+26400 solve using Cramer's rule
x-2y+3z=6
2x+y-z=-3
x+y+z=6
\(\begin{array}{rcrcrcr} 1\cdot x &-& 2\cdot y &+& 3\cdot z &=& 6 \\ 2\cdot x &+& 1\cdot y &-& 1\cdot z &=& -3 \\ 1\cdot x &+& 1\cdot y &+& 1\cdot z &=& 6 \\ \end{array}\\ \small{ \begin{array}{lcl} \\ \text{Determinant denominator} &=& \begin{vmatrix} 1&-2&3 \\ 2&1&-1 \\ 1&1&1 \\ \end{vmatrix}\\ \\ &=& 1\cdot 1\cdot 1 + 2\cdot 1\cdot 3 +1\cdot (-2)\cdot (-1) -1\cdot 1\cdot 3 -2\cdot (-2)\cdot 1 -1\cdot 1\cdot (-1) \\ &=& 1 + 6 + 2 -3 +4 +1 \\ &=& 11 \\ \end{array} }\)
\(\small{ \begin{array}{lcl} x &=& \dfrac{ \begin{vmatrix} 6&-2&3 \\ -3&1&-1 \\ 6&1&1 \\ \end{vmatrix} }{11}\\\\ &=&\dfrac{ 6\cdot 1\cdot 1 + (-3)\cdot 1\cdot 3 +6\cdot (-2)\cdot (-1) -6\cdot 1\cdot 3 -(-3)\cdot (-2)\cdot 1 -6\cdot 1\cdot (-1) } {11}\\ &=&\dfrac{ 6 -9 + 12 -18 -6 +6 } {11}\\ \mathbf{x} & \mathbf{=} & \mathbf{-\dfrac{9}{11}}\\ \end{array} }\)
\(\small{ \begin{array}{lcl} y &=& \dfrac{ \begin{vmatrix} 1&6&3 \\ 2&-3&-1 \\ 1&6&1 \\ \end{vmatrix} }{11}\\\\ &=&\dfrac{ 1\cdot (-3)\cdot 1 + 2\cdot 6\cdot 3 +1\cdot 6\cdot (-1) -1\cdot (-3)\cdot 3 -2\cdot 6\cdot 1 -1\cdot 6\cdot (-1) } {11}\\ &=&\dfrac{ -3 +36 -6 +9 -12 +6 } {11}\\ \mathbf{y} & \mathbf{=} & \mathbf{\dfrac{30}{11}}\\ \end{array} }\)
\(\small{ \begin{array}{lcl} z &=& \dfrac{ \begin{vmatrix} 1&-2&6 \\ 2&1&-3 \\ 1&1&6 \\ \end{vmatrix} }{11}\\\\ &=&\dfrac{ 1\cdot 1\cdot 6 + 2\cdot 1\cdot 6 +1\cdot (-2)\cdot (-3) -1\cdot 1\cdot 6 -2\cdot (-2)\cdot 6 -1\cdot 1\cdot (-3) } {11}\\ &=&\dfrac{ 6 +12 +6 -6 +24 + 3 } {11}\\ \mathbf{z} & \mathbf{=} & \mathbf{\dfrac{45}{11}}\\ \end{array} }\)
Check:
\(\begin{array}{rcrcrcr} 1\cdot ( -\dfrac{9}{11} ) &-& 2\cdot \dfrac{30}{11} &+& 3\cdot \dfrac{45}{11} &=& 6 \\ 2\cdot ( -\dfrac{9}{11} ) &+& 1\cdot \dfrac{30}{11} &-& 1\cdot \dfrac{45}{11} &=& -3 \\ 1\cdot ( -\dfrac{9}{11} ) &+& 1\cdot \dfrac{30}{11} &+& 1\cdot \dfrac{45}{11} &=& 6 \\ \end{array}\\\)
Solve the following system: WHOSE IDEA WAS IT TO USE CRAMER'S RULE?
