Let the system of linear equations \[ x + 2y + z = 2, \quad \alpha x + 3y - z = \alpha, \quad -\alpha x + y + 2z = -\alpha \] be inconsistent. Then $\alpha$ is equal to:
- \(\frac{5}{2}\)
- \(-\frac{5}{2}\)
- \(\frac{7}{2}\)
- \(-\frac{7}{2}\)
The Correct Option is D
Solution and Explanation
\(x + 2y + z = 2\)
\(αx + 3y – z = α\)
\(–αx + y + 2z = –α\)
\(Δ=\begin{vmatrix}1&2&1\\α&3&−1\\−α&1&2\end{vmatrix}=1(6+1)−2(2α−α)+1(α+3α)\)
= \(7 + 2α\)
\(Δ=0⇒α=−\frac{7}{2}\)
\(Δ_1=\begin{vmatrix}2&2&1\\α&3&−1\\−α&1&2\end{vmatrix}=14+2α≠0 for \;α=−\frac{7}{2}\)
\(α = - \frac{7}{2}\)
Hence, the correct option is (D): \(- \frac{7}{2}\)
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Concepts Used:
Linear Inequalities
The expressions where any two values are compared by the inequality symbols such as, ‘<’, ‘>’, ‘≤’ or ‘≥’ are called linear inequalities. These values could be numerical or algebraic or a combination of both expressions. A system of linear inequalities in two variables involves at least two linear inequalities in the identical variables. After solving linear inequality we get an ordered pair. So generally, in a system, the solution to all inequalities and the graph of the linear inequality is the graph representing all solutions of the system.
How to Solve Inequalities in Maths?
Follow the below steps to solve all types of inequalities:
- At the vert first, write the inequality as an equation.
- Solve the provided equation for one or more values.
- Now, display all the values obtained in the number line.
- Use open circles to show the excluded values on the number line.
- Find the interim.
- At the moment, take any random value from the interval and substitute it in the inequality equation to check whether the values reassure the inequality equation.
- Intervals that reassure the inequality equation are the solutions of the given inequality equation.