91Ó°ÊÓ

The solution of the system of linear equations in three variables:

$\begin{array}{l}{a}_{1}a+b_{1}b+c_{1}c=d_1\\ a_{2}a+b_{2}b+c_{2}c=d_2\\ a_{3}a+b_{3}b+c_{3}c=d_3\end{array}$

by Cramer’s rule is given by $\left(a,b,c\right)$where

$a=\frac{\left|\begin{array}{ccc}{d}_1& b_1& c_1\\ d_2& b_2& c_2\\ d_3& b_3& c_3\end{array}\right|}{\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|}$$b=\frac{\left|\begin{array}{ccc}{a}_1& d_1& c_1\\ a_2& d_2& c_2\\ a_3& d_3& c_3\end{array}\right|}{\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|}$$c=\frac{\left|\begin{array}{ccc}{a}_1& b_1& d_1\\ a_2& b_2& d_2\\ a_3& b_3& d_3\end{array}\right|}{\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|}$and$\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|≠0$

The given system of linear equations in three variables is:

$\begin{array}{c}3a−5b+2c=−5\\ 4a+b+3c=9\\ 2a−c=1\end{array}$

Therefore, by comparing the given system of linear equations with the

system of linear equations$\begin{array}{l}{a}_{1}a+b_{1}b+c_{1}c=d_1\\ a_{2}a+b_{2}b+c_{2}c=d_2\\ a_{3}a+b_{3}b+c_{3}c=d_3\end{array}$it can be obtained that:

$a_1=3$, $b_1=-5$, $c_1=2$, $d_1=-5$, $a_2=4$, $b_2=1$,$c_2=3$, $d_2=9$, $a_3=2$,$b_3=0$, $c_3=-1$and $d_3=1$.

The value of $a$is given by:

$\begin{array}{c}a=\frac{\left|\begin{array}{ccc}−5& −5& 2\\ 9& 1& 3\\ 1& 0& −1\end{array}\right|}{\left|\begin{array}{ccc}3& −5& 2\\ 4& 1& 3\\ 2& 0& −1\end{array}\right|}\\ =\frac{−5\left|\begin{array}{cc}1& 3\\ 0& −1\end{array}\right|−\left(−5\right)\left|\begin{array}{cc}9& 3\\ 1& −1\end{array}\right|+2\left|\begin{array}{cc}9& 1\\ 1& 0\end{array}\right|}{3\left|\begin{array}{cc}1& 3\\ 0& −1\end{array}\right|−\left(−5\right)\left|\begin{array}{cc}4& 3\\ 2& −1\end{array}\right|+2\left|\begin{array}{cc}4& 1\\ 2& 0\end{array}\right|}\\ =\frac{−5\left(−1−0\right)+5\left(−9−3\right)+2\left(0−1\right)}{3\left(−1−0\right)+5\left(−4−6\right)+2\left(0−2\right)}\\ =\frac{5−60−2}{−3−50−4}\\ =\frac{−57}{−57}\\ =1\end{array}$

The value of $b$is given by:

$\begin{array}{c}b=\frac{\left|\begin{array}{ccc}3& −5& 2\\ 4& 9& 3\\ 2& 1& −1\end{array}\right|}{\left|\begin{array}{ccc}3& −5& 2\\ 4& 1& 3\\ 2& 0& −1\end{array}\right|}\\ =\frac{3\left|\begin{array}{cc}9& 3\\ 1& −1\end{array}\right|−\left(−5\right)\left|\begin{array}{cc}4& 3\\ 2& −1\end{array}\right|+2\left|\begin{array}{cc}4& 9\\ 2& 1\end{array}\right|}{3\left|\begin{array}{cc}1& 3\\ 0& −1\end{array}\right|−\left(−5\right)\left|\begin{array}{cc}4& 3\\ 2& −1\end{array}\right|+2\left|\begin{array}{cc}4& 1\\ 2& 0\end{array}\right|}\\ =\frac{3\left(−9−3\right)+5\left(−4−6\right)+2\left(4−18\right)}{3\left(−1−0\right)+5\left(−4−6\right)+2\left(0−2\right)}\\ =\frac{−36−50−28}{−3−50−4}\\ =\frac{−114}{−57}\\ =2\end{array}$

The value of $c$is given by:

$\begin{array}{c}c=\frac{\left|\begin{array}{ccc}3& −5& −5\\ 4& 1& 9\\ 2& 0& 1\end{array}\right|}{\left|\begin{array}{ccc}3& −5& 2\\ 4& 1& 3\\ 2& 0& −1\end{array}\right|}\\ =\frac{3\left|\begin{array}{cc}1& 9\\ 0& 1\end{array}\right|−\left(−5\right)\left|\begin{array}{cc}4& 9\\ 2& 1\end{array}\right|+\left(−5\right)\left|\begin{array}{cc}4& 1\\ 2& 0\end{array}\right|}{3\left|\begin{array}{cc}1& 3\\ 0& −1\end{array}\right|−\left(−5\right)\left|\begin{array}{cc}4& 3\\ 2& −1\end{array}\right|+2\left|\begin{array}{cc}4& 1\\ 2& 0\end{array}\right|}\\ =\frac{3\left(1−0\right)+5\left(4−18\right)−5\left(0−2\right)}{3\left(−1−0\right)+5\left(−4−6\right)+2\left(0−2\right)}\\ =\frac{3−70+10}{−3−50−4}\\ =\frac{−57}{−57}\\ =1\end{array}$

The values of $a$, $b$ and $c$ are 1, 2 and 1 respectively.

The solution of the given system of equations is $\left(1,2,1\right)$.