{x-2 y+3 z = 6
2 x+y-z = -3
x+y+z = 6
Express the system in matrix form:
(1 | -2 | 3
2 | 1 | -1
1 | 1 | 1)(x
y
z) = (6
-3
6)
Solve the system with Cramer's rule:
x = 6 | -2 | 3
-3 | 1 | -1
6 | 1 | 1/1 | -2 | 3
2 | 1 | -1
1 | 1 | 1 and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/1 | -2 | 3
2 | 1 | -1
1 | 1 | 1 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/1 | -2 | 3
2 | 1 | -1
1 | 1 | 1
Use cofactor expansion about the 1^st row. 1 | -2 | 3
2 | 1 | -1
1 | 1 | 1 = 1 | -1
1 | 1+2 2 | -1
1 | 1+3 2 | 1
1 | 1:
x = 6 | -2 | 3
-3 | 1 | -1
6 | 1 | 1/1 | -1
1 | 1+2 2 | -1
1 | 1+3 2 | 1
1 | 1 and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/1 | -1
1 | 1+2 2 | -1
1 | 1+3 2 | 1
1 | 1 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/1 | -1
1 | 1+2 2 | -1
1 | 1+3 2 | 1
1 | 1
2 | -1
1 | 1 = 2×1-(-1) = 3:
x = 6 | -2 | 3
-3 | 1 | -1
6 | 1 | 1/(1 | -1
1 | 1+2×3+3 2 | 1
1 | 1) and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/(1 | -1
1 | 1+2×3+3 2 | 1
1 | 1) and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/(1 | -1
1 | 1+2×3+3 2 | 1
1 | 1)
2 | 1
1 | 1 = 2×1-1×1 = 1:
x = 6 | -2 | 3
-3 | 1 | -1
6 | 1 | 1/(1 | -1
1 | 1+2×3+3) and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/(1 | -1
1 | 1+2×3+3) and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/(1 | -1
1 | 1+2×3+3)
1 | -1
1 | 1 = 1×1-(-1) = 2:
x = 6 | -2 | 3
-3 | 1 | -1
6 | 1 | 1/(2+2×3+3) and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/(2+2×3+3) and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/(2+2×3+3)
2×3 = 6:
x = 6 | -2 | 3
-3 | 1 | -1
6 | 1 | 1/(2+6+3) and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/(2+6+3) and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/(2+6+3)
2+6+3 = 11:
x = 6 | -2 | 3
-3 | 1 | -1
6 | 1 | 1/11 and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
Use cofactor expansion about the 1^st row. 6 | -2 | 3
-3 | 1 | -1
6 | 1 | 1 = 6 1 | -1
1 | 1+3 -3 | 1
6 | 1+2 -3 | -1
6 | 1:
x = 6 1 | -1
1 | 1+3 -3 | 1
6 | 1+2 -3 | -1
6 | 1/11 and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
1 | -1
1 | 1 = 1×1-(-1) = 2:
x = (3 -3 | 1
6 | 1+2 -3 | -1
6 | 1+2 6)/11 and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
-3 | -1
6 | 1 = -(-6)-3×1 = 3:
x = (3 -3 | 1
6 | 1+3 2+6 2)/11 and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
-3 | 1
6 | 1 = -3×1-1×6 = -9:
x = (6×2+2×3+3×-9)/11 and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
6×2 = 12:
x = (12+2×3-9×3)/11 and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
2×3 = 6:
x = (12+6-9×3)/11 and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
3 (-9) = -27:
x = (12+6+-27)/11 and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
12+6-27 = -9:
x = -9/11 and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
Use cofactor expansion about the 1^st row. 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1 = -3 | -1
6 | 1-6 2 | -1
1 | 1+3 2 | -3
1 | 6:
x = (-9)/11 and y = -3 | -1
6 | 1-6 2 | -1
1 | 1+3 2 | -3
1 | 6/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
2 | -1
1 | 1 = 2×1-(-1) = 3:
x = (-9)/11 and y = (-3 | -1
6 | 1-63+3 2 | -3
