Find the Determinant

Problem

det(6)

Solution

1. Expand along the fifth column because it contains four zeros and one entry of 1 at position (3,5)

D=(−1)(3+5)⋅1⋅det(6)

2. Simplify the sign and the scalar factor.

(−1)8=1

D=det(6)

3. Expand along the third row of the new 4×4 matrix because it contains three zeros and one entry of 3 at position (3,1)

D=(−1)(3+1)⋅3⋅det(3)

4. Simplify the sign and the scalar factor.

(−1)4=1

D=3⋅det(3)

5. Calculate the 3×3 determinant using the rule of Sarrus or expansion along the first column.

det(3)=3*((−4)*(2)−(1)*(3))−0*((2)*(2)−(4)*(3))+2*((2)*(1)−(4)*(−4))

6. Evaluate the arithmetic inside the determinant calculation.

3*(−8−3)+2*(2+16)

3*(−11)+2*(18)

−33+36=3

7. Multiply by the external factor of 3 found in step 4.

D=3⋅3=9

Final Answer

det(6)=9

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