Learning Lab

Example 1

Solve the following for \(x,\,y\) and \(z:\) \[\begin{align*} x+2y-z & =-3 & \left(1\right)\\ 2x-3y+2z & =13 & \left(2\right)\\ -x+5y-4z & =-19. & \left(3\right) \end{align*}\]

Solution: The augmented matrix is: \[\begin{align*} \left[\begin{array}{ccc} 1 & 2 & -1\\ 2 & -3 & 2\\ -1 & 5 & -4 \end{array}\left|\begin{array}{c} -3\\ 13\\ -19 \end{array}\right.\right]. \end{align*}\] We now use row operations to get zeroes in row \(2\) and row \(3\) of column \(1\). We do this by multiplying row \(1\) \(\left(R_{1}\right)\) by 2 and subtracting it from row \(2\) \(\left(R_{2}\right)\) to get:22 We will work from left to right.

\[\begin{align*} \begin{array}{c} \\ R_{2}'=R_{2}-2R_{1}\\ \\ \end{array}\left[\begin{array}{ccc} 1 & 2 & -1\\ 0 & -7 & 4\\ -1 & 5 & -4 \end{array}\left|\begin{array}{c} -3\\ 19\\ -19 \end{array}\right.\right]. \end{align*}\]

Now, to get a zero in row \(3\) column \(1,\) we add rows \(1\) and 3 to get: \[\begin{align*} \begin{array}{c} \\ \\ R_{3}'=R_{3}+R_{1} \end{array}\left[\begin{array}{ccc} 1 & 2 & -1\\ 0 & -7 & 4\\ 0 & 7 & -5 \end{array}\left|\begin{array}{c} -3\\ 19\\ -22 \end{array}\right.\right]. \end{align*}\] Next we get a zero in row \(3\), column \(1\) by adding rows \(2\) and \(3:\) \[\begin{align*} \begin{array}{c} \\ \\ R_{3}'=R_{3}+R_{2} \end{array}\left[\begin{array}{ccc} 1 & 2 & -1\\ 0 & -7 & 4\\ 0 & 0 & -1 \end{array}\left|\begin{array}{c} -3\\ 19\\ -3 \end{array}\right.\right]. \end{align*}\] Now we can start the back substitution. From row \(3\) we have \(-1z=-3\) so \(z=3.\) From row \(2\) we have \(-7y+4z=19\). Substituting for \(z\) we get \[\begin{align*} -7y+4z & =19\\ -7y+4\left(3\right) & =19\\ -7y & =7\\ y & =-1. \end{align*}\] From row \(1\) we have \(x+2y-z=-3\). Substituting for \(y\) and \(z\) gives \[\begin{align*} x+2y-z & =-3\\ x+2\left(-1\right)-3 & =-3\\ x-5 & =-3\\ x & =2. \end{align*}\] The solution to the simultaneous equations is therefore, \(x=2,\,y=-1,\,z=3\).

Note that you can check your solution by substituting the values for \(x,y\) and \(z\) in the original equations.

For example the left hand side of equation \(\left(1\right)\) is \[\begin{align*} x+2y-z & =2+2\left(-1\right)-3\\ & =2-2-3\\ & =-3\\ & =RHS. \end{align*}\] The left hand side of equation \(\left(2\right)\) is \[\begin{align*} 2x-3y+2z & =2(2)-3\left(-1\right)+2\left(3\right)\\ & =4+3+6\\ & =13\\ & =RHS. \end{align*}\] Finally, the left hand side of equation \(\left(3\right)\) is \[\begin{align*} -x+5y-4z & =-2+5\left(-1\right)-4\left(3\right)\\ & =-2-5-12\\ & =-19\\ & =RHS. \end{align*}\]

Example 2

Solve the following simultaneous equations for \(x,y\) and \(z\) \[\begin{align*} 3x-y+2z & =3\\ 2y-5z & =-1\\ x+y+z & =4. \end{align*}\]

