Consider the system of linear equations -x + y + 2z = 0, 3x - ay + 5z = 1, 2x - 2y - az = 7. Let S1

Q: Why do we use Cramer's Rule instead of Gaussian elimination?

Cramer's Rule is often more direct for analyzing systems with parameters, as it provides explicit expressions for the determinants $\Delta$ and $\Delta_i$ that clearly show the conditions for inconsistency or infinite solutions.

Q: What if $\Delta_1 = 0$ for a value of $a$?

If $\Delta_1 = 0$, you must check $\Delta_2$ and $\Delta_3$ as well. If all $\Delta_i = 0$, then the system has infinitely many solutions. If even one is non-zero, it's inconsistent.

Q: Is it possible for the system to have a unique solution when $\Delta = 0$?

No, if $\Delta = 0$, the system cannot have a unique solution. It must be either inconsistent or have infinitely many solutions.

Question

Consider the system of linear equations $-x + y + 2z = 0$, $3x - ay + 5z = 1$, $2x - 2y - az = 7$. Let $S_1$ be the set of all $a \in \mathbf{R}$ for which the system is inconsistent and $S_2$ be the set of all $a \in \mathbf{R}$ for which the system has infinitely many solutions. If $n(S_1)$ and $n(S_2)$ denote the number of elements in $S_1$ and $S_2$ respectively, then

Options:

(A) $n(S_1) = 2, n(S_2) = 2$

(B) $n(S_1) = 1, n(S_2) = 0$

(C) $n(S_1) = 2, n(S_2) = 0$

(D) $n(S_1) = 0, n(S_2) = 2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are presented with a system of linear equations containing a parameter $a$:

$$-x + y + 2z = 0$$
$$3x - ay + 5z = 1$$
$$2x - 2y - az = 7$$

Our objective is to identify the set $S_1$ of values for $a$ that result in an inconsistent system (no solution) and the set $S_2$ of values that result in infinitely many solutions.

### The Gatekeeper: $\Delta$

To analyze the system, we calculate the coefficient determinant, $\Delta$. If $\Delta 
eq 0$, the system possesses a unique solution. If $\Delta = 0$, the system is either inconsistent or has infinitely many solutions.

The determinant is defined as:

$$\Delta = \begin{vmatrix} -1 & 1 & 2 \ 3 & -a & 5 \ 2 & -2 & -a \end{vmatrix}$$

Expanding along the first row:

$$\Delta = -1(a^2 + 10) - 1(-3a - 10) + 2(-6 + 2a)$$

Simplifying the expression, we obtain:

$$\Delta = -a^2 + 7a - 12 = -(a-3)(a-4)$$

Setting $\Delta = 0$ yields the critical values **$a = 3$** and **$a = 4$**.

### The Crossroads: Inconsistency vs. Infinity

To distinguish between inconsistency and infinite solutions, we examine the determinants $\Delta_1, \Delta_2,$ and $\Delta_3$. If $\Delta = 0$ and at least one $\Delta_i 
eq 0$, the system is **inconsistent**. If $\Delta = 0$ and all $\Delta_i = 0$, the system has **infinitely many solutions**.

We calculate $\Delta_1$ by replacing the first column with the constants $(0, 1, 7)$:

$$\Delta_1 = \begin{vmatrix} 0 & 1 & 2 \ 1 & -a & 5 \ 7 & -2 & -a \end{vmatrix}$$

Expanding this determinant:

$$\Delta_1 = 0 - 1(-a - 35) + 2(-2 + 7a) = a + 35 - 4 + 14a = 15a + 31$$

### The Verdict

We now test our critical values $a = 3$ and $a = 4$ in the expression for $\Delta_1$:

For $a = 3$:
$$\Delta_1 = 15(3) + 31 = 76 
eq 0$$

For $a = 4$:
$$\Delta_1 = 15(4) + 31 = 91 
eq 0$$

Since $\Delta = 0$ and $\Delta_1 
eq 0$ for both values, the system is **inconsistent** for both $a=3$ and $a=4$.

Therefore, the set of values for inconsistency is **$S_1 = \{3, 4\}$**, implying **$n(S_1) = 2$**. As no values of $a$ satisfy the condition for infinitely many solutions, **$S_2 = \emptyset$**, implying **$n(S_2) = 0$**.

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Visualized Solution

System of Linear Equations

Cramer's Rule and $Δ$

Setting up $Δ$

Expanding $Δ$

Factorizing $Δ$

Checking $Δ_{1}, Δ_{2}, Δ_{3}$

Setting up and Expanding $Δ_{1}$

Testing $a = 3$ in $Δ_{1}$

Testing $a = 4$ in $Δ_{1}$

Final Conclusion

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

Analyzing the Setup

The Gatekeeper: $Δ$

The Crossroads

The Verdict

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Visualized Solution

System of Linear Equations

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