Question

The following system of linear equations is,

$$\begin{gathered} 2\mathrm{x}+\; 3\mathrm{y}+\; 2\mathrm{z}=\; 9 \\ 3\mathrm{x}+\; 2\mathrm{y}+\; 2\mathrm{z}=\; 9 \\ \mathrm{x}-\mathrm{y}+\; 4\mathrm{z}=\; 8 \end{gathered}$$

• A. does not have any solution
• B. has a unique solution
• C. has a solution $(\alpha ,\beta ,\gamma )$ satisfying $\alpha +{\beta }^2+{\gamma }^3=12$
• D. has infinitely many solutions

Answer 1

The correct option is B

has a unique solution

Explanation for correct option:

Solve the given set of equations

Given equations:

$$\begin{gathered} 2\mathrm{x}+\; 3\mathrm{y}+\; 2\mathrm{z}=\; 9 \\ 3\mathrm{x}+\; 2\mathrm{y}+\; 2\mathrm{z}=\; 9 \\ \mathrm{x}-\mathrm{y}+\; 4\mathrm{z}=\; 8 \end{gathered}$$

Find out the determinant of the equations,

$\left|\begin{array}{ccc}2& 3& 2\\ 3& 2& 2\\ 1& -1& 4\end{array}\right|=2(8+2)-3(12-2)+2(-3-2)=20-30-10=-20$

We know that when the determinant is not equal to zero then it has unique solution.

Here, we are getting $-20$ which is not equal to zero.

Therefore, the given set of equations has unique solution, so, option (B) is the correct answer.