Question

Let ${\lambda }_1:IfAX=P$ has infinite solution &$cf\ne 0$ then one solution of $BX=Qis(0,0,0)$ ${\lambda }_2:$ If a system has infinite solutions then one solution must be trivial, then?

• A. ${\lambda }_1$ &${\lambda }_2$ both are false
• B. ${\lambda }_1$ is true & ${\lambda }_2$ is false
• C. ${\lambda }_1{\lambda }_2$ both are true but ${\lambda }_2$ is not the correct explanation ${\lambda }_1$
• D. ${\lambda }_1{\lambda }_2$ both are true but ${\lambda }_2$is the correct explanation of ${\lambda }_1$

Answer 1

Answer: (A)

${\lambda }_1$ &${\lambda }_2$ both are false

$A=\left|\begin{array}{ccc}a& b& ax+b\\ b& c& bx+c\\ ax+b& bx+c& 0\end{array}\right|$

$R_1\to (R_{1}x+R_2)-R_3$

$=\left|\begin{array}{ccc}0& 0& a{x}^2+bx+bx+c\\ b& c& bx+c\\ ax+b& bx+c& 0\end{array}\right|$

$=(a{x}^2+2bx+c)(b^{2}x+bc-acx-bc)$

$=(a{x}^2+2bx+c)(b^2-4ac)x$

$=$Total product$=(-)\times \left(f\right)=-1$

Determinant A is negative