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

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

### 2 thoughts on “Solve 3x+3y-z=11 , 2x+y-2z=9 and 4x+3y+2z=25 by using Cramer’s rule<br /><br />”

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

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

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

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

2. ✵$$\textbf\red{Given:}$$✵

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

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

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

$$\textbf\red{To find:}$$

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

$$\textbf\red{Solution:}$$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$\textsf{By Cramer’s rule}$$

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

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

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

$$→\mathsf{y=\dfrac{4}{2}=2}$$

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

$$→\mathsf{z=\dfrac{6}{2}=3}$$

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