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Let $A$ be a $5\times 4$ matrix with real entries such that the space of all solutions of the linear system $AX^t = (1,2,3,4,5)^t$ is given by$\{(1+2s, 2+3s, 3+4s, 4+5s)^t :s\in \mathbb{R}\}$ where $t$ denotes the transpose of a matrix. Then what would be the rank of $A$?

Here is my attempt

Number of linearly independent solution of a non homogeneous system of linear equations is given by $n-r+1$ where n refers to number of unknowns and $r$ denotes rank of the coefficient matrix $A$. Based on this fact, we may write $n-r+1 =1 $. Since there seems only one linearly independent solution (Here i am confused). Am i right? or How can i do it right? thanks

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thanks for editing sir –  srijan May 12 '12 at 4:35
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3 Answers

up vote 2 down vote accepted

The number of leading variables plus the number of parameters in a consistent system equals the number of unknowns. Here you have a system with $4$ unknowns and one parameter, so the number of "leading variables" is $3$. The number of leading variables equals the rank of the matrix, so the rank is $3$.

(It makes no sense to say "there is only one linearly independent solution"; every nonzero solution, by itself, is linearly independent; what you have "only one" of is parameters.)

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What if here have consistent homogeneous system of linear equation? –  srijan May 12 '12 at 4:50
    
Things are clear to now. Thank you so much sir for clearing my doubt. –  srijan May 12 '12 at 5:01
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Let $b=(1,2,3,4,5)^T$, $x_0 = (1,2,3,4)^T$, and $v_0 = (2,3,4,5)^T$. We know $A x_0 = b$, and that $x'$ is another solution iff $x'-x_0 \in \mathbb{sp}\{v_0\}$. Consequently we have $\ker A = \mathbb{sp}\{v_0\}$, and hence $\dim \ker A = 1$. Since $A : \mathbb{R}^4 \rightarrow \mathbb{R}^5$, we have $$\dim \ker A + \mathrm{rank} A = 4.$$ Hence the rank is 3.

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Rank theorem says that if $A$ be the coefficient matrix of a consistent system of linear equations with $n$ variables then

number of free variables (parameters) = $n$ - $rank(A)$

By using this we have $1 = 4 - rank(A)$

Thus $rank(A) = 3$ thanks to Dr Arturo sir for clearing my doubt.

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