Let A be $5\times4$ matrix with real entries such that space of all solutions of the system $AX^{t}=[1,2,3,4,5]^{t}$ is given by $\{[1+2s,2+3s,3+4s,4+5s]^{t}:s\in \mathbb{R}\}$ then what is the rank of the matrix $A$.?
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2 Answers
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Since the null space is one dimensional (based on the fact that the solution set has one parameter $s$), by rank-nullity theorem, the rank will be $3$.
added explanation:
The solution set for the given for the given system $Ax=b$ can be written as $$\begin{bmatrix} 1\\2\\3\\4 \end{bmatrix} +s\begin{bmatrix} 2\\3\\4\\5 \end{bmatrix} $$ The first vector is a particular solution, whereas the second vector represents the solution for the homogeneous system $Ax=0$. This is how you know that null space is spanned by ONE vector, hence is one dimensional.
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$\begingroup$ How null space has dimension one? $\endgroup$ Commented May 11, 2015 at 16:22
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$\begingroup$ how exactly we can say that null space has one dim...please explane a little more. $\endgroup$ Commented May 11, 2015 at 16:26
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$\begingroup$ @YOGESH please see added explanation. $\endgroup$– Anurag ACommented May 11, 2015 at 16:30
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$\begingroup$ my confusion is that how exactly we can say that last expression is solution of corresponding hom system?/ $\endgroup$ Commented May 11, 2015 at 16:44
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You can find the rank of a matrix by using elementary row operations to put the matrix into row-reduced echelon form and then, the rank will simply be equal to the number of non-zero rows. Hope that helps you
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$\begingroup$ But matrix is not given in exactly... $\endgroup$ Commented May 11, 2015 at 16:24