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Solve by Cramer's Rule

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Exercise 1

 

\begin{cases} 2x-y+z \\ 2y-z \\ -x+y \end{cases}\begin{matrix} =3 \\ =1 \\ =1 \end{matrix}

 

Exercise 2

 

\begin{cases} x+y+z \\ x-y+z \\ -x+y+z \end{cases}\begin{matrix} =1 \\ =1 \\ =1 \end{matrix}

 

Exercise 3

 

\begin{cases} x+y+z \\ x-2y+3z \\ x+z \end{cases}\begin{matrix} =1 \\ =2 \\ =5 \end{matrix}

 

Exercise 4

 

\begin{cases} 2x-y-2z \\ -x+y+z \\ x-2y+z \end{cases}\begin{matrix} =-2 \\ =0 \\ =8 \end{matrix}

 

Exercise 5

 

\begin{cases} x+2y-2z \\ 4x-y+z \\ -2x+y+z \end{cases}\begin{matrix} =10 \\ =4 \\ =-2 \end{matrix}

 

Exercise 6

 

\begin{cases} 2x-y+z-2t \\ 2x+2y-3z+t \\ -x+y-z \\ 4x-3y+2z-3t \end{cases}\begin{matrix} =-5 \\ =-1 \\ =-1 \\ =-8 \end{matrix}

 

 

Solution of exercise 1

 

\begin{cases} 2x-y+z \\ 2y-z \\ -x+y \end{cases}\begin{matrix} =3 \\ =1 \\ =1 \end{matrix}

D=\begin{vmatrix} 2 & -1 & 1 \\ 0 & 2 & -1 \\ -1 & 1 & 0 \end{vmatrix}

\left| D \right| =2\begin{vmatrix} 2 & -1 \\ 1 & 0 \end{vmatrix}-(-1)\begin{vmatrix} 0 & -1 \\ -1 & 0 \end{vmatrix}+1\begin{vmatrix} 0 & 2 \\ -1 & 1 \end{vmatrix}

\left| D \right| =2[0-(-1)]+[0-1]+[0-(-2)]

\left| D \right| =2-1+2

\left| D \right| =3

 

{ D }_{ x }=\begin{vmatrix} 3 & -1 & 1 \\ 1 & 2 & -1 \\ 1 & 1 & 0 \end{vmatrix}

\frac { { \left| D_{ x } \right| } }{ \left| D \right| } =\frac { \begin{vmatrix} 3 & -1 & 1 \\ 1 & 2 & -1 \\ 1 & 1 & 0 \end{vmatrix} }{ 3 }

\left| D_{ x } \right| =3[0-(-1)]-(-1)[0-(-1)]+1[1-2]

\left| D_{ x } \right| =3+1-1=3

\frac { { \left| D_{ x } \right| } }{ \left| D \right| } =\frac { 3 }{ 3 } =1

 

{ D }_{ y }=\begin{vmatrix} 2 & 3 & 1 \\ 0 & 1 & -1 \\ -1 & 1 & 0 \end{vmatrix}

\frac { { \left| D_{ y } \right| } }{ \left| D \right| } =\frac { \begin{vmatrix} 2 & 3 & 1 \\ 0 & 1 & -1 \\ -1 & 1 & 0 \end{vmatrix} }{ 3 }

\left| D_{ y } \right| =2[0-(-1)]-3[0-1]+1[0-(-1)]

\left| D_{ y } \right| =2+3+1=6

\frac { { \left| D_{ y } \right| } }{ \left| D \right| } =\frac { 6 }{ 3 } =2

 

{ D }_{ z }=\begin{vmatrix} 2 & -1 & 1 \\ 0 & 2 & -1 \\ -1 & 1 & 0 \end{vmatrix}

\frac { { \left| { D }_{ z } \right| } }{ \left| D \right| } =\frac { \begin{vmatrix} 2 & -1 & 3 \\ 0 & 2 & 1 \\ -1 & 1 & 1 \end{vmatrix} }{ 3 }

\left| { D }_{ z } \right| =2[2-1]-(-1)[0-1]+3[0-(-2)]

\left| { D }_{ z } \right| =2+1+6=9

\frac { { \left| { D }_{ z } \right| } }{ \left| D \right| } =\frac { 9 }{ 3 } =3

Solution of exercise 2

 

\begin{cases} x+y+z \\ x-y+z \\ -x+y+z \end{cases}\begin{matrix} =1 \\ =1 \\ =1 \end{matrix}

