Consider the following system of linear equations: 2x - y + 3z = 1 4x - 2y + 6z = 2 -6x + 3y - 9z = alpha Notice that the second and the third columns of the coefficient matrix are linearly dependent. For how many values of alpha does this system of equations have infinitely many solutions? A. 0 B. 1 C. 2 D. 3

GATE 2003 · Engineering Mathematics · System of Equations · medium

Answer: Exactly 1 value of alpha (alpha = -3) gives infinitely many solutions. Answer: B.

1. Identify the scalar relationship between rows: Row 1: [2, -1, 3 | 1]. Row 2: [4, -2, 6 | 2] = 2 * Row 1. This is satisfied since 2*1=2. Row 3: [-6, 3, -9 | alpha] = -3 * Row 1 in the coefficient part (-3*[2,-1,3] = [-6,3,-9]). For the augmented column: -3 * 1 = -3. So alpha must equal -3.
2. State the condition for infinite solutions: Infinite solutions require rank(A) = rank([A|b]). Here rank(A) = 1. For rank([A|b]) = 1, all rows of [A|b] must be proportional. Row 3 of [A|b] being proportional to Row 1 requires alpha = -3. For any other alpha, rank([A|b]) = 2 > rank(A) = 1, meaning no solution.