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+17 votes

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 ______________

asked in Linear Algebra by Veteran (96.1k points) | 3.1k views
Could we have done it like this:

step 1:

we first  find the solution of the first 3 equations.

step 2:

we then substitute the solution in the 4th equation to see iff it satisfies or not.

is this procedure correct?

5 Answers

+25 votes
Best answer
Since, equation $(2)$ - equation $(1)$ produces equation $(4)$, we have $3$ independent equations in $3$ variables, hence unique solution.

So, answer is $1.$
answered by Boss (11.4k points)
edited by
I could not understand this.. Can u please elaborate more
And solution is x=13,y=-19,z=9 only single solution
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.
+18 votes

sorry for my handwriting!

answered by Active (1.8k points)
how did you make R4 all zeros?
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.

rank of augmented matrix(AB) >= rank of A.

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

+4 votes
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.
answered by Boss (24.6k points)
should we first always reduce the equations to no. of variables in equation and then solve augmented matrix ?

@ Is it fruitful or can I use it for general method as satbir said in his comment. Can it harm solution for any example??

0 votes
rank(Augmented Matrix) = rank(Matrix) = no of unknowns. Hence it has a unique solution
answered by Loyal (9.3k points)
–1 vote
Can someone plz find the rank of thie matrix using row transformations.I m not able to do so.Plz help
answered by Active (1k points)
rank(A) = rank(AB) = n (no. of unknowns) =3
Even i am getting rank as 4 although its not possible.
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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