D=\begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ -1 & 1 & 1 \end{vmatrix}

\left| D \right| =1\begin{vmatrix} -1 & 1 \\ 1 & 1 \end{vmatrix}-1\begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix}+1\begin{vmatrix} 1 & -1 \\ -1 & 1 \end{vmatrix}

\left| D \right| =1[-1-1]-1[1-(-1)]+1[1-1]

\left| D \right| =-2-2+0

\left| D \right| =-4

 

{ D }_{ x }=\begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & 1 \end{vmatrix}

\frac { { \left| D_{ x } \right| } }{ \left| D \right| } =\frac { \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & 1 \end{vmatrix} }{ -4 }

\left| D_{ x } \right| =1[-1-1]-1[1-1]+1[1+1]

\left| D_{ x } \right| =-2-0+2=0

\frac { { \left| D_{ x } \right| } }{ \left| D \right| } =\frac { 0 }{ -4 } =0

 

{ D }_{ y }=\begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ -1 & 1 & 1 \end{vmatrix}

\frac { { \left| D_{ y } \right| } }{ \left| D \right| } =\frac { \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ -1 & 1 & 1 \end{vmatrix} }{ -4 }

\left| D_{ y } \right| =1[1-1]-1[1-(-1)]+1[1-(-1)]

\left| D_{ y } \right| =0-2+2=0

\frac { { \left| D_{ y } \right| } }{ \left| D \right| } =\frac { 0 }{ -4 } =0

 

{ D }_{ z }=\begin{vmatrix} 2 & -1 & 1 \\ 0 & 2 & -1 \\ -1 & 1 & 0 \end{vmatrix}

\frac { { \left| { D }_{ z } \right| } }{ \left| D \right| } =\frac { \begin{vmatrix} 2 & -1 & 3 \\ 0 & 2 & 1 \\ -1 & 1 & 1 \end{vmatrix} }{ -4 }

\left| { D }_{ z } \right| =1[-1-1]-1[1-(-1)]+1[1-1]

\left| { D }_{ z } \right| =-2-2+0=-4

\frac { { \left| { D }_{ z } \right| } }{ \left| D \right| } =\frac { -4 }{ -4 } =1

Solution of exercise 3

 

\begin{cases} x+y+z \\ x-2y+3z \\ x+z \end{cases}\begin{matrix} =1 \\ =2 \\ =5 \end{matrix}

D=\begin{vmatrix} 1 & 1 & 1 \\ 1 & -2 & 3 \\ 1 & 0 & 1 \end{vmatrix}

\left| D \right| =1\begin{vmatrix} -2 & 3 \\ 0 & 1 \end{vmatrix}-1\begin{vmatrix} 1 & 3 \\ 1 & 1 \end{vmatrix}+1\begin{vmatrix} 1 & -2 \\ 1 & 0 \end{vmatrix}

\left| D \right| =1[-2-0]-1[1-3]+1[0-(-2)]

\left| D \right| =-2+2+2

\left| D \right| =2

 

{ D }_{ x }=\begin{vmatrix} 1 & 1 & 1 \\ 2 & -2 & 3 \\ 5 & 0 & 1 \end{vmatrix}

\frac { { \left| D_{ x } \right| } }{ \left| D \right| } =\frac { \begin{vmatrix} 1 & 1 & 1 \\ 2 & -2 & 3 \\ 5 & 0 & 1 \end{vmatrix} }{ 2 }

\left| D_{ x } \right| =1[-2-0]-1[2-15]+1[0-(-10)]

\left| D_{ x } \right| =-2+13+10=21

\frac { { \left| D_{ x } \right| } }{ \left| D \right| } =\frac { 21 }{ 2 }

 

{ D }_{ y }=\begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 5 & 1 \end{vmatrix}

\frac { { \left| D_{ y } \right| } }{ \left| D \right| } =\frac { \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 5 & 1 \end{vmatrix} }{ 2 }

\left| D_{ y } \right| =1[2-15]-1[1-3]+1[5-2]

\left| D_{ y } \right| =-13+2+3=-8

\frac { { \left| D_{ y } \right| } }{ \left| D \right| } =\frac { -8 }{ 2 } =-4

 

{ D }_{ z }=\begin{vmatrix} 1 & 1 & 1 \\ 1 & -2 & 2 \\ 1 & 0 & 5 \end{vmatrix}

\frac { { \left| { D }_{ z } \right| } }{ \left| D \right| } =\frac { \begin{vmatrix} 1 & 1 & 1 \\ 1 & -2 & 2 \\ 1 & 0 & 5 \end{vmatrix} }{ 2 }

