne0$$, then $$\operatorname{rank}A=3$$ and the unique solution exists;
• if $$\det A=0$$, then $$\operatorname{rank}A\le2$$ and further comparison with the rank of the augmented matrix $$\left[A\;|\;\mathbf b\right]$$ decides consistency.

So we must first evaluate $$\det A$$. Expanding along the first row we write

$$ \det A= (-1)\Bigl[(-a)(-a)-5(-2)\Bigr] -\;1\Bigl[3(-a)-5\cdot2\Bigr] +\;2\Bigl[3(-2)-(-a)\cdot2\Bigr]. $$

Simplifying each bracket one by one:

$$(-a)(-a)-5(-2)=a^2+10,$$

$$3(-a)-5\cdot2=-3a-10,$$

$$3(-2)-(-a)\cdot2=-6+2a.$$

Substituting all these back gives

$$ \det A =(-1)(a^2+10) -\bigl(-3a-10\bigr) +2(-6+2a). $$

Now we remove the brackets carefully:

$$ \det A =-a^2-10 +3a+10 -12+4a. $$

Combining like terms we have

$$ \det A =-a^2+7a-12. $$

Factoring the quadratic polynomial,

$$ -a^2+7a-12 =-(a^2-7a+12) =-(a-3)(a-4). $$

Thus

$$ \det A = 0 \quad\Longleftrightarrow\quad (a-3)(a-4)=0 \quad\Longleftrightarrow\quad a=3\;\text{or}\;a=4. $$

For any $$a\ne3,4$$ the determinant is non-zero, $$\operatorname{rank}A=3=\operatorname{rank}[A|\mathbf b]$$ and the system is uniquely solvable, so such $$a$$ belong to neither $$S_1$$ nor $$S_2$$. Hence the only candidates for inconsistency or infinite solutions are $$a=3$$ and $$a=4$$. We now test them one by one.

Case 1: $$a=3$$

Substituting $$a=3$$ we obtain

$$ \begin{aligned} -x+y+2z &= 0,\\ 3x-3y+5z &= 1,\\ 2x-2y-3z &= 7. \end{aligned} $$

Writing the augmented matrix and performing elementary row operations:

$$ \left[ \begin{array}{ccc|c} -1 & 1 & 2 & 0\\ 3 & -3 & 5 & 1\\ 2 & -2 & -3 & 7 \end{array} \right] \;\xrightarrow{R_1\leftarrow-\,R_1}\; \left[ \begin{array}{ccc|c} 1 & -1 & -2 & 0\\ 3 & -3 & 5 & 1\\ 2 & -2 & -3 & 7 \end{array} \right] $$

$$ R_2\leftarrow R_2-3R_1,\; R_3\leftarrow R_3-2R_1\;\Longrightarrow\; \left[ \begin{array}{ccc|c} 1 & -1 & -2 & 0\\ 0 & 0 & 11 & 1\\ 0 & 0 & 1 & 7 \end{array} \right]. $$

Since the third column now carries two contradictory equations, $$11z=1$$ and $$z=7,$$ subtracting them yields $$0=76/11,$$ a clear impossibility. Therefore $$\operatorname{rank}A=2$$ but $$\operatorname{rank}[A|\mathbf b]=3,$$ making the system inconsistent. Hence $$a=3\in S_1.$$

Case 2: $$a=4$$

Substituting $$a=4$$ we obtain

$$ \begin{aligned} -x+y+2z &= 0,\\ 3x-4y+5z &= 1,\\ 2x-2y-4z &= 7. \end{aligned} $$

Again we row-reduce the augmented matrix:

$$ \left[ \begin{array}{ccc|c} -1 & 1 & 2 & 0\\ 3 & -4 & 5 & 1\\ 2 & -2 & -4 & 7 \end{array} \right] \;\xrightarrow{R_1\leftarrow-\,R_1}\; \left[ \begin{array}{ccc|c} 1 & -1 & -2 & 0\\ 3 & -4 & 5 & 1\\ 2 & -2 & -4 & 7 \end{array} \right] $$

$$ R_2\leftarrow R_2-3R_1\Longrightarrow \left[ \begin{array}{ccc|c} 1 & -1 & -2 & 0\\ 0 & -1 & 11 & 1\\ 2 & -2 & -4 & 7 \end{array} \right], \qquad R_3\leftarrow R_3-2R_1\Longrightarrow \left[ \begin{array}{ccc|c} 1 & -1 & -2 & 0\\ 0 & -1 & 11 & 1\\ 0 & 0 & 0 & 7 \end{array} \right]. $$

The last row represents the impossible statement $$0=7,$$ so the system is again inconsistent. Here too $$\operatorname{rank}A=2$$ while $$\operatorname{rank}[A|\mathbf b]=3,$$ hence $$a=4\in S_1.$$

Final conclusions

We have found

$$S_1=\{3,4\},\qquad S_2=\varnothing.$$

Thus the numbers of elements are

$$nS_1=2,\qquad nS_2=0.$$

Hence, the correct answer is Option A.