## Linear Algebra and Its Applications (5th Edition)

Yes, the three lines will intersect in the unique point $(\frac{-13}{7} , \frac{-5}{7})$.
For three lines to intersect in a common point, we have to find values of the variables $x_{1}$ and $x_{2}$ which will satisfy the equations of all three lines. Essentially, this means that we find a solution to the following system of equations : $x_{1} - 4x_{2} = 1$ ; $2x_{1} - x_{2} = -3$ If the values of $x_{1}$ and $x_{2}$ obtained from the solution of the above system, satisfies the equation of the third line also, we can conclude that the three lines will intersect in the unique point $(x_{1} , x_{2})$. Taking the system of equations : $x_{1} - 4x_{2} = 1$ ; $2x_{1} - x_{2} = -3$ Multiply the first equation by $-2$ and add to the second equation to get : $x_{1} - 4x_{2} = 1$ ; $7x_{2} = -5$ So, $x_{2} = \frac{-5}{7}$ Substituting this value in $x_{1} - 4x_{2} = 1$, we get $x_{1} = 1 + 4x_{2}$ or, $x_{1} = 1 + 4(\frac{-5}{7})$ or, $x_{1} = \frac{-13}{7}$ Using the values $x_{1} = \frac{-13}{7}$ and $x_{2} = \frac{-5}{7}$ in the equation of the third line, $-x_{1} - 3x_{2} = 4$ or, $-(\frac{-13}{7}) - 3( \frac{-5}{7}) = 4$ or, $(\frac{13}{7}) + (\frac{15}{7}) = 4$ or, $(\frac{28}{7}) = 4$ which is true. Thus the three lines will intersect in the unique point $(\frac{-13}{7} , \frac{-5}{7})$.