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Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case :

A = \(\begin{bmatrix} 1&a & bc \\[0.3em] 1 & b & ca \\[0.3em] 1 & c & ab \end{bmatrix}\)

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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 the matrix can be obtained for a 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

A = \(\begin{bmatrix} 1&a & bc \\[0.3em] 1 & b & ca \\[0.3em] 1 & c & ab \end{bmatrix}\) 

⇒ M11\(\begin{bmatrix} b&ca \\[0.3em] c & ab \\[0.3em] \end{bmatrix}\)

M11 = b × ab – c × ca 

M11 = ab2 – ac2

⇒ M21\(\begin{bmatrix} a&bc \\[0.3em] c & ab \\[0.3em] \end{bmatrix}\)

M21 = a × ab – c × bc 

M21 = a2b – c2b

⇒ M31\(\begin{bmatrix} a&bc \\[0.3em] b & ca \\[0.3em] \end{bmatrix}\)

M31 = a × ca – b × bc 

M31 = a2c – b2

C11 = (–1)1+1 × M11 

= 1 × (ab2 – ac2

= ab2 – ac2 

C21 = (–1)2+1 × M21 

= –1 × (a2b – c2b) 

= c2b – a2

C31 = (–1)3+1 × M31 

= 1 × (a2c – b2c) 

= a2c – b2

Now expanding along the first column we get 

|A| = a11 × C11 + a21× C21+ a31× C31 

= 1× (ab2 – ac2) + 1 × (c2b – a2b) + 1× (a2c – b2c) 

= ab2 – ac2 + c2b – a2b + a2c – b2c

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