Question

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 unique solution is
(a) \[\left( -1,3 \right)\]
(b) \[\left( -3,1 \right)\]
(c) \[\left( 2,4 \right)\]
(d) \[\left( -4,2 \right)\]

Answer 1

Hint: Here, we will convert this equation into matrix form and determinant of matrix will be not equal to zero because solution of the given three equation are unique i.e. given as $ \left| \begin{matrix}
   {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
   {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|\ne 0 $ . Then we will apply the row transformation method in such a way that we get only one $ \alpha $ and $ \beta $ which will become easier to solve. Then on solving, we will get $ \alpha +\beta $ with some value. So, we will add the option values and compare it with the determinant answer. Thus, we will get an answer.

Complete step-by-step answer:
Here, we are given with three equations having unique solutions i.e.
\[\left( 1+\alpha \right)x+\beta y+z=2\] ……………..(1)
\[\alpha x+\left( 1+\beta \right)y+z=3\] ……………….(2)
\[\alpha x+\beta y+2z=2\] …………………..(3)
Now, we know that determinant of three equation is not equal to 0. So, first we will convert three equations into matrix form. Now, we will write this coefficient of all three equations in matrix form i.e. $ \left| \begin{matrix}
   {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
   {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right| $
We will fill the first matrix with coefficients of the variables A, B and C. Second matrix we will fill as A, B and C. Then in the third matrix, we will fill in the constant terms from the equations formed.
From equations 1, 2 and 3, we can write matrix as
 $ \left| \begin{matrix}
   \left( 1+\alpha \right) & \beta & 1 \\
   \alpha & \left( 1+\beta \right) & 1 \\
   \alpha & \beta & 2 \\
\end{matrix} \right|\ne 0 $
Now, we will use the row transformation method. We have to perform operations like addition, subtraction, multiplication, or division to make any of the coefficients equals to zero so it can be easier to solve.
So, here we are performing operation on Row2 i.e. $ Row1=Row1-Row2 $ .So, by doing this we will get matrix as
 $ \left| \begin{matrix}
   \left( 1+\alpha \right)-\alpha & \beta -\left( 1+\beta \right) & 1-1 \\
   \alpha & \left( 1+\beta \right) & 1 \\
   \alpha & \beta & 2 \\
\end{matrix} \right|\ne 0 $
On further solving, we get as
 $ \left| \begin{matrix}
   1 & -1 & 0 \\
   \alpha & \left( 1+\beta \right) & 1 \\
   \alpha & \beta & 2 \\
\end{matrix} \right|\ne 0 $
Now, we will perform operation on Row 2 i.e. \[Row2=Row2-Row3\] . So, we will get matrix as
 $ \left| \begin{matrix}
   1 & -1 & 0 \\
   \alpha -\alpha & \left( 1+\beta \right)-\beta & 1-2 \\
   \alpha & \beta & 2 \\
\end{matrix} \right|\ne 0 $
On solving, we get matrix as
 $ \left| \begin{matrix}
   1 & -1 & 0 \\
   0 & 1 & -1 \\
   \alpha & \beta & 2 \\
\end{matrix} \right|\ne 0 $
Now, we will find the determinant of the above matrix of the third row. It is calculated as
\[\alpha \left| \begin{matrix}
   -1 & 0 \\
   1 & -1 \\
\end{matrix} \right|-\beta \left| \begin{matrix}
   1 & 0 \\
   0 & -1 \\
\end{matrix} \right|+2\left| \begin{matrix}
   1 & -1 \\
   0 & 1 \\
\end{matrix} \right|\ne 0\]
On solving this, we get as
\[\alpha \left( -1\times -1-0 \right)-\beta \left( -1\times 1-0 \right)+2\left( 1\times 1-0 \right)\ne 0\]
\[\alpha +\beta +2\ne 0\]
So, we can write this as
\[\alpha +\beta \ne -2\]
 Now, we will see the options and check which option is not equal to \[-2\] . That is out answer.
Taking option (a): \[\left( -1,3 \right)\] . On adding the roots, we get \[-1+3=-2\] . This is equal to \[-2\] , so it is not an answer.
Taking option (b): \[\left( -3,1 \right)\] . On adding the roots, we get \[-3+1=-2\] . This is equal to \[-2\] , so it is not an answer.
Taking option (c): \[\left( 2,4 \right)\] . On adding the roots, we get \[2+4=6\] . This is not equal to \[-2\] so, it is the answer.
Taking option (d): \[\left( -4,2 \right)\] . On adding the roots, we get \[-4+2=-2\] . This is equal to \[-2\] , so it is not an answer.
Hence, option (c) is the correct answer.

Note: Sometime students make mistake in writing $ \left| \begin{matrix}
   \left( 1+\alpha \right) & \beta & 1 \\
   \alpha & \left( 1+\beta \right) & 1 \\
   \alpha & \beta & 2 \\
\end{matrix} \right|=0 $ instead of $ \left| \begin{matrix}
   \left( 1+\alpha \right) & \beta & 1 \\
   \alpha & \left( 1+\beta \right) & 1 \\
   \alpha & \beta & 2 \\
\end{matrix} \right|\ne 0 $ . Because of this sign mistake, the sum of roots we will be equal to \[-2\] . So, while checking the option, there will be multiple answers correct which are actually wrong. So, please be careful while using the sign to avoid mistakes. Also, determinants of equations having unique solutions not equal to 0 should be known to students. So, be clear with concepts and then solve it.