Gujarati
Hindi
3 and 4 .Determinants and Matrices
normal

Number of value of $'a'$ for which the system of equations,$A^2 x + (2 - a) y = 4 + a^2$ $a x + (2 a - 1) y = a^5 - 2$ possess no solution is

A

$0$

B

$1$

C

$2$

D

infinite

Solution

$\frac{{{a^2}}}{a}$ $=$ $\frac{{2\,\, – \,\,a}}{{2\,a\,\, – \,\,1}}$ $\ne\, \frac{{4\,\, + \,\,{a^2}}}{{{a^5}\,\, – \,\,2}}$ .

$={- 1, 1}$

Standard 12
Mathematics

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