If then show that |3A| = 27|A|



We know that a determinant of a 3 x 3 matrix is calculated as



= 1[1(4) - 2(0)] – 0 + 1[0-0]


= 1[4 - 0] – 0 + 0


= 4


|A|= 4


LHS: |3A|






= 3[3(12) - 0(6)] – 0 + 3[0 – 0]


= 3(36) – 0 + 0


= 108


|3A| = 108 ----LHS


RHS: 27|A|


27|A| = 27(4)


= 108


27|A| = 108 ----RHS


LHS=RHS


Hence proved.


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