Let the system of linear equations x + y + αz = 2, 3x + y + z = 4, x + 2z = 1 have a unique solution

Q: Wait, why do we need the determinant to be non-zero for a unique solution?

Great question! In linear algebra, a system of equations can be represented as AX = B. If the determinant of A is zero, the system either has no solution or infinitely many solutions. For a unique solution to exist, the matrix A must be invertible, which only happens when its determinant is non-zero. That is why we enforce \Delta \neq 0.

Q: Can I just use the slope formula for collinearity instead of the determinant?

Absolutely! You can calculate the slope between P1 and P2 and set it equal to the slope between P2 and P3. Both methods are mathematically equivalent. However, the determinant method is often faster and less prone to algebraic errors when dealing with variables like \alpha.

Q: Why did we reject \alpha = -3 in the beginning?

We didn't necessarily reject it as a value, but rather identified it as a forbidden value. The problem states that the system has a unique solution. If \alpha were -3, the determinant would be zero, and the system would not have a unique solution. Therefore, any valid \alpha must satisfy \alpha \neq -3.

Question

Let the system of linear equations $x + y + \alpha z = 2$, $3x + y + z = 4$, $x + 2z = 1$ have a unique solution $(x^*, y^*, z^*)$. If $(\alpha, x^*)$, $(y^*, \alpha)$ and $(x^*, -y^*)$ are collinear points, then the sum of absolute values of all possible values of $\alpha$ is :

Options:

(A) 4

(B) 3

(C) 2

(D) 1

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorer of mathematics! Today, we are going to unravel a problem that beautifully bridges the gap between algebraic systems and geometric intuition.

We are presented with a system of three linear equations:
$$x + y + \alpha z = 2$$
$$3x + y + z = 4$$
$$x + 2z = 1$$

Our mission is to find the sum of the absolute values of all possible values of $\alpha$ that satisfy a specific geometric condition. Let us embark on this journey step by step.

### The Gatekeeper of Uniqueness

Before we dive into solving the system, we must ensure that a unique solution actually exists. In the world of linear algebra, the determinant of the coefficient matrix acts as a gatekeeper.

If the determinant, which we call $\Delta$, is zero, the system collapses into a state of either no solution or infinite solutions. We calculate $\Delta$ as:

$$\Delta = \begin{vmatrix} 1 & 1 & \alpha \ 3 & 1 & 1 \ 1 & 0 & 2 \end{vmatrix}$$

Expanding this along the first row, we get:
$$\Delta = 1(2 - 0) - 1(6 - 1) + \alpha(0 - 1)$$
$$\Delta = 2 - 5 - \alpha = -3 - \alpha$$

For a unique solution, we must have $\Delta 
eq 0$, which implies **$\alpha 
eq -3$**. Keep this constraint in your back pocket; it is our safety net.

### The Algebraic Dance

Now, let us find the unique solution $(x^*, y^*, z^*)$. We can use the substitution method, which is often the most intuitive approach.

From the third equation, $x + 2z = 1$, we can express $x$ as $x = 1 - 2z$. Substituting this into the second equation:
$$3(1 - 2z) + y + z = 4$$
$$3 - 6z + y + z = 4 \Rightarrow y = 1 + 5z$$

Now, we have both $x$ and $y$ in terms of $z$. Let us substitute these into the first equation:
$$(1 - 2z) + (1 + 5z) + \alpha z = 2$$

Notice the magic here: the constant terms $1 + 1$ on the left side sum to $2$, which cancels out the $2$ on the right side. We are left with:
$$3z + \alpha z = 0 \Rightarrow z(3 + \alpha) = 0$$

Since we know $\alpha 
eq -3$, the term $(3 + \alpha)$ cannot be zero. Therefore, $z^* = 0$. With $z^* = 0$, finding $x^*$ and $y^*$ is trivial:
$$x^* = 1 - 2(0) = 1$$
$$y^* = 1 + 5(0) = 1$$

Our unique solution is $(1, 1, 0)$.

### The Geometric Symphony

Now, we turn to the geometry. We are given three points: $P_1 = (\alpha, x^*)$, $P_2 = (y^*, \alpha)$, and $P_3 = (x^*, -y^*)$.

Substituting our values $x^* = 1$ and $y^* = 1$, these points become $P_1 = (\alpha, 1)$, $P_2 = (1, \alpha)$, and $P_3 = (1, -1)$. For these points to be collinear, the area of the triangle formed by them must be zero.

Mathematically, this is expressed as the determinant of their coordinates being zero:

$$\begin{vmatrix} \alpha & 1 & 1 \ 1 & \alpha & 1 \ 1 & -1 & 1 \end{vmatrix} = 0$$

Expanding this determinant, we get:
$$\alpha(\alpha - (-1)) - 1(1 - 1) + 1(-1 - \alpha) = 0$$
$$\alpha(\alpha + 1) - 0 + (-1 - \alpha) = 0$$
$$\alpha^2 + \alpha - 1 - \alpha = 0$$

The $\alpha$ terms cancel out, leaving us with $\alpha^2 - 1 = 0$. Thus, $\alpha = 1$ or $\alpha = -1$. Both values satisfy our initial constraint $\alpha 
eq -3$.

### The Final Victory

The problem asks for the sum of the absolute values of all possible values of $\alpha$. We have found $\alpha = 1$ and $\alpha = -1$.

The sum of their absolute values is:
$$|1| + |-1| = 1 + 1 = 2$$

The final answer is **2**. We have successfully navigated the algebra and the geometry to arrive at our answer.

Select Answer:

Visualized Solution

System of Linear Equations

Condition for Unique Solution

Calculating the Determinant

Solving the System: Substitution

Finding $y$ in terms of $z$

Substituting into the First Equation

Determining $z^{*}$

Calculating $x^{}$ and $y^{}$

Identifying the Collinear Points

Collinearity Condition

Expanding the Collinearity Determinant

Solving for $α$

Final Conclusion

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Setup

The Gatekeeper of Uniqueness

The Algebraic Dance

The Geometric Symphony

The Final Victory

Similar Questions

The sum of distinct values of $λ$ for which the system of equations $(λ - 1) x + (3 λ + 1) y + 2 λ z = 0 (λ - 1) x + (4 λ - 2) y + (λ + 3) z = 0 2 x + (3 λ + 1) y + 3 (λ - 1) z = 0$ has non-zero solutions, is

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The system of linear equations $λ x + 2 y + 2 z = 5$ , $2 λ x + 3 y + 5 z = 8$ , $4 x + λ y + 6 z = 10$ has:

If the system of equations $3 x + y + 4 z = 3$ , $2 x + α y - z = - 3$ , $x + 2 y + z = 4$ has no solution, then the value of $α$ is equal to :

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Consider the following system of equations $α x + 2 y + z = 1$ , $2 α x + 3 y + z = 1$ , $3 x + α y + 2 z = β$ For some $α, β \in R$ . Then which of the following is NOT correct.

Select Answer:

Visualized Solution

System of Linear Equations

Condition for Unique Solution

Calculating the Determinant

Solving the System: Substitution

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z

Substituting into the First Equation

Determining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z∗

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∗ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y∗

Identifying the Collinear Points

Collinearity Condition

Expanding the Collinearity Determinant

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Final Conclusion

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Setup

The Gatekeeper of Uniqueness

The Algebraic Dance

The Geometric Symphony

The Final Victory

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