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The following system of equations

• $x_1 + x_2 + 2x_3 = 1$
• $x_1 + 2x_2 + 3x_3 = 2$
• $x_1 + 4x_2 + αx_3 = 4$

has a unique solution. The only possible value(s) for $α$ is/are

1. $0$
2. either $0$ or $1$
3. one of $0, 1$, or $-1$
4. any real number

### 1 comment

Since no option is matching then if we have to mark the closest one it would be C) (although 0,1,-1 are not the ONLY possible values)

$A.X = B \\ \implies \begin{bmatrix} 1 &1 &2 \\ 1&2 &3 \\ 1& 4 & \alpha \end{bmatrix} \begin{bmatrix} x\\y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\2 \\4 \end{bmatrix}$

So, $X = A^{-1} B$

$A^{-1} =\dfrac{adj(A)}{\det(A)}$

$\because \left[adj(A) = \text{Transporse (cofactor(A)), cofactor(A)} = (-1)^{i+j} \;\text{minor(A)} \right]$

$adj(A) = \begin{bmatrix} 2\alpha -12 &8-\alpha &-1 \\ 3 - \alpha&\alpha - 2 &-1 \\ 2& -3 & 1 \end{bmatrix}$

$\det(A) = 2\alpha - 12 + 3 - \alpha + 4 = \alpha - 5$

Now, $A^{-1}B = \frac{1}{\alpha - 5} \begin{bmatrix} 0\\ \alpha -5 \\ 0 \end{bmatrix} = \begin{bmatrix} 0\\ 1 \\ 0 \end{bmatrix}$

So, the solution is independent of $\alpha$ and any real value of $\alpha$ would suffice. (This can be seen even by observing the equations carefully- each equation value is dependent only on $x_2$). But a value of $5$ would cause the determinant to be $0$ and multiple solutions to the given equation. So, any real value except $5$ should be the answer- none of the choices is correct.

by

@Happy Mittal

Here you are saying that there will be no solution for aplha=5

But putting alpha=5 makes the rank[AB] =rank[A] < n (no of variables )

Won't it turn out to be infinite number of solutions

actually it's a underdetermined system as  said in his comment,we have no. of equations less than no. of variables which implies we have infinite solution.

let's suppose in a 2d space i.e x-y dimension we have been given only 1 equation let's say x-y=6 i.e, we have infinite solutions of (x,y).

Since system has unique solution, r(A)= r(AB)= 3. (no of unknown variables) where A is given matrix and AB is augmented matrix. If we perform elementary operations:-
R2-> R2-R1, R3->R3-R1 and lastly R3->R3- 3R2 we get below matrix:
1 1 2: 1
0 1 1: 1
0 0 α-5: 0

Therefore  α cannot be equal to 5 as this will make r(A)= r(AB)=2 which will not have unique solution as rank is less than number of unknown variables. So the answer should be any real number other than 5.

Nice approach.
Last operation is

R3->R3- 3R2
i didn't understand what difference that it make
with last operation : R3->R3-3R2      // gives  $\alpha \neq 5$
and with : R3->R3-3R1                    // gives $\alpha \neq 8$

For unique solution:

rank(A)=rank(AB)=no. of unknown=3 (here)

therefore,

1 1 2 1

1 23  2

1 4 α  4

3rd element of last should not be equal to zero

so, after transforming row echelon form, we have 3rd element in last row as : α-2-3=α-5 !=0

so, α != 5,,i.e. α can have any value except 5.

Hence, None of ans matches.

by