Answer to Question #311837 in Linear Algebra for Showi

Question

Solve the following equation using inverse method

[1 3 0] [x] [4]

[0 0.5 1] [y]=[1]

[0.5 0 1] [z] [3]

Answer 1

Solution

Given that

$\begin{bmatrix} 1 & 3 & 0\\ 0 & 0.5 & 1\\ 0.5 & 0&1\\ \end{bmatrix}\times\begin{bmatrix} x \\ y \\ z\\ \end{bmatrix}=\begin{bmatrix} 4 \\ 1 \\ 3\\ \end{bmatrix}$

This can be written as

$AX=B$

$A^{-1}AX=A^{-1}B$

$IX=A^{-1}B$

$X=A^{-1}B$

Now

$A=\begin{bmatrix} 1 & 3 & 0\\ 0 & 0.5 & 1\\ 0.5 & 0&1\\ \end{bmatrix}$

\begin{gathered} \Rightarrow & \det \left( A \right) = 1\left( {0.5 - 0} \right) - 3\left( {0 - 0.5} \right) + 0\left( {0 - 0.25} \right) \\ \Rightarrow & \det \left( A \right) = 0.5 + 1.5 = 2 \\ \end{gathered} \

Then 

{A^{ - 1}} = \frac{{AdjA}}{{\det \left( A \right)}}\

{A^{ - 1}} = \frac{\begin{bmatrix} 0.5 & -3 & 3\\ 0.5 & 1 & -1\\ -0.25 & 1.5&0.5\\ \end{bmatrix}}{{(2)}}\

${A^{ - 1}} = \begin{bmatrix} 0.25 & -1.5 & 1.5\\ 0.25 & 0.5 & -0.5\\ -0.125 & 0.75&0.25\\ \end{bmatrix}$

Therefore,

$X=A^{-1}B$

$X=$ $\begin{bmatrix} 0.25 & -1.5 & 1.5\\ 0.25 & 0.5 & -0.5\\ -0.125 & 0.75&0.25\\ \end{bmatrix}\begin{bmatrix} x \\ y \\ z\\ \end{bmatrix}$

$\begin{bmatrix} x \\ y \\ z\\ \end{bmatrix}=$ $\begin{bmatrix} 4\\ 0\\ 1\\ \end{bmatrix}$

Hence

$x=4,\ y=0\ z=1$