Question

Solve 3x+3y-z=11 , 2x+y-2z=9 and 4x+3y+2z=25 by using Cramer's rule

Answer 1

Solve the equation by Cramer's Rule

3x + 3y - z = 11

2x + y - 2z = 9

4x + 3y + 2z = 25

$\large\underline{\bf{Solution-}}$

The matrix form of the above equation is

Let us assume that

$\rm :\longmapsto\:\begin{gathered}\sf A=\left[\begin{array}{ccc}3&3& - 1\\2&1&-2\\4&3&2\end{array}\right]\end{gathered}$

$\rm :\longmapsto\:\begin{gathered}\sf B=\left[\begin{array}{c}11\\9\\25\end{array}\right]\end{gathered}$

$\rm :\longmapsto\:\begin{gathered}\sf X=\left[\begin{array}{c}x\\y\\z\end{array}\right]\end{gathered}$

So that,

$\bf\implies \:AX = B$

Consider,

$\rm :\longmapsto\: |A| = \: \begin{gathered}\sf \left | \begin{array}{ccc}3&3&- 1\\2&1& - 2\\4&3&2\end{array}\right | \end{gathered}$

$\rm \: \: = \: 3(2 + 6) - 3(4 + 8) - 1(6 - 4)$

$\rm \: \: = \: 3(8) - 3(12) - 1(2)$

$\rm \: \: = \: 24 - 36 - 2$

$\rm \: \: = \: 24 - 38$

$\rm \: \: = \: - 14$

$\bf\implies \:|A| = - 14 \ne \: 0$

Hence,

System of equations is consistent having unique solution.

Consider,

$\rm :\longmapsto\:D_1 = \: \begin{gathered}\sf \left | \begin{array}{ccc}11&3&- 1\\9&1& - 2\\25&3&2\end{array}\right | \end{gathered}$

$\rm \: \: = \: 11(2 + 6) - 3(18 + 50) - 1(27 - 25)$

$\rm \: \: = \: 11(8) - 3(68) - 1(2)$

$\rm \: \: = \: 88 - 204 - 2$

$\rm \: \: = \: 88 - 206$

$\rm \: \: = \: - 118$

$\bf\implies \:D_1 = - 118$

Consider,

$\rm :\longmapsto\: D_2 = \: \begin{gathered}\sf \left | \begin{array}{ccc}3&11&- 1\\2&9& - 2\\4&25&2\end{array}\right | \end{gathered}$

$\rm \: \: = \: 3(18 + 50) - 11(4 + 8) - 1(50 - 36)$

$\rm \: \: = \: 3(68) - 11(12) - 1(14)$

$\rm \: \: = \: 204 - 132 - 14$

$\rm \: \: = \: 58$

$\bf\implies \:D_2 = 58$

Consider,

$\rm :\longmapsto\: D_3 = \: \begin{gathered}\sf \left | \begin{array}{ccc}3&3&11\\2&1& 9\\4&3&25\end{array}\right | \end{gathered}$

$\rm \: \: = \: 3(25 - 27) - 3(50 - 36) + 11(6 - 4)$

$\rm \: \: = \: 3( - 2) - 3(14) + 11(2)$

$\rm \: \: = \: - 6 - 42 + 22$

$\rm \: \: = \: - 48 + 22$

$\rm \: \: = \: - 26$

$\bf\implies \:D_3 = - 26$

Now,

We know that

$\rm :\longmapsto\:x = \dfrac{D_1}{|A|} = \dfrac{ - 118}{ - 14} = \dfrac{59}{7}$

$\rm :\longmapsto\:y = \dfrac{D_2}{|A|} = \dfrac{58}{ - 14} = - \dfrac{29}{7}$

$\rm :\longmapsto\:z = \dfrac{D_3}{|A|} = \dfrac{ - 26}{ - 14} = \dfrac{13}{7}$

Answer 2

✵$\textbf\red{Given:}$✵

$\mathsf{x+y+z=6}$

$\mathsf{3x+3y+z=12}$

$\mathsf{2x+3y+2z=14}$

$\textbf\red{To find:}$

$\textsf{Solution\: of \:the\: given\: system \:of\: equations \:by \:Cramer's \:rule}$

$\textbf\red{Solution:}$

$\begin{gathered}\mathsf{\triangle=\left|\begin{array}{ccc}1&1&1\\3&3&1\\2&3&2\end{array}\right|}\end{gathered}$

$\mathsf{\triangle=1(6-3)-1(6-2)+1(9-6)}$

$\mathsf{\triangle=3-4+3}$

$\mathsf{\triangle=2}$

$⟹ \begin{gathered}\mathsf{{\triangle}_x=\left|\begin{array}{ccc}6&1&1\\12&3&1\\14&3&2\end{array}\right|}\end{gathered}$

$\mathsf{{\triangle}_x=6(6-3)-1(24-14)+1(36-42)}$

$\mathsf{{\triangle}_x=18-10-6}$

$\mathsf{{\triangle}_x=2}$

$\begin{gathered}\mathsf{{\triangle}_y=\left|\begin{array}{ccc}1&6&1\\3&12&1\\2&14&2\end{array}\right|}\end{gathered}$

$\mathsf{{\triangle}_y=1(24-14)-6(6-2)+1(42-24)}$

$\mathsf{{\triangle}_y=10-24+18}$

$\mathsf{{\triangle}_y=4}$

$\mathsf{{\triangle}_z=1(42-36)-1(42-24)+6(9-6)}$

$\mathsf{{\triangle}_z=6-18+18}$

$\mathsf{{\triangle}_z=6}$

$\textsf{By Cramer's rule}$

$\mathsf{x=\dfrac{\triangle_x}{\triangle}}$

$\mathsf{x=\dfrac{2}{2}=1}$

$\mathsf{y=\dfrac{\triangle_y}{\triangle}}$

$→\mathsf{y=\dfrac{4}{2}=2}$

$→\mathsf{z=\dfrac{\triangle_z}{\triangle}}$

$→\mathsf{z=\dfrac{6}{2}=3}$

➪$\therefore\textsf{The solution is x=1, y=2 and z=3}$