1.7k views

Consider the following system of linear equations $$\left( \begin{array}{ccc} 2 & 1 & -4 \\ 4 & 3 & -12 \\ 1 & 2 & -8 \end{array} \right) \left( \begin{array}{ccc} x \\ y \\ z \end{array} \right) = \left( \begin{array}{ccc} \alpha \\ 5 \\ 7 \end{array} \right)$$ Notice that the second and the third columns of the coefficient matrix are linearly dependent. For how many values of $\alpha$, does this system of equations have infinitely many solutions?

1. $0$
2. $1$
3. $2$
4. $3$
+8

So my answer is same as answer provided by @adactive18 ji. But @Keith Kr ji is getting one more value of alpha( although i have not verified his answer). So i am thinking, Can we get more values of alpha (by using some other transformation) ?

+6
Yes, @Keith+Kr has calculated wrong value a = 1/5. You are correct
+4
+3
yes alpha is 1/5

Determinant=$0.$ Therefore apply reduction method on $(A|B)$

$R_{2}\leftarrow R_{2}-2R_{1},R_{3}\leftarrow R_{3}-0.5R_{1},R_{3}\leftarrow R_{3}-1.5R_{2},$

obtain the resultant matrix

$\large\begin{pmatrix}2&1&{-4}&a\\4&3&{-12}&5\\1&2&{-8}&7\end{pmatrix}\rightarrow\begin{pmatrix}1&0.5&{-2}&0.5a\\0&1&{-4}&5-2a\\0&0&{-0}&2+1.5a\end{pmatrix}$

or infinitely many solutions, we must have $2+1.5a=0\; i.e., a=\dfrac{-4}{3}$ so for only $1$ value of $a,$ this system has infinitely many solutions. So option (B) is correct.

edited
+1
Why not D?? Because rank of A|B will be 2 independent of the value of alpha. And r(A)=r(A|B)<unknown, system has infinte solutions. I think answer should be infinetly many
0
See r(A) is clearly 2. In order to be inf. many solutions, you have to make r(AB)=2 , and only 1 value of alpha is making r(AB)=2.
0
i think alpha can take any value.. rank of AB will always be 2.
0
If alpha takes any value then the system can become inconsistent since rank of the matrix wont be same as rank of augmented matrix.
0
How is system becoming incosistent??? Take a=3,4 or any value.
0
For a =3,4 rank of the matrix is 2 while rank of the augmented matrix is 3. And for the system to be consistent the matrix and augmented matrix must have same rank.
0
But there are two dependent colums right??? So shouldn't the rank of (A|B) also become 2??
0
Both the 3x3 submatrices of AB has determinant 0.

−12     5

−8       7

This submatrix has det not equal to 0 so we can take any value of alpha.
0

check alpha value 5/3 and 1/5 which also give infite solution so it indicate alpha have infinite solution(it is also indicated by  Rahul Jain25)

+4

@Vaishali.

Try to use the row reduced row echelon form for finding rank of matrix. Not sure but there is only 1 linearly independent column. So, only 1 value of alpha.

Better to go for row-reduced echelon form.

See here

After using that method, getting following matrix:

$\begin{bmatrix} 1 & 2 &-8 &|7 \\ 0& 1 & -4 &|\frac{23}{5} \\ 0& 0 & 0 &|\frac{14-\alpha}{3}-\frac{23}{5} \end{bmatrix}$

Hence, only 1 value.

+2

Yes , only one value of alpha and the value of alpha is 1/5  not -4/3.

Given system of equations is in form AX=B.

For infinitely many solutions here, effectively we have only two variables i.e. w=(y-4z). but they should be consistent i.e. rank(A)=rank(AB)<3(no. of variables)

Since column2 and column3 are dependent so make column3 all zero. So we got rank(A)=2

Now for the augmented matrix, AB put all zero column at last because it's of no use it will give 0 determinant value of all 3*3 submatrices.

After excluding that dependent column, we got a 3*3 matrix, solving it gives alpha= .2 this is single value.

Or see the image below

Option B ...

Note: The row rank and the column rank of a matrix A are equal. Here column rank is easy to find for A. And then for AB we must get rank 2 for having infinite solutions.

P.S. -- Ignore that "3 equation and 2 variable" in the image.

edited
+1
This answer is as per the information given in the question.
0
R u using cammer's rule ?? is there any procedure to say system is inconsistent by using determinant ??
+1
If you have two dependent equations, then for both the equations, we should not get different RHS constant values. To satisfy this, if rank(A)<n, then rank(AB)<n.