Let Mij and Cij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.
Also, Cij = (–1)i+j × Mij
\( M_{11}=bc-f^2\)
\( M_{21}=hc-fg\)
M31 = h × f – b × g
M31 = hf – bg
C11 = (–1)1+1 × M11
= 1 × (bc– f2)
= bc– f2
C21 = (–1)2+1 × M21
= –1 × (hc – fg)
= fg – hc
C31 = (–1)3+1 × M31
= 1 × (hf – bg)
= hf – bg
Now expanding along the first column we get
|A| = a11 × C11 + a21× C21+ a31× C31
= a× (bc– f2) + h× (fg – hc) + g× (hf – bg)
= abc– af2 + hgf – h2c +ghf – bg2
Answered by Sakshi | 1 year agoEvaluate
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