Matrices & Determinants
System of Linear Equations
Grade 12

Question:

<p>The system of linear equations<br>\(x + y + z = 6\)<br>\(4x + \lambda y - \lambda z = \lambda - 2\)<br>\(3x + 2y - 4z = -5\)<br>has a solution. Then \(\lambda\) equals</p>
<p>1</p>
<p>2</p>
<p>3</p>
<p>4</p>

Step-by-Step Solution

Key Concept: For a system of linear equations to have a solution, the augmented matrix must have the same rank as the coefficient matrix. Use the condition that the determinant of the coefficient matrix equals zero (for infinite solutions) or apply consistency conditions.
<p><strong>Step 1:</strong> Write the coefficient matrix and apply Gaussian elimination or determinant method.</p><p>Coefficient matrix A = [1, 1, 1; 4, λ, -λ; 3, 2, -4]</p><p><strong>Step 2:</strong> For the system to have a solution, use row operations on the augmented matrix [A|B]:</p><p>R₂ - 4R₁: [0, λ-4, -λ-4 | λ-26]</p><p>R₃ - 3R₁: [0, -1, -7 | -23]</p><p><strong>Step 3:</strong> From the third row: -y - 7z = -23, so y + 7z = 23</p><p><strong>Step 4:</strong> From the second row: (λ-4)y - (λ+4)z = λ-26</p><p><strong>Step 5:</strong> For consistency, substitute y = 23 - 7z into the second row equation and solve for λ:</p><p>(λ-4)(23-7z) - (λ+4)z = λ-26</p><p>Expanding: 23λ - 92 - 7λz + 28z - λz - 4z = λ - 26</p><p>23λ - 92 - 8λz + 24z = λ - 26</p><p>22λ - 8λz + 24z = 66</p><p><strong>Step 6:</strong> For this to hold for the system to be consistent, equating coefficients of z: 24 - 8λ = 0, giving λ = 3</p><p>Verify: 22(3) - 66 = 0 ✓</p><p>∴ Answer: λ = 3 (Option D)</p>
Correct Answer: D

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