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 ____

Answer

To determine the values of `alpha` and `beta` such that the given system of equations has infinite solutions, we can use the condition that the determinant of the coefficient matrix must be zero, and the ratios of the constants must also be equal.

The given system of equations is:

1. \( x + y - z = 2 \) (Equation 1)
2. \( x + 2y + \alpha z = 1 \) (Equation 2)
3. \( 2x - y + z = \beta \) (Equation 3)

### Step 1: Write the augmented matrix
The augmented matrix for the system is:

\[ \begin{bmatrix} 1 & 1 & -1 & | & 2 \\ 1 & 2 & \alpha & | & 1 \\ 2 & -1 & 1 & | & \beta \end{bmatrix} \]

### Step 2: Calculate the determinant of the coefficient matrix
The coefficient matrix is:

\[ \begin{bmatrix} 1 & 1 & -1 \\ 1 & 2 & \alpha \\ 2 & -1 & 1 \end{bmatrix} \]

To find the determinant, we can use the formula for a 3x3 matrix:

\[ \text{det} = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix:

- \( a = 1, b = 1, c = -1 \)
- \( d = 1, e = 2, f = \alpha \)
- \( g = 2, h = -1, i = 1 \)

Calculating the determinant:

\[ \text{det} = 1(2 \cdot 1 - \alpha \cdot (-1)) - 1(1 \cdot 1 - \alpha \cdot 2) - 1(1 \cdot (-1) - 2 \cdot 2) \]

This simplifies to:

\[ \text{det} = 1(2 + \alpha) - 1(1 - 2\alpha) - 1(-1 - 4) \]

\[ = 2 + \alpha - 1 + 2\alpha + 5 \]

\[ = 3 + 3\alpha \]

Setting the determinant equal to zero for infinite solutions:

\[ 3 + 3\alpha = 0 \]

\[ \alpha = -1 \]

### Step 3: Substitute \(\alpha\) into the second condition
Now, we need to ensure that the ratios of the constants are equal. This means:

\[ \frac{2 - 1}{1 - 2} = \frac{1 - \beta}{-1 - 2} \]

Substituting \(\alpha = -1\):

\[ \frac{1}{-1} = \frac{1 - \beta}{-3} \]

This simplifies to:

\[ -1 = \frac{1 - \beta}{-3} \]

Cross-multiplying gives:

\[ 3 = 1 - \beta \]

\[ \beta = 1 - 3 = -2 \]

### Step 4: Calculate \(\alpha + \beta\)
Now we have \(\alpha = -1\) and \(\beta = -2\):

\[ \alpha + \beta = -1 + (-2) = -3 \]

Thus, the final answer is:

\[ \alpha + \beta = -3 \]