Consider the following system of linear equations : $$2x_1 - x_2 + 3x_3 = 1$$ $$3x_1 + 2x_2 + 5x_3 = 2$$ $$-x_1+4x_2+x_3 = 3$$ The system of equations has
@Angkit rank(r)>n this case will never arise
@Jhaiyam then I must ask you how can you numbers of stars in the sky:) LOL
determinant is only defined for square matrix and augmented matrix is not a square matrix.
Determinant of matrix =14 which is non zero
If The determinant of the coefficient matrix is non zero then definitely the system of given equation has a unique solution
so option B
Method-1:
2
3
-1
5
1
In this matrix we can clearly see that three columns are Linearly independent and 3 L.I columns in R3 space
So It will fill the space and Ax=b
So there will be unique solution
If the column space is filled and Ax=0 then there will be a trivial solution
Method-2: Converting into echelon form
After converting into echelon form we obtain augmented matrix as
0
7
32
46
We can clear see all the columns have pivot and there is no [00..00|b] form
Method-3: By using rank
rank[A] =3 and rank[A|b]=3 and number of columns =3
Therefore unique solution is possible
Answer is option-B
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