Search any question & find its solution
Question:
Answered & Verified by Expert
The rank of the matrix $A=\left[\begin{array}{cccc}1 & 1 & 1 & 3 \\ 2 & 2 & -1 & 3 \\ 1 & 1 & -1 & 1\end{array}\right]$ is
Options:
Solution:
1692 Upvotes
Verified Answer
The correct answer is:
2
The given matrix is $A=\left[\begin{array}{cccc}1 & 1 & 1 & 3 \\ 2 & 2 & -1 & 3 \\ 1 & 1 & -1 & 1\end{array}\right]$
$\mathrm{R}_2 \rightarrow 2 \mathrm{R}_1-\mathrm{R}_2, \mathrm{R}_3 \rightarrow \mathrm{R}_1-\mathrm{R}_3$
$A=\left[\begin{array}{llll}1 & 1 & 1 & 3 \\ 0 & 0 & 3 & 3 \\ 0 & 0 & 2 & 2\end{array}\right]$
$\mathrm{R}_2 \rightarrow \frac{\mathrm{R}_2}{3}, R_3 \rightarrow \frac{\mathrm{R}_3}{2}$
$\approx\left[\begin{array}{llll}1 & 1 & 1 & 3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1\end{array}\right]$
Rank $(A)=$ No of non-zero rows $=2$
$\approx\left[\begin{array}{llll}1 & 1 & 1 & 3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0\end{array}\right]$
$\mathrm{R}_2 \rightarrow 2 \mathrm{R}_1-\mathrm{R}_2, \mathrm{R}_3 \rightarrow \mathrm{R}_1-\mathrm{R}_3$
$A=\left[\begin{array}{llll}1 & 1 & 1 & 3 \\ 0 & 0 & 3 & 3 \\ 0 & 0 & 2 & 2\end{array}\right]$
$\mathrm{R}_2 \rightarrow \frac{\mathrm{R}_2}{3}, R_3 \rightarrow \frac{\mathrm{R}_3}{2}$
$\approx\left[\begin{array}{llll}1 & 1 & 1 & 3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1\end{array}\right]$
Rank $(A)=$ No of non-zero rows $=2$
$\approx\left[\begin{array}{llll}1 & 1 & 1 & 3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0\end{array}\right]$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.