Simplify the Matrix

Problem

[[(2−y)/6,0.0,x/6]]*[[3.2,0.8,0.0],[0.8,3.2,0.0],[0.0,0.0,1.2]]*[[(2−y)/6],[0.0],[x/6]]

Solution

1. Multiply the row vector by the 3×3 matrix.

[[(2−y)/6,0.0,x/6]]*[[3.2,0.8,0.0],[0.8,3.2,0.0],[0.0,0.0,1.2]]=[[(3.2*(2−y))/6,(0.8*(2−y))/6,(1.2*x)/6]]

2. Simplify the resulting row vector components.

[[(3.2*(2−y))/6,(0.8*(2−y))/6,(1.2*x)/6]]=[[(6.4−3.2*y)/6,(1.6−0.8*y)/6,0.2*x]]

3. Multiply the simplified row vector by the final column vector.

[[(6.4−3.2*y)/6,(1.6−0.8*y)/6,0.2*x]]*[[(2−y)/6],[0.0],[x/6]]=((6.4−3.2*y)*(2−y))/36+0+(0.2*x2)/6

4. Expand the numerator of the first term.

(6.4−3.2*y)*(2−y)=12.8−6.4*y−6.4*y+3.2*y2

(6.4−3.2*y)*(2−y)=3.2*y2−12.8*y+12.8

5. Combine the terms into a single expression.

(3.2*y2−12.8*y+12.8)/36+(1.2*x2)/36

(3.2*y2−12.8*y+12.8+1.2*x2)/36

6. Divide the entire numerator by 4 to simplify the fraction.

(0.8*y2−3.2*y+3.2+0.3*x2)/9

Final Answer

[[(2−y)/6,0.0,x/6]]*[[3.2,0.8,0.0],[0.8,3.2,0.0],[0.0,0.0,1.2]]*[[(2−y)/6],[0.0],[x/6]]=(0.3*x2+0.8*y2−3.2*y+3.2)/9

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