Question

Solve this:

Q. 13. Let S be the set of all real values of k for which the system of linear equations.
x + y +z =2
2x + y - z =3
3x +2y +kz =4
has a unique solution. Then S is :
Option
1. an empty set
2. equal to {0}
3. equal to R
4. equal to R - {0}

Answer 1

Dear Student

$$\begin{gathered} Given \\ x+y+z=2 \\ 2x+y-z=3and \\ 3x+2y+kz=4 \\ Thesystemoflinearequationshasauniquesolutionthen \\ \left|\begin{array}{ccc}1& 1& 1\\ 2& 1& -1\\ 3& 2& k\end{array}\right|\ne 0 \\ \Rightarrow 1\left(k+2\right)-1\left(2k+3\right)+1\left(4-3\right)\ne 0 \\ \Rightarrow k+2-2k-3+1\ne 0 \\ \Rightarrow k\ne 0 \\ So,thevaluesofkareR-\left\{0\right\} \\ \therefore S=R-\left\{0\right\} \end{gathered}$$

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