Solution: The augmented matrix is \[\begin{align*} \left[\begin{array}{ccc} 3 & -1 & 2\\ 0 & 2 & -5\\ 1 & 1 & 1 \end{array}\left|\begin{array}{c} 3\\ -1\\ 4 \end{array}\right.\right]. \end{align*}\] We first interchange rows \(1\)and \(3.\) This puts a one in the top left corner of the matrix and makes the arithmetic easier: \[\begin{align*} \begin{array}{c} R_{1}'=R_{3}\\ \\ R_{3}'=R_{1} \end{array}\left[\begin{array}{ccc} 1 & 1 & 1\\ 0 & 2 & -5\\ 3 & -1 & 2 \end{array}\left|\begin{array}{c} 4\\ -1\\ 3 \end{array}\right.\right]. \end{align*}\] We now want a zero in the first position of row \(3\). We subtract \(3R_{1}\) from \(R_{3}\)to get: \[\begin{align*} \begin{array}{c} \\ \\ R_{3}'=R_{3}-3R_{1} \end{array}\left[\begin{array}{ccc} 1 & 1 & 1\\ 0 & 2 & -5\\ 0 & -4 & -1 \end{array}\left|\begin{array}{c} 4\\ -1\\ -9 \end{array}\right.\right]. \end{align*}\] Now we want a zero in row \(3\)column \(2\). We get this by adding \(2R_{2}\) to \(R_{3}\): \[\begin{align*} \begin{array}{c} \\ \\ R_{3}'=R_{3}+2R_{2} \end{array}\left[\begin{array}{ccc} 1 & 1 & 1\\ 0 & 2 & -5\\ 0 & 0 & -11 \end{array}\left|\begin{array}{c} 4\\ -1\\ -11 \end{array}\right.\right]. \end{align*}\] From row \(3\), \(-11z=-11\) so \(z=1.\) From row \(2\) and substituting \(z=1\) we get \[\begin{align*} 2y-5z & =-1\\ 2y-5\left(1\right) & =-1\\ 2y & =4\\ y & =2. \end{align*}\] From row \(3\) and substituting for \(x\)and \(y\) we have \[\begin{align*} 3x-y+2z & =3\\ 3x-2+2\left(1\right) & =3\\ 3x & =3\\ x & =1. \end{align*}\] So the solution is \(x=1,\,y=2\) and \(z=1.\)

Example 3

Solve the equations: \[\begin{align*} 2x-2y+2z & =-6\\ 3x+4y+z & =-2\\ 2x+6y+3z & =2. \end{align*}\]

Solution: The augmented matrix is \[\begin{align*} \left[\begin{array}{ccc} 2 & -2 & 2\\ 3 & 4 & 1\\ 2 & 6 & 3 \end{array}\left|\begin{array}{c} -6\\ -2\\ 2 \end{array}\right.\right]. \end{align*}\]

The first step is to try and get a 1 in the top left hand corner of the matrix. In example 2 we did that by interchanging rows. In this case, that won’t work. However row \(1\) can be divided by \(2\) to get \[\begin{align*} \begin{array}{c} R_{1}'=R_{1}/2\\ \\ \\ \end{array}\left[\begin{array}{ccc} 1 & -1 & 1\\ 3 & 4 & 1\\ 2 & 6 & 3 \end{array}\left|\begin{array}{c} -3\\ -2\\ 2 \end{array}\right.\right]. \end{align*}\] Now we proceed in the usual way. In this example we do not put in all the comments. \[\begin{align*} \begin{array}{c} \\ R_{2}'=R_{2}-3R_{1}\\ \\ \end{array}\left[\begin{array}{ccc} 1 & -1 & 1\\ 0 & 7 & -2\\ 2 & 6 & 3 \end{array}\left|\begin{array}{c} -3\\ 7\\ 2 \end{array}\right.\right] \end{align*}\] \[\begin{align*} \begin{array}{c} \\ \\ R_{3}'=R_{3}-2R_{1} \end{array}\left[\begin{array}{ccc} 1 & -1 & 1\\ 0 & 7 & -2\\ 0 & 8 & 1 \end{array}\left|\begin{array}{c} -3\\ 7\\ 8 \end{array}\right.\right] \end{align*}\] \[\begin{align*} \begin{array}{c} \\ \\ R_{3}'=7R_{3}-8R_{2} \end{array}\left[\begin{array}{ccc} 1 & -1 & 1\\ 0 & 7 & -2\\ 0 & 0 & 23 \end{array}\left|\begin{array}{c} -3\\ 7\\ 0 \end{array}\right.\right]. \end{align*}\]

From \(R_{3}\)we have \(23z=0\) and so \(z=0.\) Substituting this into \(R_{2}\) gives \[\begin{align*} 7y-2z & =7\\ 7y-2\left(0\right) & =7\\ 7y & =7\\ y & =1. \end{align*}\] Substituting \(z=0,\,y=1\) into \(R_{1}\) we get \[\begin{align*} x-y+z & =-3\\ x-1+0 & =-3\\ x & =-2. \end{align*}\] The solution is \(x=-2,\,y=1,\,z=0\).

