Solving Systems of Equations with Row Operations

Question

x + 2z = 3

2x + y - 3z = 4

3x - 4y + 5z = -12

Answer 1

Hi Nathan

You have a 3 x 3 in x, y, and z the coefficients are inside the brackets the constants outside the brackets

⌈1 0 2⌉ 3

Ι 2 1 -3Ι 4

⌊3 -4 5⌋ 12

Note the y term is zero in the first row and has a coefficient of 1 in the second row so you can multiply the second row by positive 4 to eliminate it between rows 2 and 3 to yield another equation in x and z see below.

x = 1

y = 5

z = 1

Lets number the equations

Equation 1

x + 2z = 4

Equation 2

2x + y - 3z = 4

Equation 3

3x - 4y + 5z = -12

Since the first equation has no y term

Eliminate y from the other two equations, and solve for x and z first

Multiply Equation 2 by 4 and combine it with Equation 3

8x + 4y - 12z = 16 (this is Equation 2 multiplied by 4)

3x - 4y + 5z = -12

Combining the two above gives

11x - 7z = 4

Combine the equation above with Equation 1 to eliminate x

x + 2z = 3 Equation 1

11x - 7z = 4

Multiply Equation 1 by -11

-11x - 22z = -33 (this is Equation 1 multiplied by -11)

11x - 7z = 4

This leaves

-29z = -29

Divide both sides by - 29 to give

z = 1

Substitute this value back into Equation 1 to solve for x

x + 2z = 3 Equation 1

x +2(1) = 3

x + 2 = 3

Subtract 2 from both sides of the equation

x = 3 -2

x = 1

Substitute x = 1 and z = 1 into Equation 2 to solve for y

Equation 2

2x + y - 3z = 4

2(1) + y - 3(1) = 4

2 + y - 3 = 4

Combine like terms

y - 1 = 4

Add 1 to both sides of the equation

y = 4 + 1

y = 5

Finally check all the values in Equation 3

Equation 3

3x - 4y + 5z = -12

3(1) -4(5) +5(1) = -12

3 - 20 + 5 = =12

-17 + 5 = -12

-12 = -12

You can check the other equations as well. I hope you find this useful, leave a message if you have any questions

Answer 2

Here are your valid row operations:

R1: Swap row a with b (here order does not matter)

R2: multiply a scalar m to all elements of row a

R3: Add each elements of row a to its corresponding element in row b (this is just vector addition, but order matters, as we are only changing the elements of row b)

Ultimately we want to reduce this to Row echelon form, and then back substitute. To help organize things.You can combine operation 2 with operation 3, and call this operation R4.

R4: perform R2 and R3 with rows a and b, and scalar m, then revert row a back to "normal", by doing R2 with 1/m on a.

Then label your rows 1,2,3 in the order given. How I would start is by doing

R4: m = -2 on a=1 b=2, This makes row 2 have 0 in the x place, while keeping row 1 the same. Likewise:

R4: m = -3 on a=1 b=3, This makes row 3 have 0 in the x place. Show that this is true.

Now if we want to eliminate the y element in row 3. Show that to do this you must perform:

R4: m = +4 on a=2 b=3

This nicely gets equation -29z = -29, or z = 1. From here you may back solve using back substitution.

Hope this helps and is devoid of algebra errors.