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Official PYQ · 2003Q41Linear Algebra: Matrices, Determinants, System of Linear EquationsMCQ1 mark

GATE CS 2003 Q41 — Linear Algebra: Matrices, Determinants, System of Linear Equations

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Consider the following system of linear equations (21−443−1212−8)(xyz)=(α57)\left( \begin{array}{ccc} 2 & 1 & -4 \\ 4 & 3 & -12 \\ 1 & 2 & -8 \end{array} \right) \left( \begin{array}{ccc} x \\ y \\ z \end{array} \right) = \left( \begin{array}{ccc} \alpha \\ 5 \\ 7 \end{array} \right) Notice that the second and the third columns of the coefficient matrix are linearly dependent. For how many values of α\alpha, does this system of equations have infinitely many solutions?

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Answer: B

To find the number of values of α\alpha for which the system has infinitely many solutions, we perform row operations on the augmented matrix [A∣B][A \mid B]:

[A∣B]=(21−4∣α43−12∣512−8∣7)[A \mid B] = \begin{pmatrix} 2 & 1 & -4 & \mid & \alpha \\ 4 & 3 & -12 & \mid & 5 \\ 1 & 2 & -8 & \mid & 7 \end{pmatrix}

Swap the first and third rows (R1↔R3R_1 \leftrightarrow R_3) to simplify calculations:

(12−8∣743−12∣521−4∣α)\begin{pmatrix} 1 & 2 & -8 & \mid & 7 \\ 4 & 3 & -12 & \mid & 5 \\ 2 & 1 & -4 & \mid & \alpha \end{pmatrix}

Apply row operations R2←R2−4R1R_2 \leftarrow R_2 - 4R_1 and R3←R3−2R1R_3 \leftarrow R_3 - 2R_1:

(12−8∣70−520∣−230−312∣α−14)\begin{pmatrix} 1 & 2 & -8 & \mid & 7 \\ 0 & -5 & 20 & \mid & -23 \\ 0 & -3 & 12 & \mid & \alpha - 14 \end{pmatrix}

Now, perform R3←R3−35R2R_3 \leftarrow R_3 - \frac{3}{5}R_2 to eliminate the second element in the third row:

(12−8∣70−520∣−23000∣α−14−35(−23))\begin{pmatrix} 1 & 2 & -8 & \mid & 7 \\ 0 & -5 & 20 & \mid & -23 \\ 0 & 0 & 0 & \mid & \alpha - 14 - \frac{3}{5}(-23) \end{pmatrix}

Simplifying the last entry in the augmented column:

α−14+695=α−705+695=α−15\alpha - 14 + \frac{69}{5} = \alpha - \frac{70}{5} + \frac{69}{5} = \alpha - \frac{1}{5}

For the system to have infinitely many solutions, the rank of the coefficient matrix AA must equal the rank of the augmented matrix [A∣B][A \mid B], and both must be less than the number of variables (3). Since rank(A)=2\text{rank}(A) = 2, we must have rank([A∣B])=2\text{rank}([A \mid B]) = 2. This requires:

α−15=0  ⟹  α=15\alpha - \frac{1}{5} = 0 \implies \alpha = \frac{1}{5}

Thus, there is exactly 1 unique value of α\alpha for which the system has infinitely many solutions.

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