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Mathematics Question on Matrices and Determinants

If the system of equations x+4y−z=λx + 4y - z = \lambda, 7x+9y+μz=−37x + 9y + \mu z = -3, 5x+y+2z=−15x + y + 2z = -1 has infinitely many solutions, then (2μ+3λ)(2\mu + 3\lambda) is equal to:

A

2

B

-3

C

3

D

-2

Answer

-3

Explanation

Solution

Given the system of equations:

x+4y−z=λ,7x+9y+μz=−3,5x+y+2z=−1,x + 4y - z = \lambda, \quad 7x + 9y + \mu z = -3, \quad 5x + y + 2z = -1,

we form the coefficient matrix: Δ=∣14−1 79μ 512∣.\Delta = \begin{vmatrix} 1 & 4 & -1 \\\ 7 & 9 & \mu \\\ 5 & 1 & 2 \end{vmatrix}.

For the system to have infinitely many solutions, the determinant of this matrix must be zero: Δ=1⋅∣9μ 12∣−4⋅∣7μ 52∣−1⋅∣79 51∣.\Delta = 1 \cdot \begin{vmatrix} 9 & \mu \\\ 1 & 2 \end{vmatrix} - 4 \cdot \begin{vmatrix} 7 & \mu \\\ 5 & 2 \end{vmatrix} - 1 \cdot \begin{vmatrix} 7 & 9 \\\ 5 & 1 \end{vmatrix}.

Calculating each term: Δ=1⋅(18−μ)−4⋅(14−5μ)−(7−45).\Delta = 1 \cdot (18 - \mu) - 4 \cdot (14 - 5\mu) - (7 - 45).

Simplifying: Δ=18−μ−4(14−5μ)−(−38).\Delta = 18 - \mu - 4(14 - 5\mu) - (-38). Δ=18−μ−56+20μ+38.\Delta = 18 - \mu - 56 + 20\mu + 38. Δ=19μ+0.\Delta = 19\mu + 0.

For the determinant to be zero (infinitely many solutions): 19μ=0  ⟹  μ=0.19\mu = 0 \implies \mu = 0. Substituting μ=0\mu = 0 into the augmented matrix and setting Δx=Δy=Δz=0\Delta_x = \Delta_y = \Delta_z = 0,

we find: Δx=∣λ4−1 −390 −112∣=0.\Delta_x = \begin{vmatrix} \lambda & 4 & -1 \\\ -3 & 9 & 0 \\\ -1 & 1 & 2 \end{vmatrix} = 0.

Expanding: λ⋅∣90 12∣−4⋅∣−30 −12∣−1⋅∣−39 −11∣.\lambda \cdot \begin{vmatrix} 9 & 0 \\\ 1 & 2 \end{vmatrix} - 4 \cdot \begin{vmatrix} -3 & 0 \\\ -1 & 2 \end{vmatrix} - 1 \cdot \begin{vmatrix} -3 & 9 \\\ -1 & 1 \end{vmatrix}.

Calculating each term: λ(18)−4(−6)−(−12)=0.\lambda (18) - 4(-6) - (-12) = 0. Simplifying: 18λ+24+12=0.18\lambda + 24 + 12 = 0. 18λ=−36  ⟹  λ=−1.18\lambda = -36 \implies \lambda = -1.

Finally, calculating 2μ+3λ2\mu + 3\lambda: 2μ+3λ=2(0)+3(−1)=−3.2\mu + 3\lambda = 2(0) + 3(-1) = -3.

Therefore: −3.-3.