in Linear Algebra edited by
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27 votes
27 votes

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
in Linear Algebra edited by
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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)
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3 Answers

36 votes
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. 

Ref: http://www.mathwords.com/i/inverse_of_a_matrix.htm

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4 Comments

@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  

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@

please read above comment very carefully.

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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).

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15 votes
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.
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3 Comments

Nice approach.
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Last operation is

R3->R3- 3R2
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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 $\alpha \neq 8$
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1 vote
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

 

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.

 

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