Key Concepts and Formulas

• System of Linear Equations: For a system of three linear equations in three variables $x, y, z$: $$\begin{align*} a_1x + b_1y + c_1z &= d_1 \\ a_2x + b_2y + c_2z &= d_2 \\ a_3x + b_3y + c_3z &= d_3 \end{align*}$$
• Determinant of Coefficient Matrix ($\Delta$): $$\Delta = \left| {\begin{matrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{matrix}} \right|$$
• Determinants $\Delta_1, \Delta_2, \Delta_3$: These are obtained by replacing the 1st, 2nd, and 3rd columns of $\Delta$ respectively with the column of constant terms $(d_1, d_2, d_3)^T$.
• Condition for Infinite Solutions: A system of linear equations has infinite solutions if and only if: $\Delta = 0 \quad \text{AND} \quad \Delta_1 = 0 \quad \text{AND} \quad \Delta_2 = 0 \quad \text{AND} \quad \Delta_3 = 0$ If $\Delta = 0$ but at least one of $\Delta_1, \Delta_2, \Delta_3$ is non-zero, then the system has no solution.

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Step-by-Step Solution

The given system of linear equations is:

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

Step 1: Calculate the Determinant of the Coefficient Matrix ($\Delta$) and find $\alpha$.

For the system to have infinite solutions, the determinant of the coefficient matrix, $\Delta$, must be zero. The coefficient matrix $A$ is:

$$A = \left[ {\begin{matrix} 1 & 1 & -1 \\ 1 & 2 & \alpha \\ 2 & -1 & 1 \end{matrix}} \right]$$

Now, we calculate its determinant:

\Delta = \left| {\begin{matrix} 1 & 1 & -1 \\ 1 & 2 & \alpha \\ 2 & -1 & 1 \end{vmatrix}} \right|

To simplify the calculation, we can expand along the first row:

\Delta = 1 \cdot \left| {\begin{matrix} 2 & \alpha \\ -1 & 1 \end{vmatrix}} \right| - 1 \cdot \left| {\begin{matrix} 1 & \alpha \\ 2 & 1 \end{vmatrix}} \right| + (-1) \cdot \left| {\begin{matrix} 1 & 2 \\ 2 & -1 \end{vmatrix}} \right| $$\Delta = 1 \cdot ((2)(1) - (\alpha)(-1)) - 1 \cdot ((1)(1) - (\alpha)(2)) - 1 \cdot ((1)(-1) - (2)(2))$$ $$\Delta = (2 + \alpha) - (1 - 2\alpha) - 1(-1 - 4)$$ $$\Delta = 2 + \alpha - 1 + 2\alpha + 5$$ $$\Delta = 3\alpha + 6$$

For infinite solutions, we must have $\Delta = 0$:

$$3\alpha + 6 = 0 \implies 3\alpha = -6 \implies \alpha = -2$$

Reasoning: Setting $\Delta = 0$ is a necessary condition for a system to have either infinite solutions or no solutions. If $\Delta \neq 0$, the system would have a unique solution, which contradicts the problem statement.

Step 2: Calculate $\Delta_2$ (or any other $\Delta_i$) and find $\beta$.

Now that we have $\alpha = -2$, we substitute this value into the system. The constant terms are $d_1 = 2, d_2 = 1, d_3 = \beta$. For infinite solutions, all determinants $\Delta_1, \Delta_2, \Delta_3$ must also be zero. We will calculate $\Delta_2$. To form $\Delta_2$, we replace the second column of the coefficient matrix $A$ with the constant terms $(2, 1, \beta)^T$, using $\alpha = -2$ for the third column:

\Delta_2 = \left| {\begin{matrix} 1 & 2 & -1 \\ 1 & 1 & -2 \\ 2 & \beta & 1 \end{vmatrix}} \right|

Now, we calculate this determinant by expanding along the first row:

\Delta_2 = 1 \cdot \left| {\begin{matrix} 1 & -2 \\ \beta & 1 \end{vmatrix}} \right| - 2 \cdot \left| {\begin{matrix} 1 & -2 \\ 2 & 1 \end{vmatrix}} \right| + (-1) \cdot \left| {\begin{matrix} 1 & 1 \\ 2 & \beta \end{vmatrix}} \right| $$\Delta_2 = 1 \cdot ((1)(1) - (-2)(\beta)) - 2 \cdot ((1)(1) - (-2)(2)) - 1 \cdot ((1)(\beta) - (1)(2))$$ $$\Delta_2 = 1 \cdot (1 + 2\beta) - 2 \cdot (1 + 4) - 1 \cdot (\beta - 2)$$ $$\Delta_2 = (1 + 2\beta) - 2 \cdot (5) - (\beta - 2)$$ $$\Delta_2 = 1 + 2\beta - 10 - \beta + 2$$ $$\Delta_2 = \beta - 7$$

For infinite solutions, $\Delta_2$ must be zero:

$$\beta - 7 = 0 \implies \beta = 7$$

Reasoning: Since $\Delta = 0$, if any of $\Delta_1, \Delta_2, \Delta_3$ were non-zero, the system would have no solution. Therefore, for infinite solutions, all $\Delta_i$ must be zero. We used $\Delta_2 = 0$ to find the value of $\beta$. (It can be verified that $\Delta_1=0$ and $\Delta_3=0$ also hold for $\alpha=-2$ and $\beta=7$).

Step 3: Calculate $\alpha + \beta$.

We found $\alpha = -2$ and $\beta = 7$. Therefore,

$$\alpha + \beta = -2 + 7 = 5$$

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Common Mistakes & Tips

• Distinguishing Infinite Solutions from No Solutions: Both conditions require $\Delta = 0$. The key difference is that for infinite solutions, ALL $\Delta_1, \Delta_2, \Delta_3$ must also be zero. For no solutions, $\Delta = 0$ but at least one $\Delta_i \neq 0$.
• Determinant Calculation Errors: Be meticulous with signs and arithmetic when calculating $3 \times 3$ determinants. Using row or column operations to introduce zeros can simplify calculations and reduce errors.
• Correct Formation of $\Delta_i$ Matrices: Ensure you replace the correct column of the coefficient matrix with the constant terms for each $\Delta_i$.

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Summary

To find the values of $\alpha$ and $\beta$ for which the given system of linear equations has infinite solutions, we applied Cramer's Rule conditions. First, we calculated the determinant of the coefficient matrix ($\Delta$) and set it to zero, which yielded $\alpha = -2$. Next, we substituted this value of $\alpha$ into the determinant $\Delta_2$ (formed by replacing the second column of coefficients with the constant terms) and set it to zero, which gave $\beta = 7$. Finally, we calculated the required sum $\alpha + \beta$. Based on our calculations, $\alpha + \beta = 5$.

The final answer is $\boxed{1}$.