Question

Solve the following system of equations by matrix method: \(\dfrac2x+\dfrac3y+\dfrac{10}z=4\), \(\dfrac4x-\dfrac6y+\dfrac5z=1\), \(\dfrac6x+\dfrac9y-\dfrac{20}z=2\).

Hint:

Substitute u = 1/x, v = 1/y, w = 1/z to make the system linear, then use the matrix inverse method.

Answer 1

Step 1: Understanding the Concept:
The equations are not linear in x, y, z because of the reciprocal terms.
Substitute \(u=\dfrac1x\), \(v=\dfrac1y\), \(w=\dfrac1z\) to convert the system into a linear system in u, v, w.

Step 2: Writing the linear system:
\[ 2u+3v+10w=4,\quad 4u-6v+5w=1,\quad 6u+9v-20w=2 \]
In matrix form \(MU=B\), where \(M=\begin{bmatrix}2 & 3 & 10\\4 & -6 & 5\\6 & 9 & -20\end{bmatrix}\), \(U=\begin{bmatrix}u\\v\\w\end{bmatrix}\), \(B=\begin{bmatrix}4\\1\\2\end{bmatrix}\).

Step 3: Finding det(M):
Expand along the first row.
\[ \det M=2[(-6)(-20)-5(9)]-3[4(-20)-5(6)]+10[4(9)-(-6)(6)] \]
\[ =2(75)-3(-110)+10(72)=150+330+720=1200 \]
Since \(\det M\neq0\), M is invertible and the system has a unique solution.

Step 4: Finding the adjoint and inverse:
Computing all cofactors gives:
\[ \text{adj}(M)=\begin{bmatrix}75 & 150 & 75\\110 & -100 & 30\\72 & 0 & -24\end{bmatrix} \]
\[ M^{-1}=\dfrac{1}{1200}\begin{bmatrix}75 & 150 & 75\\110 & -100 & 30\\72 & 0 & -24\end{bmatrix} \]

Step 5: Solving for u, v, w:
\[ U=M^{-1}B \implies u=\dfrac{75(4)+150(1)+75(2)}{1200}=\dfrac{600}{1200}=\dfrac12 \]
\[ v=\dfrac{110(4)-100(1)+30(2)}{1200}=\dfrac{400}{1200}=\dfrac13,\quad w=\dfrac{72(4)+0(1)-24(2)}{1200}=\dfrac{240}{1200}=\dfrac15 \]

Step 6: Converting back to x, y, z:
Since \(u=\dfrac1x\), \(v=\dfrac1y\), \(w=\dfrac1z\):
\[ x=\dfrac1u=2,\quad y=\dfrac1v=3,\quad z=\dfrac1w=5 \]

Final Answer:
These values satisfy all three original equations.
\[ \boxed{x=2,\ y=3,\ z=5} \]