Chapter SH, Problem 3.22EP

To determine

To solve: the given system of equation by using Cramer’s rule.

Answer to Problem 3.22EP

  $x=4,y=-2,z=-3$ .

Explanation of Solution

Given:

The given system of equation is $\begin{array}{l}x+y+z=-3\\ 3x-2y=16\\ 2x-y+3z=-5\end{array}$ .

Concept used:

Cramer’s rule: For a $3\times 3$ matrix,

  $x=\frac{D_x}{D}=\frac{\left|\begin{array}{lll}{d}_1\hfill & b_1\hfill & c_1\hfill \\ d_2\hfill & b_2\hfill & c_2\hfill \\ d_3\hfill & b_3\hfill & c_3\hfill \end{array}\right|}{\left|\begin{array}{lll}{a}_1\hfill & b_1\hfill & c_1\hfill \\ a_2\hfill & b_2\hfill & c_2\hfill \\ a_3\hfill & b_3\hfill & c_3\hfill \end{array}\right|}$;$y=\frac{D_y}{D}=\frac{\left|\begin{array}{lll}{a}_1\hfill & d_1\hfill & c_1\hfill \\ a_2\hfill & d_2\hfill & c_2\hfill \\ a_3\hfill & d_3\hfill & c_3\hfill \end{array}\right|}{\left|\begin{array}{lll}{a}_1\hfill & b_1\hfill & c_1\hfill \\ a_2\hfill & b_2\hfill & c_2\hfill \\ a_3\hfill & b_3\hfill & c_3\hfill \end{array}\right|}$;$z=\frac{D_z}{D}=\frac{\left|\begin{array}{lll}{a}_1\hfill & b_1\hfill & d_1\hfill \\ a_2\hfill & b_2\hfill & d_2\hfill \\ a_3\hfill & b_3\hfill & d_3\hfill \end{array}\right|}{\left|\begin{array}{lll}{a}_1\hfill & b_1\hfill & c_1\hfill \\ a_2\hfill & b_2\hfill & c_2\hfill \\ a_3\hfill & b_3\hfill & c_3\hfill \end{array}\right|}$

Calculation:

The givensystem of equation is $\begin{array}{l}x+y+z=-3\\ 3x-2y=16\\ 2x-y+3z=-5\end{array}$ .

Now, change the equation in matrix form, then $\left|\begin{array}{l}123|-3\\ 3-20|16\\ 2-13|-5\end{array}\right|$

Now, the determinant is

  $x=\frac{D_x}{D}=\frac{\left|\begin{array}{lll}{d}_1\hfill & b_1\hfill & c_1\hfill \\ d_2\hfill & b_2\hfill & c_2\hfill \\ d_3\hfill & b_3\hfill & c_3\hfill \end{array}\right|}{\left|\begin{array}{lll}{a}_1\hfill & b_1\hfill & c_1\hfill \\ a_2\hfill & b_2\hfill & c_2\hfill \\ a_3\hfill & b_3\hfill & c_3\hfill \end{array}\right|}$$=\frac{\left|\begin{array}{ccc}-3& 1& 1\\ 16& -2& 0\\ -5& -1& 3\end{array}\right|}{\left|\begin{array}{ccc}1& 1& 1\\ 3& -2& 0\\ 2& -1& 3\end{array}\right|}$

  $=\frac{-3\left|\begin{array}{cc}-2& 0\\ -1& 3\end{array}\right|-1\left|\begin{array}{cc}16& 0\\ -5& 3\end{array}\right|+1\left|\begin{array}{cc}16& -2\\ -5& -1\end{array}\right|}{\left|\begin{array}{cc}-2& 0\\ -1& 3\end{array}\right|-1\left|\begin{array}{cc}3& 0\\ 2& 3\end{array}\right|+1\left|\begin{array}{cc}3& -2\\ 2& -1\end{array}\right|}$

  $=\frac{-3(-2\times 3-0\times (-1))-1(16\times 3-0\times (-5))+1(16\times (-1)-(-2)\times (-5))}{1(-2\times 3-0\times (-1))-1(3\times 3-0\times (-5))+1(3\times (-1)-(-2)\times 2)}$

