Linear Algebra

Linear Algebra

Professor Ostrovskiy

We are given:

2×1+ 8×2 + 3×3 = 2

x1 + 3×2 + 2×3= 5

2×1 + 7×2 + 4×3= 8

as our system. Converting this to a matrix yields: 

2 8 3 | 2

1              3 2 | 5

2              7 4 | 8

Next we will get this matrix into Reduced Echelon Form:

1. Subtract the first row from the third row (R3 = R3 – R1)

2 8 3  | 2

1 3 2  | 5

0 -1 1 | 6

2. Subtract the first row and multiply it by (½) from the second row (R2 = R2 – (½)R1)

2  8  3 | 2

0 -1 .5 | 4

0 -1 1  | 6

3. Lastly, we will subtract the second row from the third row (R3 = R3 – R2)

Thus giving us our matrix in Reduced Echelon Form:

2  8  3 | 2

0 -1 .5 | 4

0  0 .5 | 2

Our system will now look like: 

2×1+ 8×2 + 3×3 = 2

         -x2 + .5×3= 4

       .5×3= 2

Using back substitution:

We can easily solve for x3 as x3= (2/.5) -> x3 = 4

To solve for x2 simply substitute x3= 4 into the second row, giving us -x2 + .5(4) = 4. 

Further simplifies to -x2 + 2 = 4 -> -x2 = 2. 

Giving us x2= -2.

To solve for x1, we would substitute x2 and x3into the first row leaving us with:

2×1 + 8(-2) + 3(4) = 2

Simplifying into:

2×1 -16 + 12 = 2

Further simplifying into: 

2×1 -4 = 2

Then using basic algebra, we come to the conclusion that:

2×1 = 6 -> x1 = (6/2) -> x1 = 3

After everything, we are left with

 x1 = 3, x2= -2, and x3 = 4.

We can further verify this by substituting these back into the original system.

2×1+ 8×2 + 3×3 = 2 -> 2(3) + 8(-2) + 3(4) = 2 -> 6 – 16 + 12 = 2 -> 2 = 2, which is correct.

x1 + 3×2 + 2×3= 5 -> 3 + 3(-2) + 2(4) = 5 -> 3 -6 + 8 = 5 -> 5 = 5, which is correct

2×1 + 7×2 + 4×3= 8 -> 2(3) + 7(-2) + 4(4) = 8 -> 6 -14 + 16 = 8 -> 8 = 8, which is correct.

So, we have concluded that x1 = 3, x2= -2, and x3 = 4,  for this system.