Question

Two lines are represented by equations \(5x - 7y - 2 = 0\) and \(7x - 5y - 3 = 0\) in the xy-plane. Using matrix method, find their point of intersection.

Hint:

Write the system as \(AX = B\) and use \(X = A^{-1}B\), where \(|A| = 24\).

Answer 1

Step 1: Write the system as a matrix equation:
The equations are \(5x - 7y = 2\) and \(7x - 5y = 3\).
This is \(AX = B\) with \(A = \begin{bmatrix} 5 & -7 \\ 7 & -5 \end{bmatrix}\), \(X = \begin{bmatrix} x \\ y \end{bmatrix}\) and \(B = \begin{bmatrix} 2 \\ 3 \end{bmatrix}\).

Step 2: Check that A is invertible:
\[ |A| = (5)(-5) - (-7)(7) = -25 + 49 = 24 \neq 0 \]
So \(A^{-1}\) exists and the system has a unique solution.

Step 3: Find the adjoint:
For a \(2 \times 2\) matrix, swap the diagonal entries and change the signs of the other two.
\[ \text{adj}\,A = \begin{bmatrix} -5 & 7 \\ -7 & 5 \end{bmatrix}, \quad A^{-1} = \frac{1}{24}\begin{bmatrix} -5 & 7 \\ -7 & 5 \end{bmatrix} \]

Step 4: Find X:
\[ X = A^{-1}B = \frac{1}{24}\begin{bmatrix} -5 & 7 \\ -7 & 5 \end{bmatrix}\begin{bmatrix} 2 \\ 3 \end{bmatrix} = \frac{1}{24}\begin{bmatrix} -10 + 21 \\ -14 + 15 \end{bmatrix} = \frac{1}{24}\begin{bmatrix} 11 \\ 1 \end{bmatrix} \]
So \(x = \dfrac{11}{24}\) and \(y = \dfrac{1}{24}\).

Step 5: Check in the equations:
\(5\left(\frac{11}{24}\right) - 7\left(\frac{1}{24}\right) = \frac{55 - 7}{24} = 2\).
\(7\left(\frac{11}{24}\right) - 5\left(\frac{1}{24}\right) = \frac{77 - 5}{24} = 3\). Both equations hold.

Final Answer:
The two lines meet at the point \(\left(\dfrac{11}{24}, \dfrac{1}{24}\right)\). \[ \boxed{\left(\dfrac{11}{24}, \dfrac{1}{24}\right)} \]