Question

Solve the system of equations using matrices. Use the Gaussian elimination method with back substitution

Answer 1

The first step is to write the system of equations as an augmented matrix:

`\begin{bmatrix}{4} & {-5} & +4 & {-67} \\ {5} & {-4} & -3 & {-12} \\ {1} & {-5} & +0 & {-36} \\ {} & {} & {} & {}\end{bmatrix}`

The next step is to let the first number at the mid-row to zero, then let

R2-5/4*R1->R2

`\begin{bmatrix}{4} & {-5} & +4 & {-67} \\ {0} & {9/4} & -8 & {287/4} \\ {1} & {-5} & +0 & {-36} \\ {} & {} & {} & {}\end{bmatrix}`

Now, let's convert the first number in the last row to 0, then:

R3-1/4*R1->R3

`\begin{bmatrix}{4} & {-5} & +4 & {-67} \\ {0} & {9/4} & -8 & {287/4} \\ {0} & {-15/4} & -1 & {-77/4} \\ {} & {} & {} & {}\end{bmatrix}`

Let's convert the first number in the first row to 1, then:

1/4*R1->R1

`\begin{bmatrix}{1} & {-5/4} & +1 & {-67/4} \\ {0} & {9/4} & -8 & {287/4} \\ {0} & {-15/4} & -1 & {-77/4} \\ {} & {} & {} & {}\end{bmatrix}`

Now, let's convert the second number in the last row to zero:

R3-(-5/3)R2->R3

`\begin{bmatrix}{1} & {-5/4} & +1 & {-67/4} \\ {0} & {9/4} & -8 & {287/4} \\ {0} & {0} & -43/3 & {301/3} \\ {} & {} & {} & {}\end{bmatrix}`

Now let's convert the third number in the last row to 1:

(-3/43)R3->R3

`\begin{bmatrix}{1} & {-5/4} & +1 & {-67/4} \\ {0} & {9/4} & -8 & {287/4} \\ {0} & {0} & +1 & {-7} \\ {} & {} & {} & {}\end{bmatrix}`

And finally, convert the second number in the mid-row to 1:

(4/9)*R2->R2

`\begin{bmatrix}{1} & {-5/4} & +1 & {-67/4} \\ {0} & {1} & -32/9 & {287/9} \\ {0} & {0} & +1 & {-7} \\ {} & {} & {} & {}\end{bmatrix}`

Now, we obtained the next equations:

`\begin{gathered} x-(5)/(4)y+z=-67/4\text{ Equation (1)} \\ y-(32)/(9)z=287/9\text{ Equation (2)} \\ z=-7\text{ Equation (3)} \end{gathered}`

Replace equation 3 into equation 2 and solve for y:

`\begin{gathered} y-(32)/(9)\cdot(-7)=(287)/(9) \\ y+(224)/(9)=(287)/(9) \\ y=(287)/(9)-(224)/(9) \\ y=(63)/(9) \\ y=7 \end{gathered}`

Now, replace x and y in equation 1 and solve for x:

`\begin{gathered} x-(5)/(4)\cdot7+(-7)=-(67)/(4) \\ x-(35)/(4)-7=-(67)/(4) \\ x=-(67)/(4)+(35)/(4)+7 \\ x=(-32)/(4)+7 \\ x=-8+7 \\ x=-1 \end{gathered}`

Therefore, the solution set is: (-1, 7, -7)