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

- $0$
- either $0$ or $1$
- one of $0, 1$, or $-1$
- any real number

### 1 comment

## 2 Answers

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

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.

### 17 Comments

after applying row operations, we get the matrix as :

$\left( \begin{array}{ccc} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & \alpha-5 \end{array} \middle\vert \begin{array}{c} 1\\1\\0\end{array}\right)$

If we put $\alpha=5$, then we have 3 equations like this :

$x_1+x_2+2x_3 = 1$

$x_2 + x_3 = 1$

$0 = 0$

Third equation is trivially true, so we have 2 equations in 3 variables, and thus we get values of two variables in terms of third variable, which we can put anything, so we have infinite many solutions.

Had we got third equation as something like

$0 = 1$

Then there is no solution to this system of equations.

actually it's a underdetermined system as Happy Mittal 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).

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.