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Let $P = \begin{bmatrix}1 & 1 & -1 \\2 & -3 & 4 \\3 & -2 & 3\end{bmatrix}$  and $Q = \begin{bmatrix}-1 & -2 &-1 \\6 & 12 & 6 \\5 & 10 & 5\end{bmatrix}$ be two matrices.

Then the rank of $ P+Q$ is ___________ .
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$P +Q = \begin{bmatrix}0 & -1 &-2 \\8 & 9 & 10 \\8 & 8 & 8\end{bmatrix}$

$\det(P+Q) = 0$, So Rank cannot be $3$, but there exists a $2*2$ submatrix such that determinant of submatrix is not $0$.

So, $\text{Rank}(P+Q) = 2$
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$P +Q = \begin{bmatrix}0 & -1 &-2 \\8 & 9 & 10 \\8 & 8 & 8\end{bmatrix}$

Applying $R_2 \leftarrow R_2+R_1$, we get

$P +Q = \begin{bmatrix}0 & -1 &-2 \\8 & 8 & 8 \\8 & 8 & 8\end{bmatrix}$

Since there are only $2$ independent rows $\implies Rank(P+Q) = 2$
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