First, write the coefficient matrix A and the constant vector b:
A=800−120053,b=1657
Then, calculate the determinant of A, denoted by |A|:
∣A∣=800−120053=(8)(2)(3)−(0)(−1)(5)−(0)(0)(−1)=48
Next, replace the first column of A with b and calculate the determinant, denoted by |A₁|:
A1=1657−120053,∣A1∣=1657−120053=(16)(2)(3)−(5)(−1)(0)−(7)(−1)(5)=96
Similarly, replace the second and third columns of A with b and calculate the determinants, denoted by |A₂| and |A₃|:
A2=8001657053,∣A2∣=8001657053=(8)(5)(3)−(16)(0)(3)−(0)(7)(0)=120
A3=800−1201657,∣A3∣=800−1201657=(8)(2)(7)−(16)(−1)(0)−(−1)(5)(0)=112
Finally, use Cramer’s rule to find the values of x₁, x₂ and x₃:
x1=∣A∣∣A1∣=4896=2
x2=∣A∣∣A2∣=48120=2.5
x3=∣A∣∣A3∣=48112=2.333...
Therefore, the solution is x₁=2, x₂=2.5 and x₃=2.333…
(b)-x1 + 3x2 + 2x3 =24
X₁+ x3=6
5x2-x3=8
First, write the coefficient matrix A and the constant vector b:
A=−11030521−1,b=2468
Then, calculate the determinant of A, denoted by |A|:
∣A∣=−11030521−1=(−1)(−1+10)−(3)(−1+0)+(2)(−5+0)=9+3−10=2
Next, replace the first column of A with b and calculate the determinant, denoted by |A₁|:
A1=246830521−1,∣A1∣=246830521−1=(24)(−1+5)−(3)(−6+8)+(2)(−30+40)=96−6+20=110
Similarly, replace the second and third columns of A with b and calculate the determinants, denoted by |A₂| and |A₃|:
A2=−110246821−1,∣A2∣=−110246821−1=(−1)(−6+8)−(24)(−1+0)+(2)(−8+0)=−2+24−16=6
A3=−1103052468,∣A3∣=−1103052468=(−1)(0+30)−(3)(−8+0)+(24)(−5+0)=−30+24−120=−126
Finally, use Cramer’s rule to find the values of x₁, x₂ and x₃:
x1=∣A∣∣A1∣=2110=55
x2=∣A∣∣A2∣=26=3
x3=∣A∣∣A3∣=2−126=−63
Therefore, the solution is x₁=55, x₂=3 and x₃=-63.
© 4x+3y-2z=1
x + 2y=6
3x+z=4
First, write the coefficient matrix A and the constant vector b:
A=413320−201,b=164
Then, calculate the determinant of A, denoted by |A|:
∣A∣=413320−201=(4)(2+0)−(3)(0+0)+(−2)(0−6)=8+12=20
Next, replace the first column of A with b and calculate the determinant, denoted by |A₁|:
A1=164320−201,∣A1∣=164320−201=(1)(2+0)−(3)(0+0)+(−2)(0−24)=2+48=50
Similarly, replace the second and third columns of A with b and calculate the determinants, denoted by |A₂| and |A₃|:
A2= :
A2=413164−201,∣A2∣=413164−201=(4)(6+0)−(1)(0+0)+(−2)(−12+4)=24+16=40
A3=413320164,∣A3∣=413320164=(4)(8+0)−(3)(4+18)+(1)(0−12)=32−66−12=−46
Finally, use Cramer’s rule to find the values of x, y and z:
x=∣A∣∣A1∣=2050=2.5
y=∣A∣∣A2∣=2040=2
z=∣A∣∣A3∣=20−46=−2.3
Therefore, the solution is x=2.5, y=2 and z=-2.3.
(d) -x+y+z=a
x-y+z=b
x+y-z = C
First, write the coefficient matrix A and the constant vector b:
A=−1111−1111−1,b=abc
Then, calculate the determinant of A, denoted by |A|:
∣A∣=−1111−1111−1=(−1)(−2+1)−(1)(−2+1)+(1)(−2+1)=−3+3+3=3
Next, replace the first column of A with b and calculate the determinant, denoted by |A₁|:
A1=
$$A_2=\begin{bmatrix}4&1&-2\\1&6&0\\3&4&1\end{bmatrix}, |A_2|=\begin{vmatrix}4&1&-2\\1&6&0\\3&4&1\end{vmatrix}=(4)(6+0)-(1)(0+0)+(-2)(-12+4)=24+16=40$$
$$A_3=\begin{bmatrix}4&3&1\\1&2&6\\3&0&4\end{bmatrix}, |A_3|=\begin{vmatrix}4&3&1\\1&2&6\\3&0&4\end{vmatrix}=(4)(8+0)-(3)(4+18)+(1)(0-12)=32-66-12=-46$$
Finally, use Cramer's rule to find the values of x, y and z:
$$x=\fracA=\frac{50}{20}=2.5$$
$$y=\fracA_2A=\frac{40}{20}=2$$
$$z=\frac=\frac{-46}{20}=-2.3$$
Therefore, the solution is x=2.5, y=2 and z=-2.3.
(d) -x+y+z=a
x-y+z=b
x+y-z = C
First, write the coefficient matrix A and the constant vector b:
$$A=\begin{bmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{bmatrix}, b=\begin{bmatrix}a\\b\\c\end{bmatrix}$$
Then, calculate the determinant of A, denoted by |A|:
$$|A|=\begin{vmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{vmatrix}=(-1)(-2+1)-(1)(-2+1)+(1)(-2+1)=-3+3+3=3$$
Next, replace the first column of A with b and calculate the determinant, denoted by |A₁|:
$$A_1=\begin{bmatrix}a&1&1\\b&-1