00:01
We're going to solve the system given here, for which we write the general solution using two parameters s and t, and that means we must find two variables in terms of the other two, and that's because this linear system has no unique solution.
00:22
If we calculate the determinant of the coefficient matrix, which is 5, negative 6, 3, 2, first row, then negative 1, 1, 2, 3, second row, 4, negative 5, 5, 5, third row, and negative 2, 2, 4, 6, fourth row.
00:42
So it's a 4 by 4 matrix whose determinant is equal to zero.
00:47
So for that reason this linear system has infinitely many solutions, but we can find two free variables and put the other two in terms of these two free variables.
00:59
So let's do that by using the second and third equations.
01:05
So the second and third equations are negative x1 plus x2 plus 2x3 plus 3x4 equal 5 and 4x1 plus, sorry, minus 5x2 plus 5x3 plus 5x4 equal 5.
01:48
Now we multiply the first equation by 4, so we get negative 4x1 plus 4x2 plus 8x3 plus 12x4 equal 20.
02:11
And the second equation stays the same, 4x1 minus 5x2 plus 5x3 plus 5x4 equal 5.
02:25
Now we add up these two equations side by side and we get negative 4x1 plus 4x1 cancel out and 4x2 minus 5x2 is negative x2, 8x3 plus 5x3 is 13x3, then 12x4 plus 5x4 is plus 17x4 equal 25.
02:55
And from here we can solve for x2 and we get x2 equal negative 25 plus 13x3 plus 17x4.
03:16
So we have written x2 in terms of x3 and x4, which for the time being can be considered a free variable.
03:27
So we get to put x1 in terms also of x3 and x4.
03:33
Let's see that...