  $=\frac{-3(-6-0)-1(48-0)+1(-16-10)}{1(-6-0)-1(9-0)+1(-3+4)}$

  $=\frac{18-48-26}{-6-9+1}=\frac{-56}{-14}=4$

and

  $y=\frac{D_y}{D}=\frac{\left|\begin{array}{lll}{a}_1\hfill & d_1\hfill & c_1\hfill \\ a_2\hfill & d_2\hfill & c_2\hfill \\ a_3\hfill & d_3\hfill & c_3\hfill \end{array}\right|}{\left|\begin{array}{lll}{a}_1\hfill & b_1\hfill & c_1\hfill \\ a_2\hfill & b_2\hfill & c_2\hfill \\ a_3\hfill & b_3\hfill & c_3\hfill \end{array}\right|}$$=\frac{\left|\begin{array}{rrr}\hfill 1& \hfill -3& \hfill 1\\ \hfill 3& \hfill 16& \hfill 0\\ \hfill 2& \hfill -5& \hfill 3\end{array}\right|}{\left|\begin{array}{rrr}\hfill 1& \hfill 1& \hfill 1\\ \hfill 3& \hfill -2& \hfill 0\\ \hfill 2& \hfill -1& \hfill 3\end{array}\right|}$

  $=\frac{1\left|\begin{array}{cc}16& 0\\ -5& 3\end{array}\right|+3\left|\begin{array}{cc}3& 0\\ 2& 3\end{array}\right|+1\left|\begin{array}{cc}3& 16\\ 2& -5\end{array}\right|}{\begin{array}{cc}\mid -2& 0\\ -1& 3\end{array}|-1|\begin{array}{cc}3& 0\\ 2& 3\end{array}|+1|\begin{array}{cc}3& -2\\ 2& -1\end{array}\mid }$

  $=\frac{-1(16\times 3-0\times (-5))+3(3\times 3-0\times 2)+1(3\times (-5)-2\times 16)}{1(-2\times 3-0\times (-1))-1(3\times 3-0\times (-5))+1(3\times (-1)-(-2)\times 2)}$

  $=\frac{-1(-48-0)+3(9-0)+1(-15-32)}{1(-6-0)-1(9-0)+1(-3+4)}$

  $=\frac{48+27-47}{-6-9+1}=\frac{28}{-14}=-2$

And

  $z=\frac{D_z}{D}=\frac{\left|\begin{array}{lll}{a}_1\hfill & b_1\hfill & d_1\hfill \\ a_2\hfill & b_2\hfill & d_2\hfill \\ a_3\hfill & b_3\hfill & d_3\hfill \end{array}\right|}{\left|\begin{array}{lll}{a}_1\hfill & b_1\hfill & c_1\hfill \\ a_2\hfill & b_2\hfill & c_2\hfill \\ a_3\hfill & b_3\hfill & c_3\hfill \end{array}\right|}$$=\frac{\left|\begin{array}{ccc}1& 1& -3\\ 3& -2& 16\\ 2& -1& -5\end{array}\right|}{\left|\begin{array}{ccc}1& 1& 1\\ 3& -2& 0\\ 2& -1& 3\end{array}\right|}$

  $=\frac{1\left|\begin{array}{cc}-2& 16\\ -1& -5\end{array}\right|-1\left|\begin{array}{cc}3& 16\\ 2& -5\end{array}\right|-3\left|\begin{array}{cc}3& -2\\ 2& -1\end{array}\right|}{\left|\begin{array}{cc}-2& 0\\ -1& 3\end{array}\right|-1\left|\begin{array}{cc}3& 0\\ 2& 3\end{array}\right|+1\left|\begin{array}{cc}3& -2\\ 2& -1\end{array}\right|}$

  $\begin{array}{l}=\frac{1(-2\times (-5)-16\times (-1))-1(3\times -5-0\times 2)-3(3\times (-1)-2\times (-2))}{1(-2\times 3-0\times (-1))-1(3\times 3-0\times (-5))+1(3\times (-1)-(-2)\times 2)}\\ =\frac{-1(10-16)-1(-15-0)-3(-3-4)}{1(-6-0)-1(9-0)+1(-3+4)}\\ =\frac{6+15+21}{-6-9+1}=\frac{42}{-14}=-3\end{array}$

Hence, from Cramer’s rule, the solution of the given system of equation is $x=4,y=-2,z=-3$ .