\left| { D }_{ z } \right| =1[-10-0]-1[5-2]+1[0-(-2)]

\left| { D }_{ z } \right| =-10-3+2=-11

\frac { { \left| { D }_{ z } \right| } }{ \left| D \right| } =\frac { -11 }{ 2 }

Solution of exercise 4

 

\begin{cases} 2x-y-2z \\ -x+y+z \\ x-2y+z \end{cases}\begin{matrix} =-2 \\ =0 \\ =8 \end{matrix}

{ D }=\begin{vmatrix} 2 & -1 & -2 \\ -1 & 1 & 1 \\ 1 & -2 & 1 \end{vmatrix}

\left| D \right| =2\begin{vmatrix} 1 & 1 \\ -2 & 1 \end{vmatrix}-(-1)\begin{vmatrix} -1 & 1 \\ 1 & 1 \end{vmatrix}-2\begin{vmatrix} -1 & 1 \\ 1 & -2 \end{vmatrix}

\left| D \right| =2[1-(-2)]+1[-1-1]-2[2-1]

\left| D \right| =6-2-2

\left| D \right| =2

 

{ D }_{ x }=\begin{vmatrix} -2 & -1 & -2 \\ 0 & 1 & 1 \\ 8 & -2 & 1 \end{vmatrix}

\frac { { \left| D_{ x } \right| } }{ \left| D \right| } =\frac { \begin{vmatrix} -2 & -1 & -2 \\ 0 & 1 & 1 \\ 8 & -2 & 1 \end{vmatrix} }{ 2 }

\left| D_{ x } \right| =-2[1-(-2)]-(-1)[0-8]-2[0-8]

\left| D_{ x } \right| =-6-8+16=2

\frac { { \left| D_{ x } \right| } }{ \left| D \right| } =\frac { 2 }{ 2 } =1

 

{ D }_{ y }=\begin{vmatrix} 2 & -2 & -2 \\ -1 & 0 & 1 \\ 1 & 8 & 1 \end{vmatrix}

\frac { { \left| D_{ y } \right| } }{ \left| D \right| } =\frac { \begin{vmatrix} 2 & -2 & -2 \\ -1 & 0 & 1 \\ 1 & 8 & 1 \end{vmatrix} }{ 2 }

\left| D_{ y } \right| =2[0-8]-(-2)[-1-1]-2[-8-0]

\left| D_{ y } \right| =-16-4+16=-4

\frac { { \left| D_{ y } \right| } }{ \left| D \right| } =\frac { -4 }{ 2 } =-2

 

{ D }_{ z }=\begin{vmatrix} 2 & -1 & -2 \\ -1 & 1 & 0 \\ 1 & -2 & 8 \end{vmatrix}

\frac { { \left| { D }_{ z } \right| } }{ \left| D \right| } =\frac { \begin{vmatrix} 2 & -1 & -2 \\ -1 & 1 & 0 \\ 1 & -2 & 8 \end{vmatrix} }{ 2 }

\left| { D }_{ z } \right| =2[8-0]-(-1)[-8-0]-2[2-1]

\left| { D }_{ z } \right| =16-8-2=6

\frac { { \left| { D }_{ z } \right| } }{ \left| D \right| } =\frac { 6 }{ 2 } =3

Solution of exercise 5

 

\begin{cases} x+2y-2z \\ 4x-y+z \\ -2x+y+z \end{cases}\begin{matrix} =10 \\ =4 \\ =-2 \end{matrix}

\left| D \right| =\begin{vmatrix} 1 & 2 & -2 \\ 4 & -1 & 1 \\ -2 & 1 & 1 \end{vmatrix}

\left| D \right| =1\begin{vmatrix} -1 & 1 \\ 1 & 1 \end{vmatrix}-2\begin{vmatrix} 4 & 1 \\ -2 & 1 \end{vmatrix}-2\begin{vmatrix} 4 & -1 \\ -2 & 1 \end{vmatrix}

\left| D \right| =1[-1-1]-2[4-(-2)]-2[4-2]

\left| D \right| =-2-12-4

\left| D \right| =-18

 

{ D }_{ x }=\begin{vmatrix} 10 & 2 & -2 \\ 4 & -1 & 1 \\ -2 & 1 & 1 \end{vmatrix}

