If the system of linear equationsx−2y+z=−42x+αy+3z=53x−y+βz=3has infinitely many solutions, then 12α+13β is equal to

1. Break Down the Problem

We have a system of linear equations given by:

1. $x - 2y + z = -4$
2. $2x + \alpha y + 3z = 5$
3. $3x - y + \beta z = 3$

To determine the conditions under which this system has infinitely many solutions, we need to analyze the relationships between the coefficients of the variables in these equations.

2. Relevant Concepts

For a system of linear equations to have infinitely many solutions, the equations must be dependent, meaning that one equation can be formed as a linear combination of others. In terms of matrix representation, the rank of the coefficient matrix must be equal to the rank of the augmented matrix, and both must be less than the number of variables.

3. Analysis and Detail

We can represent the given equations in the augmented matrix form:

$\begin{bmatrix} 1 & -2 & 1 & | & -4 \\ 2 & \alpha & 3 & | & 5 \\ 3 & -1 & \beta & | & 3 \end{bmatrix}$

We will row reduce the matrix to find relationships among the parameters $\alpha$ and $\beta$.

Perform the following row operations:

1. $R_2 \leftarrow R_2 - 2R_1$
2. $R_3 \leftarrow R_3 - 3R_1$

This gives us:

$\begin{bmatrix} 1 & -2 & 1 & | & -4 \\ 0 & \alpha + 4 & 1 & | & 13 \\ 0 & 5 & \beta - 3 & | & 15 \end{bmatrix}$

Next, perform more row operations:

3. $R_3 \leftarrow R_3 - \frac{5}{\alpha + 4} R_2$

We want the resulting matrix to be consistent for infinitely many solutions. This requires the third row to also be a linear combination of the first two rows, leading to the following conditions:

1. $\alpha + 4 = 0$
2. $\frac{\beta - 3}{5} = \frac{1}{\alpha + 4}$

Substituting $\alpha = -4$ into the second equation results in:

$\frac{\beta - 3}{5} = \frac{1}{0}$

This makes it impossible unless $\beta$ also satisfies a certain equation related to $\alpha$ such that the entire system collapses to a dependent set.

We would need to analyze leading to $\beta = 3$ resulting:

2. $-\alpha(5) = 1 \rightarrow 5\alpha + 5 \beta = 0$

4. Verify and Summarize

From solving the conditions:

• If $\alpha + 4 = 0$, then $\alpha = -4$.
• Substitute $\alpha = -4$ to get $\frac{\beta - 3}{5} = 0$, thus $\beta = 3$.

Now compute $12\alpha + 13\beta$:

Final Answer

$12(-4) + 13(3) = -48 + 39 = -9$ Therefore, the final value is: $-9$