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If the ranks of the matrices $A=\left[\begin{array}{ccc}1 & 0 & 1 \\ 2 & 1 & 2 \\ 1 & 0 & -1\end{array}\right]$ and $B=\left[\begin{array}{cccc}1 & 2 & 3 & 4 \\ 2 & 4 & 6 & -8\end{array}\right]$ are $r_1$ and $r_2$ respectively then $r_1-r_2=$
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The correct answer is:
1
$A=\left[\begin{array}{ccc}1 & 0 & 1 \\ 2 & 1 & 2 \\ 1 & 0 & -1\end{array}\right]$
$$
|\mathrm{A}|=1(-1-0)+1(0-1)=-1-1=-2=0
$$
$\therefore$ Rank of $\mathrm{A}\left(\mathrm{r}_1\right)=3$
$$
\begin{aligned}
& \mathrm{B}=\left[\begin{array}{cccc}
1 & 2 & 3 & 4 \\
2 & 4 & 6 & -8
\end{array}\right] \\
& \because\left|\begin{array}{cc}
3 & 4 \\
6 & -8
\end{array}\right|=-24-24=-48 \neq 0
\end{aligned}
$$
$\therefore$ Rank of $B\left(r_2\right)=2$
Now, $\mathrm{r}_1-\mathrm{r}_2=3-2=1$
$$
|\mathrm{A}|=1(-1-0)+1(0-1)=-1-1=-2=0
$$
$\therefore$ Rank of $\mathrm{A}\left(\mathrm{r}_1\right)=3$
$$
\begin{aligned}
& \mathrm{B}=\left[\begin{array}{cccc}
1 & 2 & 3 & 4 \\
2 & 4 & 6 & -8
\end{array}\right] \\
& \because\left|\begin{array}{cc}
3 & 4 \\
6 & -8
\end{array}\right|=-24-24=-48 \neq 0
\end{aligned}
$$
$\therefore$ Rank of $B\left(r_2\right)=2$
Now, $\mathrm{r}_1-\mathrm{r}_2=3-2=1$
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