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Official PYQ · 1995Q1.24Linear Algebra: Matrices, Determinants, System of Linear EquationsMCQ1 mark

GATE CS 1995 Q1.24 — Linear Algebra: Matrices, Determinants, System of Linear Equations

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The rank of the following (n+1)×(n+1)(n+1) \times (n+1) matrix, where aa is a real number is [1aa2…an1aa2…an⋮⋮⋮ ⋮⋮⋮⋮ ⋮1aa2…an]\begin{bmatrix} 1 & a & a^2 & \dots & a^n \\ 1 & a & a^2 & \dots & a^n \\ \vdots & \vdots & \vdots & \: & \vdots \\ \vdots & \vdots & \vdots & \: & \vdots \\ 1 & a & a^2 & \dots & a^n \end{bmatrix}

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Answer: A

Let the given (n+1)×(n+1)(n+1) \times (n+1) matrix be AA:

A=[1aa2…an1aa2…an⋮⋮⋮⋮1aa2…an]A = \begin{bmatrix} 1 & a & a^2 & \dots & a^n \\ 1 & a & a^2 & \dots & a^n \\ \vdots & \vdots & \vdots & & \vdots \\ 1 & a & a^2 & \dots & a^n \end{bmatrix}

Observe that all the rows of the matrix AA are identical. We can perform the following elementary row operations to simplify the matrix:

Ri→Ri−R1for i=2,3,…,n+1R_i \to R_i - R_1 \quad \text{for } i = 2, 3, \dots, n+1

Applying these operations reduces the matrix to:

[1aa2…an000…0⋮⋮⋮⋮000…0]\begin{bmatrix} 1 & a & a^2 & \dots & a^n \\ 0 & 0 & 0 & \dots & 0 \\ \vdots & \vdots & \vdots & & \vdots \\ 0 & 0 & 0 & \dots & 0 \end{bmatrix}

The rank of a matrix is the number of non-zero rows in its row echelon form.

  • The first row is [1,a,a2,…,an][1, a, a^2, \dots, a^n].
  • Since the first element of this row is 11, the row is non-zero for any real value of aa (even if a=0a = 0).

Thus, there is exactly 11 linearly independent row. Therefore, the rank of the matrix is always 11, independent of the value of aa.

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