GATE CSE 2014 | Set 1 | Question: 4

Question

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 ______________

Comments

@goku4199 isn't 0|66 indicate no-solution

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Thanks for response @nobodysomebody I have edited

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Easy Trick ( Most of the time I don't recomend ) : https://gateoverflow.in/1757/gate-cse-2014-set-1-question-4?show=532675#a532675

Answer 1

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

So, answer is $1.$

Comments

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.

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Form a (A|B) matrix and then apply row elementary operation between row 4 and row 2. When you subtract it you will get 3 independent equations. 1st and 3rd row will have different equation and 2nd and 4th row will have same equation. So,There are 3 independent equation. Hence, unique solution which is 1. @anchitjindal07

Answer 2

sorry for my handwriting!

Comments

@shaurya

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

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

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One more approach is :- 

Step 1 :-  R1 ↔ R4  and R2 ↔ R3

Step 2 :-  R2 - R1 ,  R3 - 4R1 ,  R4 - 3R1       ............after this, u will get R3 and R4 as identical

Step 3 :-  R4 - R3                                           .............and done with the answer

 

We can state, it is unique as  rank(A) = rank([A∣B]) = 3

Unique means only 1 solution possible,

Hence, Ans (1)

Answer 3

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.

Comments

should we first always reduce the equations to no. of variables in equation and then solve augmented matrix ?

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

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@Satbir Does an overdetermined system(No of equations> no of variables) always have a unique solution?

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NO.

x + y =2

2x+2y =4

6x + 6y = 12

These will have infinite solution.

Answer 4

It is having UNIQUE Solution .

 

Comments

In this question many people get trapped

Augmented matrix for given equations is:

3   4   1   1

2   0   1   -1

0   7   1   7

1   1   3   0

After converting into echelon form we obtain:

3   0   0   0

2   -8   0   0

0   21   45   0

1   -1   63   0

After seeing [00..|0] we think that there are infinitely many solutions. But the catch here is there is no free variable.

So there will be unique solution 

Method-2: 

By using rank

Rank[A] = 3 

Rank[A|b] = 3 

So rank=number of columns

There will be unique solution

Answer is 1

Answer 5

Can someone plz find the rank of thie matrix using row transformations.I m not able to do so.Plz help

Comments

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.

Answer 6

rank(Augmented Matrix) = rank(Matrix) = no of unknowns. Hence it has a unique solution

Comments

Don't you think that this might be a very lengthy solution? although it is the right way of getting answers for these kind of questions