1 | 6)/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
2 | -3
1 | 6 = 2×6--3×1 = 15:
x = (-9)/11 and y = (-3 | -1
6 | 1-6×3+3×15)/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
-3 | -1
6 | 1 = -(-6)-3×1 = 3:
x = (-9)/11 and y = (3-6×3+3×15)/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
-6×3 = -18:
x = (-9)/11 and y = (3+-18+3×15)/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
3×15 = 45:
x = (-9)/11 and y = (3-18+45)/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
3-18+45 = 30:
x = (-9)/11 and y = 30/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
Use cofactor expansion about the 1^st row. 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6 = 1 | -3
1 | 6+2 2 | -3
1 | 6+6 2 | 1
1 | 1:
x = (-9)/11 and y = 30/11 and z = 1 | -3
1 | 6+2 2 | -3
1 | 6+6 2 | 1
1 | 1/11
2 | -3
1 | 6 = 2×6--3×1 = 15:
x = (-9)/11 and y = 30/11 and z = (1 | -3
1 | 6+2×15+6 2 | 1
1 | 1)/11
2 | 1
1 | 1 = 2×1-1×1 = 1:
x = (-9)/11 and y = 30/11 and z = (1 | -3
1 | 6+2×15+6)/11
1 | -3
1 | 6 = 1×6--3×1 = 9:
x = (-9)/11 and y = 30/11 and z = (9+2×15+6)/11
2×15 = 30:
x = (-9)/11 and y = 30/11 and z = (9+30+6)/11
9+30+6 = 45:
Answer: | x = (-9)/11 and y = 30/11 and z = 45/11
+26400 solve using Cramer's rule
x-2y+3z=6
2x+y-z=-3
x+y+z=6
\(\begin{array}{rcrcrcr} 1\cdot x &-& 2\cdot y &+& 3\cdot z &=& 6 \\ 2\cdot x &+& 1\cdot y &-& 1\cdot z &=& -3 \\ 1\cdot x &+& 1\cdot y &+& 1\cdot z &=& 6 \\ \end{array}\\ \small{ \begin{array}{lcl} \\ \text{Determinant denominator} &=& \begin{vmatrix} 1&-2&3 \\ 2&1&-1 \\ 1&1&1 \\ \end{vmatrix}\\ \\ &=& 1\cdot 1\cdot 1 + 2\cdot 1\cdot 3 +1\cdot (-2)\cdot (-1) -1\cdot 1\cdot 3 -2\cdot (-2)\cdot 1 -1\cdot 1\cdot (-1) \\ &=& 1 + 6 + 2 -3 +4 +1 \\ &=& 11 \\ \end{array} }\)
\(\small{ \begin{array}{lcl} x &=& \dfrac{ \begin{vmatrix} 6&-2&3 \\ -3&1&-1 \\ 6&1&1 \\ \end{vmatrix} }{11}\\\\ &=&\dfrac{ 6\cdot 1\cdot 1 + (-3)\cdot 1\cdot 3 +6\cdot (-2)\cdot (-1) -6\cdot 1\cdot 3 -(-3)\cdot (-2)\cdot 1 -6\cdot 1\cdot (-1) } {11}\\ &=&\dfrac{ 6 -9 + 12 -18 -6 +6 } {11}\\ \mathbf{x} & \mathbf{=} & \mathbf{-\dfrac{9}{11}}\\ \end{array} }\)
\(\small{ \begin{array}{lcl} y &=& \dfrac{ \begin{vmatrix} 1&6&3 \\ 2&-3&-1 \\ 1&6&1 \\ \end{vmatrix} }{11}\\\\ &=&\dfrac{ 1\cdot (-3)\cdot 1 + 2\cdot 6\cdot 3 +1\cdot 6\cdot (-1) -1\cdot (-3)\cdot 3 -2\cdot 6\cdot 1 -1\cdot 6\cdot (-1) } {11}\\ &=&\dfrac{ -3 +36 -6 +9 -12 +6 } {11}\\ \mathbf{y} & \mathbf{=} & \mathbf{\dfrac{30}{11}}\\ \end{array} }\)
\(\small{ \begin{array}{lcl} z &=& \dfrac{ \begin{vmatrix} 1&-2&6 \\ 2&1&-3 \\ 1&1&6 \\ \end{vmatrix} }{11}\\\\ &=&\dfrac{ 1\cdot 1\cdot 6 + 2\cdot 1\cdot 6 +1\cdot (-2)\cdot (-3) -1\cdot 1\cdot 6 -2\cdot (-2)\cdot 6 -1\cdot 1\cdot (-3) } {11}\\ &=&\dfrac{ 6 +12 +6 -6 +24 + 3 } {11}\\ \mathbf{z} & \mathbf{=} & \mathbf{\dfrac{45}{11}}\\ \end{array} }\)
Check:
\(\begin{array}{rcrcrcr} 1\cdot ( -\dfrac{9}{11} ) &-& 2\cdot \dfrac{30}{11} &+& 3\cdot \dfrac{45}{11} &=& 6 \\ 2\cdot ( -\dfrac{9}{11} ) &+& 1\cdot \dfrac{30}{11} &-& 1\cdot \dfrac{45}{11} &=& -3 \\ 1\cdot ( -\dfrac{9}{11} ) &+& 1\cdot \dfrac{30}{11} &+& 1\cdot \dfrac{45}{11} &=& 6 \\ \end{array}\\\)
