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36 votes

Best answer

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

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.

0

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).

0

15 votes

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.

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.

1 vote

For unique solution:

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

therefore,

1 1 2 1

1 23 2

1 4 α 4

3^{rd} element of last should not be equal to zero

so, after transforming row echelon form, we have 3^{rd} 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.