eRank

Example Find the rank of the matrix $A$

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$\displaystyle A = \left(\begin{array}{rrrr} 1&-2&5&3\\ 2&3&1&-1\\ 3&8&-3&-5 \end{array}\right)$

Answer Note that the rank of matrix is the same as the nonzero rows of the reduced matrix.

$\displaystyle A$	$\displaystyle =$	$\displaystyle \left(\begin{array}{rrrr} 1&-2&5&3\\ 2&3&1&-1\\ 3&8&-3&-5 \end{... ...left(\begin{array}{rrrr} 1&-2&5&3\\ 0&7&-9&-7\\ 0&14&-18&4 \end{array}\right)$
	$\displaystyle \stackrel{\begin{array}{cc} {}^{\frac{1}{7} \times R_{2}}\\ {}^{-2R_{2} + R_{3}} \end{array}}{\longrightarrow}$	$\displaystyle \left(\begin{array}{rrrr} 1&-2&5&3\\ 0&1&\frac{-9}{7}&-1\\ 0&0&... ...rrrr} 1&0&\frac{17}{7}&1\\ 0&1&\frac{-9}{7}&-1\\ 0&0&0&0 \end{array}\right) .$

Then the number of nonzero rows of the reduced matrix $A_{R}$ is $2$

. Thus ${\rm rank}(A) = 2$ . Furthermore, the row vector $\displaystyle{(1,0,\frac{17}{7},1)}$ of $A_{R}$ and the row vector $\displaystyle{(0,1,\frac{-9}{7},-1)}$ are bases of vector space $A$

. Therefore, the dimension of the row space is $2$

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