Question

In solving the following equations by Cramer's rule
$x+y+z=9$
$2x+5y+7z=52$
$2x+y-z=0$,
the values ${\Delta }_1,{\Delta }_2,{\Delta }_3$ are in the ratio

• A. $1:3:5$
• B. $1:3:-5$
• C. $-1:3:5$
• D. $1:-3:5$

Answer 1

The correct option is A $1:3:5$

Here $\Delta =\left|\begin{array}{ccc}1& 1& 1\\ 2& 5& 7\\ 2& 1& -1\end{array}\right|$

applying $C_2\to C_2-C_1$ and $C_3\to C_3-C_1$

$\therefore \Delta =\left|\begin{array}{ccc}1& 0& 0\\ 2& 3& 5\\ 2& -1& -3\end{array}\right|=1(-9+5)=-4$

${\Delta }_1=\left|\begin{array}{ccc}9& 1& 1\\ 52& 5& 7\\ 0& 1& -1\end{array}\right|$ Applying $C_2\to C_2+C_3$

$\therefore {\Delta }_1=\left|\begin{array}{ccc}9& 2& 1\\ 52& 12& 7\\ 0& 0& -1\end{array}\right|=-1(108-104)=-4$

${\Delta }_2=\left|\begin{array}{ccc}1& 9& 1\\ 2& 52& 7\\ 2& 0& -1\end{array}\right|$ Applying $C_1\to C_1+2C_3$

$\therefore {\Delta }_2=\left|\begin{array}{ccc}3& 9& 1\\ 16& 52& 7\\ 0& 0& -1\end{array}\right|$

$=-1(156-144)=-12$

and

${\Delta }_3=\left|\begin{array}{ccc}1& 1& 9\\ 2& 5& 52\\ 2& 1& 0\end{array}\right|$ Applying $C_1\to C_1-2C_2$

$\therefore {\Delta }_3=\left|\begin{array}{ccc}-1& 1& 9\\ -8& 5& 52\\ 0& 1& 0\end{array}\right|$

$=-1(-51+72)$

$=-20$
Clearly ${\Delta }_1:{\Delta }_2:{\Delta }_3=-4:-12:-20=1:3:5$