\frac { { \left| D_{ x } \right| } }{ \left| D \right| } =\frac { \begin{vmatrix} 10 & 2 & -2 \\ 4 & -1 & 1 \\ -2 & 1 & 1 \end{vmatrix} }{ -18 }

\left| D_{ x } \right| =10[-1-1]-2[4-(-2)]-2[4-2]

\left| D_{ x } \right| =-20-12-4=-36

\frac { { \left| D_{ x } \right| } }{ \left| D \right| } =\frac { -36 }{ -18 } =2

 

{ D }_{ y }=\begin{vmatrix} 1 & 10 & -2 \\ 4 & 4 & 1 \\ -2 & -2 & 1 \end{vmatrix}

\frac { { \left| D_{ y } \right| } }{ \left| D \right| } =\frac { \begin{vmatrix} 1 & 10 & -2 \\ 4 & 4 & 1 \\ -2 & -2 & 1 \end{vmatrix} }{ -18 }

\left| D_{ y } \right| =1[4-(-2)]-10[4-(-2)]-2[-8-(-8)]

\left| D_{ y } \right| =6-60-0=-54

\frac { { \left| D_{ y } \right| } }{ \left| D \right| } =\frac { -54 }{ -18 } =3

 

{ D }_{ z }=\begin{vmatrix} 1 & 2 & 10 \\ 4 & -1 & 4 \\ -2 & 1 & -2 \end{vmatrix}

\frac { { \left| { D }_{ z } \right| } }{ \left| D \right| } =\frac { \begin{vmatrix} 1 & 2 & 10 \\ 4 & -1 & 4 \\ -2 & 1 & -2 \end{vmatrix} }{ -18 }

\left| { D }_{ z } \right| =1[2-4]-2[-8-(-8)]+10[4-2]

\left| { D }_{ z } \right| =-2-0+20=18

\frac { { \left| { D }_{ z } \right| } }{ \left| D \right| } =\frac { 18 }{ -18 } =-1

Solution of exercise 6

 

\begin{cases} 2x-y+z-2t \\ 2x+2y-3z+t \\ -x+y-z \\ 4x-3y+2z-3t \end{cases}\begin{matrix} =-5 \\ =-1 \\ =-1 \\ =-8 \end{matrix}

D=\begin{vmatrix} 2 & -1 & 1 & -2 \\ 2 & 2 & -3 & 1 \\ -1 & 1 & -1 & 0 \\ 4 & -3 & 2 & -3 \end{vmatrix}

\left| D \right| =2\begin{vmatrix} 2 & -3 & 1 \\ 1 & -1 & 0 \\ -3 & 2 & -3 \end{vmatrix}-(-1)\begin{vmatrix} 2 & -3 & 1 \\ -1 & -1 & 0 \\ 4 & 2 & -3 \end{vmatrix}+1\begin{vmatrix} 2 & 2 & 1 \\ -1 & 1 & 0 \\ 4 & -3 & -3 \end{vmatrix}-(-2)\begin{vmatrix} 2 & 2 & -3 \\ -1 & 1 & -1 \\ 4 & -3 & 2 \end{vmatrix}

\left| D \right| =2[\{ (2)(-1)(-3)-(2)(0)(2)\} -\{ (-3)(1)(-3)-(-3)(0)(-3)\} +\{ (1)(1)(2)-(1)(-1)(-3)\} ]+1[\{ (2)(-1)(-3)-(2)(0)(2)\} -\{ (-3)(-1)(-3)-(-3)(0)(4)\} +\{ (1)(-1)(2)-(1)(-1)(4)\} ]+1[\{ (2)(1)(-3)-(2)(0)(-3)\} -\{ (2)(-1)(-3)-(2)(0)(4)\} +\{ (1)(-1)(-3)-(1)(1)(4)\} ]+2[\{ (2)(1)(2)-(2)(-1)(-3)\} -\{ (2)(-1)(2)-(2)(-1)(4)\} +\{ (-3)(-1)(-3)-(-3)(1)(4)\} ]

\left| D \right| =2[\{ 6-0\} -\{ 9-0\} +\{ 2-3\} ]+1[\{ 6-0\} -\{ -9-0\} +\{ -2+4\} ]+1[\{ -6+0\} -\{ 6-0\} +\{ 3-4\} ]+2[\{ 4-6\} -\{ -4+8\} +\{ -9+12\} ]

