Question #116194
Solve the following system of linear equations using (a) matrix inversion, and (b) Cramer's rule. [ 1 3 ox] To 0.5 1 y 10.5 0 1Z
1
Expert's answer
2020-05-19T09:02:07-0400

We are given that A=[13000.510.501]A =\begin{bmatrix} 1 & 3 & 0 \\ 0 & 0.5 & 1 \\ 0.5 & 0 & 1 \end{bmatrix}, b=[414]b = \begin{bmatrix} 4 \\ 1 \\ 4 \end{bmatrix}

(a) Using the matrix inversion

A1=1det(A)A1=110.51+30.50.5[0.50.50.25311.5310.5]T=[0.251.51.50.250.50.50.1250.750.25]A^{-1} =\frac{1}{det(A)}A^{-1}=\frac{1}{1*0.5*1 + 3*0.5*0.5} \begin{bmatrix} 0.5 & 0.5 & -0.25 \\ -3 & 1 & 1.5 \\ 3 & -1 & 0.5 \end{bmatrix}^{T} = \begin{bmatrix} 0.25 & -1.5 & 1.5 \\ 0.25 & 0.5 & -0.5 \\ -0.125 & 0.75 & 0.25 \end{bmatrix}

[xyz]=A1b=[5.50.51.25]\begin{bmatrix} x \\ y \\ z \end{bmatrix} = A^{-1} b = \begin{bmatrix} 5.5 \\ -0.5 \\ 1.25 \end{bmatrix}


(b) Using the Cramer's rule

det(A1)=det[43010.51401]=40.51+31(1+4)=11det(A_1)=det\begin{bmatrix} 4 & 3 & 0 \\ 1 & 0.5 & 1 \\ 4 & 0 & 1 \end{bmatrix} = 4 * 0.5 * 1 + 3 * 1 * (-1 + 4) = 11

det(A2)=det[1400110.541]=1(1114)+410.5=1det(A_2)=det\begin{bmatrix} 1 & 4 & 0 \\ 0 & 1 & 1 \\ 0.5 & 4 & 1 \end{bmatrix} = 1 * (1 * 1 - 1 * 4) + 4 * 1 * 0.5 = -1


det(A3)=det[13400.510.504]=10.54+310.540.50.5=2.5det(A_3)=det\begin{bmatrix} 1 & 3 & 4 \\ 0 & 0.5 & 1 \\ 0.5 & 0 & 4 \end{bmatrix} = 1 * 0.5 * 4 + 3 * 1 * 0.5 - 4 * 0.5 * 0.5 = 2.5

[xyz]=[det(A1)/det(A)det(A2)/det(A)det(A3)/det(A)]=[5.50.51.25]\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} det(A_1)/det(A) \\ det(A_2)/det(A) \\ det(A_3)/det(A) \end{bmatrix} = \begin{bmatrix} 5.5 \\ -0.5 \\ 1.25 \end{bmatrix}


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