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An ordered pair $\left ( \alpha ,\beta \right )$ for which the system of linear equations

$\left ( 1+\alpha \right )x+\beta y+z=2$

$\alpha x+\left ( 1+\beta \right )y+z=3$

$\alpha x+\beta y+2z=2$

has a unique solution, is :
Option 1)
$\left ( 2,4 \right )$

Option 2)
$\left ( -4,2 \right )$
Option 3)
$\left ( 1,-3 \right )$
Option 4)
$\left ( -3,1\right )$

Answers (1)

best_answer

Solution of a homogeneous system of linear equations -

Let $Ax=0$

If $A$ is non-singular then the system of equations will have a unique solution that is trivial soluion

-

Solution of a non-homogeneous system of linear equations by matrix method -

If $A$ is a non-singular matrix then the system of equations given by $Ax=b$ has a unique solution given by $x=A^{-1}b$

- wherein

$\\(1 +\alpha)x + \beta y + z = 0 \\ \alpha x + (1+\beta )y + z =0 \\ \alpha x + \beta y +2z = 0$

$\\D = \begin{vmatrix} 1 + \alpha &\beta &1 \\ \alpha & 1 + \beta & 1 \\ \alpha & \beta & 2 \end{vmatrix}\\ C_1\rightarrow C_1 + C_2 + C_3$

$\\D =(\alpha + \beta + 2) \begin{vmatrix} 1 &\beta &1 \\ 1 & 1 + \beta & 1 \\ 1 & \beta & 2 \end{vmatrix}\\$

$R_2 \rightarrow R_2 - R_1 \;\;\;\;\;\; R_3 \rightarrow R_3 - R_1$

$\\D =(\alpha + \beta + 2) \begin{vmatrix} 1 &\beta &1 \\ 0 & 1 & 0 \\ 0 & 0 &1 \end{vmatrix}= \alpha +\beta +2$

For a unique solution, $\alpha + \beta \neq -2$

Option 1)
$\left ( 2,4 \right )$

Option 2)

$\left ( -4,2 \right )$

Option 3)

$\left ( 1,-3 \right )$

Option 4)

$\left ( -3,1\right )$

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