Matrices & Determinants
System of Linear Equations
Grade 12
Question:
<p>If the system of linear equations <math>x + y + z = 6</math>, <math>x + 2y + 3z = 10</math>, <math>3x + 2y + \lambda z = m</math> has more than two solutions, then <math>m - \lambda^2</math> is equal to __________.</p>
Step-by-Step Solution
Key Concept: A system of linear equations has more than two solutions (infinitely many solutions) if and only if the system is consistent and the rank of the coefficient matrix equals the rank of the augmented matrix, which is less than the number of variables. This occurs when the three equations are linearly dependent.
<p><strong>Step 1:</strong> Write the augmented matrix for the system:</p><p>$$\begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 1 & 2 & 3 & | & 10 \\ 3 & 2 & \lambda & | & m \end{bmatrix}$$</p><p><strong>Step 2:</strong> Perform row operations. Subtract Row 1 from Row 2:</p><p>$$R_2 - R_1: \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 2 & | & 4 \\ 3 & 2 & \lambda & | & m \end{bmatrix}$$</p><p><strong>Step 3:</strong> Subtract 3 times Row 1 from Row 3:</p><p>$$R_3 - 3R_1: \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 2 & | & 4 \\ 0 & -1 & \lambda - 3 & | & m - 18 \end{bmatrix}$$</p><p><strong>Step 4:</strong> Add Row 2 to Row 3:</p><p>$$R_3 + R_2: \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 2 & | & 4 \\ 0 & 0 & \lambda - 1 & | & m - 14 \end{bmatrix}$$</p><p><strong>Step 5:</strong> For infinitely many solutions (more than two solutions), the system must have rank < 3. This requires the third row to be all zeros:</p><p>$$\lambda - 1 = 0 \quad \text{and} \quad m - 14 = 0$$</p><p>$$\therefore \lambda = 1 \quad \text{and} \quad m = 14$$</p><p><strong>Step 6:</strong> Calculate $m - \lambda^2$:</p><p>$$m - \lambda^2 = 14 - (1)^2 = 14 - 1 = 13$$</p><p><strong>∴ Answer: 13</strong></p>
Correct Answer: 13