Algebra

Matrices & Determinants

Medium

applied

The system of equation $x-y+z=4, x-2y-2z=9, 2x+y+\alpha z=1$ has unique solution then value(/s) of $\alpha$ is (/are) (where $\mathbb{R}$ is set of real numbers)

$\alpha\in\mathbb{R}$

$\alpha\in\mathbb{R}\setminus\{-11\}$

$\alpha\in\mathbb{R}\setminus\{11\}$

✅ Correct Option: 4

A unique solution needs the coefficient determinant non-zero.

$\Delta = 1(-2\alpha+2)+1(\alpha+4)+1(1+4)=11-\alpha$ .

So $\alpha\neq 11$ , i.e. $\alpha\in\mathbb{R}\setminus\{11\}$ .

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The system of equation x−y+z=4,x−2y−2z=9,2x+y+αz=1x-y+z=4, x-2y-2z=9, 2x+y+\alpha z=1x−y+z=4,x−2y−2z=9,2x+y+αz=1 has unique solution then value(/s) of α\alphaα is (/are) (where R\mathbb{R}R is set of real numbers)