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Consider the following system of equations:

$3x + 2y = 1$

$4x + 7z = 1$

$x + y + z = 3$

$x - 2y + 7z = 0$

The number of solutions for this system is ______________

Since, equation $(2)$ - equation $(1)$ produces equation $(4)$, we have $3$ independent equations in $3$ variables, hence unique solution.

So, answer is $1.$
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I could not understand this.. Can u please elaborate more
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And solution is x=13,y=-19,z=9 only single solution
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Finding rank of 4x4 matrix is quite time consuming .But logic should be clear.

Which three equation are used to determine the answers is not mentioned.

sorry for my handwriting!

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how did you make R4 all zeros?
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if R4 is all zeros , then there is 3x3 square  matrix inside  the augmented matrix , for which the determinant is zero .. hence rank of the augmented matrix cannot be 3.
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@shaurya

rank of augmented matrix(AB) >= rank of A.
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In the pic R= R4 - R2 ..... And To be unique solution R(A|B) = R(A) =  n where n is # of variables ...

rank(Augmented Matrix) = rank(Matrix) = no of unknowns. Hence it has a unique solution
Add first two equations and you will get $7x+2y+7z=2$

remaining equations are $x+y+z=3$ and $x-2y+7z=0$

My augmented matrix

$\left[\begin{array}{ccc|c} 1&1&1&3 \\ 7&2&7&2 \\ 1&-2&7&0 \\ \end{array} \right ]$

do $R_2-7R_1 \rightarrow R_2$ and $R_3-R_1 \rightarrow R_3$

$\left[\begin{array}{ccc|c} 1&1&1&3 \\ 0&-5&0&-19 \\ 0&-3&6&-3 \\ \end{array} \right ]$

do $5R_3 - 3R_2 \rightarrow R_3$

$\left[\begin{array}{ccc|c} 1&1&1&3 \\ 0&-5&0&-19 \\ 0&0&30&42 \\ \end{array} \right ]$

your Augmented matrix has same rank as the coefficient matrix=3=number of unknowns, so only 1 solution possible.
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Can someone plz find the rank of thie matrix using row transformations.I m not able to do so.Plz help
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rank(A) = rank(AB) = n (no. of unknowns) =3
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Even i am getting rank as 4 although its not possible.
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Rank will be 4 if you solve all 4 equations together. But note that 2 rows in Echelon form will be identical.

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