We are given that A = [ 1 3 0 0 0.5 1 0.5 0 1 ] A =\begin{bmatrix}
1 & 3 & 0 \\
0 & 0.5 & 1 \\
0.5 & 0 & 1
\end{bmatrix} A = ⎣ ⎡ 1 0 0.5 3 0.5 0 0 1 1 ⎦ ⎤ , b = [ 4 1 4 ] b = \begin{bmatrix}
4 \\
1 \\
4
\end{bmatrix} b = ⎣ ⎡ 4 1 4 ⎦ ⎤
(a) Using the matrix inversion
A − 1 = 1 d e t ( A ) A − 1 = 1 1 ∗ 0.5 ∗ 1 + 3 ∗ 0.5 ∗ 0.5 [ 0.5 0.5 − 0.25 − 3 1 1.5 3 − 1 0.5 ] T = [ 0.25 − 1.5 1.5 0.25 0.5 − 0.5 − 0.125 0.75 0.25 ] 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} A − 1 = d e t ( A ) 1 A − 1 = 1 ∗ 0.5 ∗ 1 + 3 ∗ 0.5 ∗ 0.5 1 ⎣ ⎡ 0.5 − 3 3 0.5 1 − 1 − 0.25 1.5 0.5 ⎦ ⎤ T = ⎣ ⎡ 0.25 0.25 − 0.125 − 1.5 0.5 0.75 1.5 − 0.5 0.25 ⎦ ⎤
[ x y z ] = A − 1 b = [ 5.5 − 0.5 1.25 ] \begin{bmatrix}
x \\
y \\
z
\end{bmatrix} = A^{-1} b = \begin{bmatrix}
5.5 \\
-0.5 \\
1.25
\end{bmatrix} ⎣ ⎡ x y z ⎦ ⎤ = A − 1 b = ⎣ ⎡ 5.5 − 0.5 1.25 ⎦ ⎤
(b) Using the Cramer's rule
d e t ( A 1 ) = d e t [ 4 3 0 1 0.5 1 4 0 1 ] = 4 ∗ 0.5 ∗ 1 + 3 ∗ 1 ∗ ( − 1 + 4 ) = 11 det(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 d e t ( A 1 ) = d e t ⎣ ⎡ 4 1 4 3 0.5 0 0 1 1 ⎦ ⎤ = 4 ∗ 0.5 ∗ 1 + 3 ∗ 1 ∗ ( − 1 + 4 ) = 11
d e t ( A 2 ) = d e t [ 1 4 0 0 1 1 0.5 4 1 ] = 1 ∗ ( 1 ∗ 1 − 1 ∗ 4 ) + 4 ∗ 1 ∗ 0.5 = − 1 det(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 d e t ( A 2 ) = d e t ⎣ ⎡ 1 0 0.5 4 1 4 0 1 1 ⎦ ⎤ = 1 ∗ ( 1 ∗ 1 − 1 ∗ 4 ) + 4 ∗ 1 ∗ 0.5 = − 1
d e t ( A 3 ) = d e t [ 1 3 4 0 0.5 1 0.5 0 4 ] = 1 ∗ 0.5 ∗ 4 + 3 ∗ 1 ∗ 0.5 − 4 ∗ 0.5 ∗ 0.5 = 2.5 det(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 d e t ( A 3 ) = d e t ⎣ ⎡ 1 0 0.5 3 0.5 0 4 1 4 ⎦ ⎤ = 1 ∗ 0.5 ∗ 4 + 3 ∗ 1 ∗ 0.5 − 4 ∗ 0.5 ∗ 0.5 = 2.5
[ x y z ] = [ d e t ( A 1 ) / d e t ( A ) d e t ( A 2 ) / d e t ( A ) d e t ( A 3 ) / d e t ( A ) ] = [ 5.5 − 0.5 1.25 ] \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} ⎣ ⎡ x y z ⎦ ⎤ = ⎣ ⎡ d e t ( A 1 ) / d e t ( A ) d e t ( A 2 ) / d e t ( A ) d e t ( A 3 ) / d e t ( A ) ⎦ ⎤ = ⎣ ⎡ 5.5 − 0.5 1.25 ⎦ ⎤
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