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GATE CS 1998 Q2.1 — Linear Algebra: Matrices, Determinants, System of Linear Equations

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The rank of the matrix given below is:

[1487003042313122421]\begin{bmatrix} 1 &4 &8 &7\\ 0 &0& 3 &0\\ 4 &2& 3 &1\\ 3 &12 &24 &21 \end{bmatrix}

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Answer: A

To find the rank of the matrix AA, we perform elementary row operations:

A=[1487003042313122421]A = \begin{bmatrix} 1 & 4 & 8 & 7 \\ 0 & 0 & 3 & 0 \\ 4 & 2 & 3 & 1 \\ 3 & 12 & 24 & 21 \end{bmatrix}

Observe that the fourth row R4R_4 is a scalar multiple of the first row R1R_1, specifically R4=3R1R_4 = 3R_1. Applying the row operation R4→R4−3R1R_4 \to R_4 - 3R_1 gives:

[1487003042310000]\begin{bmatrix} 1 & 4 & 8 & 7 \\ 0 & 0 & 3 & 0 \\ 4 & 2 & 3 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}

To determine if the remaining three rows are linearly independent, we compute the determinant of the 3×33 \times 3 submatrix formed by the first three columns:

M=[148003423]M = \begin{bmatrix} 1 & 4 & 8 \\ 0 & 0 & 3 \\ 4 & 2 & 3 \end{bmatrix}

Expanding along the second row:

det⁡(M)=−3∣1442∣=−3(2−16)=42≠0\det(M) = -3 \begin{vmatrix} 1 & 4 \\ 4 & 2 \end{vmatrix} = -3(2 - 16) = 42 \neq 0

Since there exists a non-zero minor of order 3, the rank of the matrix is 3.

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