\left| D \right| =2[6-9-1]+1[6+9+2]+1[-6-6-1]+2[-2-4+3]

\left| D \right| =2[-4]+1[17]+1[-13]+2[-3]

\left| D \right| =-8+17-13-6

\left| D \right| =-10

 

{ D }_{ x }=\begin{vmatrix} -5 & -1 & 1 & -2 \\ -1 & 2 & -3 & 1 \\ -1 & 1 & -1 & 0 \\ -8 & -3 & 2 & -3 \end{vmatrix}

\frac { { \left| D_{ x } \right| } }{ \left| D \right| } =\frac { \begin{vmatrix} -5 & -1 & 1 & -2 \\ -1 & 2 & -3 & 1 \\ -1 & 1 & -1 & 0 \\ -8 & -3 & 2 & -3 \end{vmatrix} }{ -10 }

\left| { D }_{ x } \right| =-5\begin{vmatrix} 2 & -3 & 1 \\ 1 & -1 & 0 \\ -3 & 2 & -3 \end{vmatrix}-(-1)\begin{vmatrix} -1 & -3 & 1 \\ -1 & -1 & 0 \\ -8 & 2 & -3 \end{vmatrix}+1\begin{vmatrix} -1 & 2 & 1 \\ -1 & 1 & 0 \\ -8 & -3 & -3 \end{vmatrix}-(-2)\begin{vmatrix} -1 & 2 & -3 \\ -1 & 1 & -1 \\ -8 & -3 & 2 \end{vmatrix}

\left| D_{ x } \right| =0

\frac { { \left| D_{ x } \right| } }{ \left| D \right| } =\frac { 0 }{ -10 } =0

 

{ D }_{ y }=\begin{vmatrix} 2 & -5 & 1 & -2 \\ 2 & -1 & -3 & 1 \\ -1 & -1 & -1 & 0 \\ 4 & -8 & 2 & -3 \end{vmatrix}

\frac { { \left| D_{ y } \right| } }{ \left| D \right| } =\frac { \begin{vmatrix} 2 & -5 & 1 & -2 \\ 2 & -1 & -3 & 1 \\ -1 & -1 & -1 & 0 \\ 4 & -8 & 2 & -3 \end{vmatrix} }{ -10 }

\left| D_{ y } \right| =2\begin{vmatrix} -1 & -3 & 1 \\ -1 & -1 & 0 \\ -8 & 2 & -3 \end{vmatrix}-(-5)\begin{vmatrix} 2 & -3 & 1 \\ -1 & -1 & 0 \\ 4 & 2 & -3 \end{vmatrix}+1\begin{vmatrix} 2 & -1 & 1 \\ -1 & -1 & 0 \\ 4 & -8 & -3 \end{vmatrix}-(-2)\begin{vmatrix} 2 & -1 & -3 \\ -1 & -1 & -1 \\ 4 & -8 & 2 \end{vmatrix}

\left| D_{ y } \right| =-10

\frac { { \left| D_{ y } \right| } }{ \left| D \right| } =\frac { -10 }{ -10 } =1

 

{ D }_{ z }=\begin{vmatrix} 2 & -1 & -5 & -2 \\ 2 & 2 & -1 & 1 \\ -1 & 1 & -1 & 0 \\ 4 & -3 & -8 & -3 \end{vmatrix}

\frac { { \left| D_{ z } \right| } }{ \left| D \right| } =\frac { \begin{vmatrix} 2 & -1 & -5 & -2 \\ 2 & 2 & -1 & 1 \\ -1 & 1 & -1 & 0 \\ 4 & -3 & -8 & -3 \end{vmatrix} }{ -10 }

\left| D_{ z } \right| =2\begin{vmatrix} 2 & -1 & 1 \\ 1 & -1 & 0 \\ -3 & -8 & -3 \end{vmatrix}-(-1)\begin{vmatrix} 2 & -1 & 1 \\ -1 & -1 & 0 \\ 4 & -8 & -3 \end{vmatrix}+(-5)\begin{vmatrix} 2 & 2 & 1 \\ -1 & 1 & 0 \\ 4 & -3 & -3 \end{vmatrix}-(-2)\begin{vmatrix} 2 & 2 & -1 \\ -1 & 1 & -1 \\ 4 & -3 & -8 \end{vmatrix}

\left| D_{ z } \right| =-20

\frac { { \left| D_{ z } \right| } }{ \left| D \right| } =\frac { -20 }{ -10 } =2

 

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.