GATE Overflow Test Series | Linear Algebra | Test 1 | Question: 2

Question

The rank of the matrix $A = \begin{bmatrix}
1 & 0 & 2 & 1\\
0 & 2 & 4 & 2\\
0 & 2 & 2 & 1
\end{bmatrix}$ is ________

Answer 1

Given that $,A = \begin{bmatrix}
 1 & 0 & 2 & 1 \\
 0 & 2 & 4 & 2 \\
 0 & 2 & 2 & 1
\end{bmatrix}$

We use elementary row operations. $R_{3} \rightarrow R_{3} - R_{2}.$

$\implies A = \begin{bmatrix}
 1 & 0 & 2 & 1 \\
 0 & 2 & 4 & 2 \\
 0 & 0 & -2 & -1
\end{bmatrix}$

Since the echelon form has pivots in the first three columns, $A$ has rank $rk(A) = 3.$ The first three columns of $A$ are linearly independent.

$\text{(OR)}$

The maximal minors have order $3,$ and we found that the one obtained by deleting the last column is $-4 \neq 0.$ Hence $rk(A) = 3.$

So, the correct answer is $3.$