Question

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

Hint:

For a system of 3 linear equations to have infinite solutions, two conditions must be met: 1) The determinant of the coefficient matrix $\Delta=0$. 2) The system must be consistent (i.e., $\Delta_x=\Delta_y=\Delta_z=0$, or show that one plane equation is a linear combination of the others).

Answer 1

Correct Answer: 5

For a system of non-homogeneous linear equations to have infinite solutions, the determinant of the coefficient matrix ($\Delta$) must be zero. 

Expanding the determinant: 
$1(2 - (-\alpha)) - 1(1 - 2\alpha) - 1(-1 - 4) = 0$. 
$2 + \alpha - 1 + 2\alpha + 5 = 0$. 
$3\alpha + 6 = 0 \implies \alpha = -2$. 
For infinite solutions, the planes must be consistent. This means that one plane equation must be a linear combination of the other two. Let $P_1, P_2, P_3$ be the three planes. Let's try to find constants $k_1, k_2$ such that $P_3 = k_1 P_1 + k_2 P_2$. 
$2x-y+z = k_1(x+y-z) + k_2(x+2y-2z)$ (using $\alpha=-2$). 
Equating the coefficients of x, y, and z: 
Coeff of x: $2 = k_1 + k_2$. 
Coeff of y: $-1 = k_1 + 2k_2$. 
Coeff of z: $1 = -k_1 - 2k_2$ (This is consistent with the y-equation). 
Solving the first two equations for $k_1$ and $k_2$: 
Subtracting the first from the second: $(-1) - (2) = (k_1+2k_2) - (k_1+k_2) \implies -3 = k_2$. 
Substitute $k_2 = -3$ into the first equation: $2 = k_1 - 3 \implies k_1 = 5$. 
Now, the same linear combination must hold for the constant terms for the system to be consistent. 
$\beta = k_1(2) + k_2(1)$. 
$\beta = 5(2) + (-3)(1) = 10 - 3 = 7$. 
So, for the system to have infinite solutions, we must have $\alpha=-2$ and $\beta=7$. 
The question asks for the value of $\alpha+\beta$. 
$\alpha + \beta = -2 + 7 = 5$.