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

Nathan G.

You're a hero in my eyes08/06/20

Brenda D.

08/06/20

Brenda D.

08/06/20

Nathan G.

Thanks Big B!08/06/20