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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. $3$
+9 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
0

Given this type of qsn. 1st we've to check whether |A| = 0 or not. as it's a Ax = b format.

Now here |A| = 0 implies there is infintely many solutions or no solution exists.How to differentiate both of them, I mean what indicates that there is infinitely many solutions & what indicates that there is no solution exists.

Since the second and third columns of the coefficient matrix are linearly deoendent, determis $0.$ So, the system of equations either has infinitely many solutions (if they are consistent) or no solution. To check for consistency, we apply reduction method on $(A\mid B)$

$R_{2}\leftarrow R_{2}-2R_{1},R_{3}\leftarrow R_{3}-0.5R_{1},R_{3}\leftarrow R_{3}-1.5R_{2}, R_1 \leftarrow R_1/2$

obtain the resultant matrix

$\large\begin{pmatrix}2&1&{-4}&\alpha\\4&3&{-12}&5\\1&2&{-8}&7\end{pmatrix}\rightarrow\begin{pmatrix}1&0.5&{-2}&0.5\alpha\\0&1&{-4}&5-2\alpha\\0&0&{-0}& -0.5+2.5\alpha\end{pmatrix}$

or infinitely many solutions, we must have $-0.5+2.5\alpha =0\; i.e., \alpha=\dfrac{1}{5}.$ So, for only $1$ value of $\alpha,$ this system has infinitely many solutions. So, option (B) is correct.

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+3
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.
+2
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.

+6

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

+1
Vaishali, it doesn't have only 2 3*3 submatrix.it actually has 4 and all of them ought to be zero. You have only checked for two. The matrix where you leave out third column is the one which will go to zero only when alpha = 0.2
0
@Sushant

0
You can't get rid of $\alpha$  in the last row of reduced augmented matrix..

One small doubt--    $|A|=0$  is this the only significance of giving 2nd and 3rd column are linearly dependent?

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.

For infinitely many solution we've to check this condition, rank(A) = rank(A|B) < n, where n is the unknown variable.

Now make the given matrix(augmented) into echleon form by Gauss-elimination method.

$\begin{bmatrix} 2 &1 &-4 &\alpha \\ 4 &3 &-12 &5 \\ 1 &2 &-8 &7 \end{bmatrix}$      R2 = r2 - 2r1

$\begin{bmatrix} 2 &1 &-4 &\alpha \\ 0 &1 &-4 &5-2\alpha \\ 1 &2 &-8 &7 \end{bmatrix}$        r3 = 2r3 -r1

$\begin{bmatrix} 2 &1 &-4 &\alpha \\ 0 &1 &-4 &5-2\alpha \\ 0 &3 &-12 &14 - \alpha \end{bmatrix}$     r3 = 3r2 - r3

$\begin{bmatrix} 2 &1 &-4 &\alpha \\ 0 &1 &-4 &5-2\alpha \\ 0 &0 &0 &1 - 5\alpha \end{bmatrix}$

Now to satisfy the above written condition for infinite solution 1 - 5$\alpha$ = 0 i.e  $\alpha$ = $\frac{1}{5}$