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If $A=\left[\begin{array}{ccc}2 & 4 & 5 \\ 4 & 8 & 10 \\ -6 & -12 & -15\end{array}\right]$. Then rank of $A$ is equal to
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$1$
$A=\left[\begin{array}{ccc}2 & 4 & 5 \\ 4 & 8 & 10 \\ -6 & -12 & -15\end{array}\right]_{3 \times 3}$
$|A|=0$, then rank cannot be 3 .
Considering a $2 \times 2$ minor, $\left[\begin{array}{ll}2 & 4 \\ 4 & 8\end{array}\right]$ its determinant is zero. Similarly considering
$\left[\begin{array}{cc}4 & 5 \\ 8 & 10\end{array}\right]$ $\left[\begin{array}{cc}4 & 8 \\ -6 & -12\end{array}\right]$ $\left[\begin{array}{cc}8 & 10 \\ -12 & 15\end{array}\right]$ $\left[\begin{array}{cc}2 & 5 \\ 4 & 10\end{array}\right]$ $\left[\begin{array}{cc}4 & 10 \\ -6 & -15\end{array}\right]$ their determinants is zero. Each rank can not be 2 . Thus rank $=1$.
$|A|=0$, then rank cannot be 3 .
Considering a $2 \times 2$ minor, $\left[\begin{array}{ll}2 & 4 \\ 4 & 8\end{array}\right]$ its determinant is zero. Similarly considering
$\left[\begin{array}{cc}4 & 5 \\ 8 & 10\end{array}\right]$ $\left[\begin{array}{cc}4 & 8 \\ -6 & -12\end{array}\right]$ $\left[\begin{array}{cc}8 & 10 \\ -12 & 15\end{array}\right]$ $\left[\begin{array}{cc}2 & 5 \\ 4 & 10\end{array}\right]$ $\left[\begin{array}{cc}4 & 10 \\ -6 & -15\end{array}\right]$ their determinants is zero. Each rank can not be 2 . Thus rank $=1$.
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