Question

1.5 Solve the equation below using Inverse method (10 marks)

$\begin{bmatrix} 1 & 3 & 0 \\ 0 & 0.5 & 1\\ 0.05 & 0 & 1\\ \end{bmatrix}$ $\begin{bmatrix} x \\ y\\ z\\ \end{bmatrix}$ =$\begin{bmatrix} 4 \\ 1\\ 3\\ \end{bmatrix}$

Answer 1

$A=\begin{bmatrix} 1 & 3 & 0\\ 0 & 0.5 & 1\\ 0.05 & 0 & 1 \end{bmatrix},X=\begin{bmatrix} x \\ y \\ z \end{bmatrix}, B=\begin{bmatrix} 4 \\ 1\\ 3 \end{bmatrix}.$

Matrix 𝑋 satisfying the equation 𝐴𝑋=𝐵

 is given by $X=A^{-1} B.$

Let us begin by finding the inverse of the 3×3 matrix .

$\text{det} A=a_{11}|A_{11}|-a_{12}|A_{12}|+a_{13}|A_{13}|=$

$=1\cdot\begin{vmatrix} 0.5& 1\\0&1\end{vmatrix}-3\cdot\begin{vmatrix} 0& 1\\0.05&1\end{vmatrix}+0\cdot\begin{vmatrix} 0& 0.5\\0.05&0\end{vmatrix}=$

$=1\cdot(0.5\cdot1-1\cdot0)-3\cdot(0\cdot1-1\cdot0.05)+0=0.65;$

$+|A_{11 } |=+\begin{vmatrix}0.5 & 1 \\0 & 1\end{vmatrix}=0.5\cdot1-1\cdot0=0.5;$

$-|A_{12} |=-\begin{vmatrix}0 & 1 \\0.05& 1\end{vmatrix}=-(0\cdot1-1\cdot0.05)=0.05;$

$+|A_{13} |=+\begin{vmatrix}0 & 0.5\\0.05 & 0\end{vmatrix}=-0.025;\\ \ \\ -|A_{21} |=-\begin{vmatrix}3 & 0 \\0& 1\end{vmatrix}=-3;\\$

$+|A_{22} |=+\begin{vmatrix}1& 0\\0.05 & 1\end{vmatrix}=1;\\ \ \\ -|A_{23} |=-\begin{vmatrix}1 & 3 \\0.05& 0\end{vmatrix}=0.15;$

$+|A_{31} |=+\begin{vmatrix}3 & 0\\0 & 1\end{vmatrix}=3;\\ \ \\ -|A_{32} |=-\begin{vmatrix}1 & 0 \\0& 1\end{vmatrix}=-1;\\ \ \\ +|A_{33} |=+\begin{vmatrix}1 & 3\\0 & 0.5\end{vmatrix}=0.5;$

$A^{-1} =\cfrac{1}{0.65}\cdot\begin{bmatrix} 0.5 & - 3 & 3\\0.05 & 1&-1\\-0.025&0.15&0.5\end{bmatrix};$

$\begin{bmatrix} x\\ y\\ z \end{bmatrix}=\cfrac{1}{0.65}\cdot\begin{bmatrix} 0.5 & - 3 & 3\\0.05 & 1&-1\\-0.025&0.15&0.5\end{bmatrix}\cdot\begin{bmatrix} 4\\ 1\\ 3 \end{bmatrix}=$

$=\cfrac{1} {0.65} \cdot\begin{bmatrix} 0.5\cdot4-3\cdot1+3\cdot3\\ 0.05\cdot4+1\cdot1-1\cdot3\\ -0.025\cdot4+0.15\cdot1+0.5\cdot3 \end{bmatrix}=$

$=\cfrac{1} {0.65} \cdot\begin{bmatrix} 8\\ - 1.8\\ 1.55 \end{bmatrix}=\begin{bmatrix} 12.31\\ - 2.77\\ 2.38 \end{bmatrix}.$