# Basic Maths - Grand Test 1 -Q7

Rank of the following matrix is ____

$\begin{bmatrix} 6 & 4& 10 & 16 \\ 1 & 2 & 3 & 4\\ 3 & 6 & 9 & 12\\ 8 & -1 & 7 & 15 \end{bmatrix}$

$(A). 4$

$(B). 3$

$(C). 2$

$(D). 1$

reshown Jun 23

+1 vote

Applying $R_2\rightarrow 6R_2-R_1;R_3\rightarrow 2R_3-R_1;R_4\rightarrow 6R_4-8R_1;$

$\begin{bmatrix} 6 &4 &10 &16 \\ 0& 8 &8 &8 \\ 0& 8 &8 &8 \\ 0& -38 &-38 &-38 \end{bmatrix}$

$R_3\rightarrow R_3-R_2;R_4\rightarrow 8R_4-38R_2;$

$\begin{bmatrix} 6 &4 &10 &16 \\ 0& 8 &8 &8 \\ 0& 0 &0 &0 \\ 0& 0 &0 &0 \end{bmatrix} \rightarrow \text{Echelon form}$

Rank = no. of nonzero rows = 2

answered Jun 24 by (21,290 points)
Sir please clear this doubt. I have read that when rows or columns are proportional then rank of the matrix is equal to 1. This means each and every row or column should be proportional to each other? Because i mislead this question by seeing that R2 and R3 are proportional and marked ans as 1.
can u give reference of that as well as, please tell which rows or column u r talking about.
1 2 3 |  Sorry, i misunderstood the concept and realized now that every row should
2 4 6 |  be proportional then only rank=1 possible.
3 6 9 |