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GO Classes DPP | GATE CS | Linear Algebra | Rank-Nullity Theorem

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5 5 votes
Suppose that $Q$ is a $6\times 8$ matrix such that every solution of the equation $Qx=0$ is of the form $x=s(1,0,-1,2,0,3,1,-2)^T+t(0,1,2,-1,1,0,4,3)^T+u(2,-1,0,1,3,-2,0,5)^T$, where $s,t,u \in \mathbb{R}$, and the three vectors are linearly independent. The rank of $Q$ is ______.

2 Answers

1 1 vote

Since every solution of $Qx=0$ is a linear combination of $3$ linearly independent vectors, the nullity of $Q$ is $3$.

By the rank-nullity theorem,

$\operatorname{rank}(Q)+\operatorname{nullity}(Q)=8$.

So,

$\operatorname{rank}(Q)+3=8$,

hence

$\operatorname{rank}(Q)=5$.

Therefore, the answer is $5$.

1 1 vote

For ( Ax = 0 ), if ( {x1.......,xk} ) are linearly independent solutions, then they form linearly independent vectors in the null space of (A). This is equivalent to eigenvectors corresponding to eigenvalue (λ=0).

Nullity =3

So rank =8-3=5
So rank =5

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