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: Substitution to linearise:
Put $u=\dfrac1x$, $v=\dfrac1y$, $w=\dfrac1z$, giving the same linear system:
$2u+3v+10w=4$, $4u-6v+5w=1$, $6u+9v-20w=2$, with coefficient matrix $M=\begin{bmatrix}2 & 3 & 10\\4 & -6 & 5\\6 & 9 & -20\end{bmatrix}$ and $\det M=1200$.

Step 2: Applying Cramer's rule for u:
Replace the first column of M by the constants column to get $M_u$.
\[ M_u=\begin{bmatrix}4 & 3 & 10\\1 & -6 & 5\\2 & 9 & -20\end{bmatrix},\quad \det M_u=600 \]
\[ u=\dfrac{\det M_u}{\det M}=\dfrac{600}{1200}=\dfrac12 \]

Step 3: Applying Cramer's rule for v and w:
Replace the second and third columns by the constants column in turn.
\[ M_v=\begin{bmatrix}2 & 4 & 10\\4 & 1 & 5\\6 & 2 & -20\end{bmatrix},\ \det M_v=400 \implies v=\dfrac{400}{1200}=\dfrac13 \]
\[ M_w=\begin{bmatrix}2 & 3 & 4\\4 & -6 & 1\\6 & 9 & 2\end{bmatrix},\ \det M_w=240 \implies w=\dfrac{240}{1200}=\dfrac15 \]

Step 4: 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:
Cramer's rule confirms the same solution obtained by the inverse matrix method.
\[ \boxed{x=2,\ y=3,\ z=5} \]