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The rank of the following $(n+1) \times (n+1)$ matrix, where $a$ is a real number is $$\begin{bmatrix} 1 & a & a^2 & \dots & a^n \\ 1 & a & a^2 & \dots & a^n \\ \vdots & \vdots & \vdots & \: & \vdots \\ \vdots & \vdots & \vdots & \: & \vdots \\ 1 & a & a^2 & \dots & a^n \end{bmatrix}$$

1. $1$
2. $2$
3. $n$
4. Depends on the value of $a$
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$\begin{bmatrix} 1 & a & a^2 & \dots & a^n \\ 1 & a & a^2 & \dots & a^n \\ \vdots & \vdots & \vdots & \: & \vdots \\ \vdots & \vdots & \vdots & \: & \vdots \\ 1 & a & a^2 & \dots & a^n \end{bmatrix}$

$R_2 \rightarrow R_2-R_1 , R_3 \rightarrow R_3-R_1 , R_4 \rightarrow R_4-R_1,$ and so on

$\begin{bmatrix} 1 & a & a^2 & \dots & a^n \\ 0 & 0 & 0 & \dots & 0 \\ \vdots & \vdots & \vdots & \: & \vdots \\ \vdots & \vdots & \vdots & \: & \vdots \\ 0 & 0 &0 & \dots &0 \end{bmatrix}$

Rank of the Matrix $=1$

Hence, option (A) 1 is the correct choice.

by Boss
edited
+1
only 1 linearly independent row hence rank=1
Ans is A.

we can eliminate all other rows using row 1. in the last only 1 row will be left.

rank = no of non zero rows = 1
by Loyal
0
Good one.

Thanks.
All the rows of the given matrix are same. So the matrix has only one independent row.
Rank of a matrix = No. of independent row (or columns) of the matrix.

i.e. Ans- A