For real numbers $$\alpha$$ and $$\beta$$, consider the following system of linear equations: $$x + y - z = 2$$, $$x + 2y + \alpha z = 1$$ and $$2x - y + z = \beta$$. If the system has infinite solutions, then $$\alpha + \beta$$ is equal to _________.

We have the system of three linear equations in the variables $$x$$, $$y$$ and $$z$$:

$$\begin{aligned} x+y-z&=2,\\ x+2y+\alpha z&=1,\\ 2x-y+z&=\beta. \end{aligned}$$

For infinitely many solutions to exist, the three equations must be dependent, that is, the rank of the coefficient matrix must be smaller than the number of variables. In particular, the determinant of the coefficient matrix must be zero.

First we write the coefficient matrix $$A$$ and compute its determinant. The matrix is

$$A=\begin{bmatrix} 1&1&-1\\ 1&2&\alpha\\ 2&-1&1 \end{bmatrix}.$$

The determinant of a $$3\times3$$ matrix $$\begin{bmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33} \end{bmatrix}$$ is given by the rule

$$\det A = a_{11}(a_{22}a_{33}-a_{23}a_{32})-a_{12}(a_{21}a_{33}-a_{23}a_{31})+a_{13}(a_{21}a_{32}-a_{22}a_{31}).$$

Applying this formula, we get

$$\begin{aligned} \det A &= 1\bigl(2\cdot1-\alpha(-1)\bigr)\;-\;1\bigl(1\cdot1-\alpha\cdot2\bigr)\;+\;(-1)\bigl(1\cdot(-1)-2\cdot2\bigr)\\ &= 1(2+\alpha)\;-\;1(1-2\alpha)\;+\;(-1)(-1-4)\\ &= (2+\alpha)\;-\;(1-2\alpha)\;+\;5\\ &= 2+\alpha-1+2\alpha+5\\ &= 6+3\alpha\\ &= 3(2+\alpha). \end{aligned}$$

For infinite solutions, $$\det A=0$$, so

$$3(2+\alpha)=0 \quad\Longrightarrow\quad \alpha=-2.$$

With $$\alpha=-2$$, the second equation becomes

$$x+2y-2z=1.$$

Now one of the three equations must be a linear combination of the other two. Let us try to express the third equation as a combination of the first two. Assume there exist real numbers $$p$$ and $$q$$ such that

$$p\,(x+y-z=2)+q\,(x+2y-2z=1)=2x-y+z=\beta.$$

Equating coefficients of $$x$$, $$y$$ and $$z$$ on both sides, we obtain the system

$$\begin{aligned} \text{(i)}&\; p+q &=& 2,\\ \text{(ii)}&\; p+2q &=& -1,\\ \text{(iii)}&\; -p-2q &=& 1. \end{aligned}$$

Solving (i) and (ii): subtract (i) from (ii) to eliminate $$p$$,

$$p+2q-(p+q)= -1-2\quad\Longrightarrow\quad q=-3.$$

Substituting $$q=-3$$ in (i) gives

$$p+(-3)=2\quad\Longrightarrow\quad p=5.$$

Checking with (iii): $$-p-2q=-5-2(-3)=1,$$ which is exactly the required coefficient of $$z$$, so the choice of $$p$$ and $$q$$ is consistent.

The right-hand side now yields

$$\beta = 2p+ q = 2\cdot5 + (-3)=10-3=7.$$

Finally, we calculate

$$\alpha+\beta = (-2)+7 = 5.$$

So, the answer is $$5$$.