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+14 votes
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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
asked in Linear Algebra by Veteran (59.4k points)
edited by | 1.5k views
0
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)

2 Answers

+18 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, $X = A^{-1} B$

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

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

$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

answered by Veteran (348k points)
edited by
+13
$\alpha$ can't be 5, because then det(A) will be 0, and there will be no unique solution (as given in question).
+3
So, any real number except 5 is the answer?
+5
yes, any real number except 5, and so none of the option is correct.
0
But what exactly happens when $\alpha = 5$? Are we getting more solutions?
+19
Yes, if you look at Q. 3 here : http://www.cse.iitd.ac.in/~mittal/gate/gate_math_2008.html

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.
0
Thanks :)
0
Answer should be any real number except Alpha as 5 .
0
Which option should we select in this question? Should we leave this question?
0
yes, just leave or choose any. In either case Marks will be given to All.
0
Okay Thanks Sir :)
+4 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.
answered by Active (1.3k points)
edited by
+1
Nice approach.
0
Last operation is

R3->R3- 3R2
0
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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