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Worked Examples · Example 41

Q.Solve the following system of equations using Cramer's rule: x+y+z=10x + y + z = 10, 2x+y=132x + y = 13, x+y−4z=0x + y - 4z = 0.

Arunachal CbseNCERTSubjective· 5mImportance★★★★★
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✓ Free question

Cramer's rule with D=5, Dx=25, Dy=15, Dz=10D=5,\ D_x=25,\ D_y=15,\ D_z=10 gives x=5, y=3, z=2x=5,\ y=3,\ z=2.

x=DxD, y=DyD, z=DzDx=\dfrac{D_x}{D},\ y=\dfrac{D_y}{D},\ z=\dfrac{D_z}{D}; DD is the coefficient determinant and Dx,Dy,DzD_x,D_y,D_z replace the respective column with the constants.

  1. Equations: x+y+z=10, 2x+y=13, x+y−4z=0x+y+z=10,\ 2x+y=13,\ x+y-4z=0, coefficient matrix [11121011−4]\begin{bmatrix}1&1&1\\2&1&0\\1&1&-4\end{bmatrix}.

  2. DD:

D=∣11121011−4∣=1(−4−0)−1(−8−0)+1(2−1)=−4+8+1=5.D = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 1 & 0 \\ 1 & 1 & -4 \end{vmatrix} = 1(-4-0)-1(-8-0)+1(2-1) = -4+8+1 = 5.

  1. DxD_x (constants in column 1):

Dx=∣1011131001−4∣=10(−4)−1(−52)+1(13)=−40+52+13=25.D_x = \begin{vmatrix} 10 & 1 & 1 \\ 13 & 1 & 0 \\ 0 & 1 & -4 \end{vmatrix} = 10(-4)-1(-52)+1(13) = -40+52+13 = 25.

  1. DyD_y (constants in column 2):

Dy=∣1101213010−4∣=1(−52)−10(−8)+1(−13)=−52+80−13=15.D_y = \begin{vmatrix} 1 & 10 & 1 \\ 2 & 13 & 0 \\ 1 & 0 & -4 \end{vmatrix} = 1(-52)-10(-8)+1(-13) = -52+80-13 = 15.

  1. DzD_z (constants in column 3):

Dz=∣11102113110∣=1(−13)−1(−13)+10(1)=−13+13+10=10.D_z = \begin{vmatrix} 1 & 1 & 10 \\ 2 & 1 & 13 \\ 1 & 1 & 0 \end{vmatrix} = 1(-13)-1(-13)+10(1) = -13+13+10 = 10.

  1. Therefore

x=255=5,y=155=3,z=105=2.x = \frac{25}{5} = 5, \qquad y = \frac{15}{5} = 3, \qquad z = \frac{10}{5} = 2.

  1. Check: 5+3+2=105+3+2=10 ✓; 2(5)+3=132(5)+3=13 ✓; 5+3−4(2)=05+3-4(2)=0 ✓.
✓Final answer

x=5,y=3,z=2x = 5,\quad y = 3,\quad z = 2

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