How to calculate the rank of a matrix?
I need to calculate the rank of the matrix $A$, shown below: $$ A= \begin{bmatrix} 3 & 2 & -1\\ 2 & -3 & -5\\ -1 & -4 &- 3 \end{bmatrix} $$
I know that I need to calculate $\det(A)$ and if $\det(A) \neq 0$ then the rank will be equal to $3$, but in this case I'm required to zero-out first column of matrix $A$ using element $a_{31} = -1$.
$\endgroup$ 04 Answers
$\begingroup$The second column - first column is the last column so the rank is $<3$.
The first two colums are linearly independent so the rank is $2$.
$\endgroup$ $\begingroup$Simply use row reduction: the rank is the number of non-zero rows after you've performed row reduction:
\begin{align}
&\begin{bmatrix}
3&2&-1\\2&-3&-5\\-1&-4&-3
\end{bmatrix}\rightsquigarrow
\begin{bmatrix}
1&4&3\\3&2&-1\\2&-3&-5
\end{bmatrix}\rightsquigarrow
\begin{bmatrix}
1&4&3\\0&-10&-10\\0&-11&-11
\end{bmatrix}\\[1ex]
\rightsquigarrow&\begin{bmatrix}
1&4&3\\0&1&1\\0&-11&-11
\end{bmatrix}\rightsquigarrow
\begin{bmatrix}
1&4&3\\0&1&1\\0&0&0
\end{bmatrix}
\end{align}
Thus, the rank is $2$.
Note that it was obvious after the second step.
$\endgroup$ $\begingroup$The key word is row echelon form.
$\endgroup$ 1 $\begingroup$You can apply linear transformations to $A$ and find an upper triangular matrix. The number of non-zero lines of that matrix will give you the rank of $A$
$\endgroup$
