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.
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