Example 4

Solve the following set of equations: \[\begin{align*} x+y+z & =3\\ 2x-y+z & =5\\ 3x+2z & =8. \end{align*}\]

Solution: The augmented matrix is \[\begin{align*} \left[\begin{array}{ccc} 1 & 1 & 1\\ 2 & -1 & 1\\ 3 & 0 & 2 \end{array}\left|\begin{array}{c} 3\\ 5\\ 8 \end{array}\right.\right]. \end{align*}\] Performing row operations we get \[\begin{align*} \begin{array}{c} \\ R_{2}'=R_{2}-2R_{1}\\ \\ \end{array}\left[\begin{array}{ccc} 1 & 1 & 1\\ 0 & -3 & -1\\ 3 & 0 & 2 \end{array}\left|\begin{array}{c} 3\\ -1\\ 8 \end{array}\right.\right] \end{align*}\] \[\begin{align*} \begin{array}{c} \\ \\ R_{3}'=R_{3}-3R_{1} \end{array}\left[\begin{array}{ccc} 1 & 1 & 1\\ 0 & -3 & -1\\ 0 & -3 & -1 \end{array}\left|\begin{array}{c} 3\\ -1\\ -1 \end{array}\right.\right] \end{align*}\] \[\begin{align*} \begin{array}{c} \\ \\ R_{3}'=R_{3}-R_{2} \end{array}\left[\begin{array}{ccc} 1 & 1 & 1\\ 0 & -3 & -1\\ 0 & 0 & 0 \end{array}\left|\begin{array}{c} 3\\ -1\\ 0 \end{array}\right.\right]. \end{align*}\] We can go no further with elementary row operations and so start back substitution. Row \(3\) tells us nothing. From row \(2\) we have \[\begin{align*} -3y-z & =-1\\ 3y+z & =1. \end{align*}\] This is an equation with two unknowns. In this case we let one of the variables, either \(y\) or \(z\) be free. That is we let \(y\) or \(z\) be equal to some parameter \(t\in\mathbb{R}\). It doesn’t matter which variable you choose to be free, though the answer will look different. We will set \(z=t\) then we have \[\begin{align*} 3y+z & =1\\ 3y+t & =1\\ 3y & =1-t\\ y & =\frac{1-t}{3}. \end{align*}\] From row \(1\) and substituting \(z=t\) and \(y=\left(1-t\right)/3\) we have \[\begin{align*} x+y+z & =3\\ x+\frac{1-t}{3}+t & =3\\ 3x+1-t+3t & =9\\ 3x & =8-2t\\ x & =\frac{8-2t}{3}. \end{align*}\] So the solution here is \[\begin{align*} x & =\frac{8-2t}{3},y=\frac{1-t}{3},z=t,\;t\in\mathbb{R}. \end{align*}\] This gives us an infinite number of solutions because \(t\) can be any real number. To check the solution you can pick any value of \(t,\)determine \(x,y\) and \(z\) and substitute in the original equations. For example let \(t=0\).33 Here we took \(t=0\) because it is easy to get the corresponding \(x,\,y\) and \(z\) values. But you can pick any value of \(t\). Once you choose \(t\), there will be a corresponding value of \(x,\,y\) and \(z.\) These \(x,\,y\) and \(z\) values will satisfy the original equations. Note that picking a value of \(t\) and checking if the corresponding \(x,\,y\) and \(z\) values will the original equation is not a proof that your solution is correct. A real proof would have to consider all values of \(t.\) However it is a useful check. Then\[\begin{align*} x & =\frac{8}{3},y=\frac{1}{3},z=0. \end{align*}\] Substituting in equation 1 gives \[\begin{align*} LHS & =x+y+z\\ & =\frac{8}{3}+\frac{1}{3}+0\\ & =\frac{9}{3}\\ & =3\\ & =RHS. \end{align*}\] You can substitute into the second and third equation to check this solution is correct.