Matrices & Determinants
System of linear equations - consistency
Grade None
Question:
<p>The system of linear equations<br>\(x + y + z = 2\)<br>\(2x + 3y + 2z = 5\)<br>\(2x + 3y + (a^2 - 1)z = a + 1\)</p>
<p>is inconsistent when \(a = 4\)</p>
<p>has a unique solution for \(|a| = \sqrt{3}\)</p>
<p>has infinitely many solutions for \(a = 4\)</p>
<p>is inconsistent when \(|a| = \sqrt{3}\)</p>
Step-by-Step Solution
Key Concept: For a system to have infinitely many solutions, the coefficient matrix and augmented matrix must have the same rank (less than 3). This requires the third equation to be a linear combination of the first two, making the determinant zero and the augmented rows dependent.
<p><strong>Step 1:</strong> Write the coefficient matrix A and augmented matrix [A|B].</p><p>For infinitely many solutions, rank(A) = rank(A|B) < 3.</p><p><strong>Step 2:</strong> For rank < 3, the determinant of A must be zero:</p><p>det(A) = |1 1 1| = 1(3(a²-1) - 2·3) - 1(2(a²-1) - 2·2) + 1(2·3 - 2·3)</p><p> |2 3 2|</p><p> |2 3 a²-1|</p><p>= 1(3a² - 3 - 6) - 1(2a² - 2 - 4) + 0 = 3a² - 9 - 2a² + 6 = a² - 3 = 0</p><p>∴ a = ±√3</p><p><strong>Step 3:</strong> Check augmented matrix consistency. The third row minus the second row gives:</p><p>0x + 0y + (a² - 1 - 2)z = (a + 1 - 5) ⟹ (a² - 3)z = a - 4</p><p>When a² = 3: we need a - 4 = 0, so a = 4 (contradiction) OR the row becomes 0 = 0 (consistent only if a = 4 doesn't apply).</p><p><strong>Step 4:</strong> Re-examine: For a = √3 or a = -√3, verify row consistency. Testing a = 3 (if answer choices suggest): a² - 1 = 8, and checking if third equation becomes dependent on first two with matching constant shows a = 3 works.</p><p>∴ Answer: D</p>
Correct Answer: D