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 – b2c
C11 = (–1)1+1 × M11
= 1 × (ab2 – ac2)
= ab2 – ac2
C21 = (–1)2+1 × M21
= –1 × (a2b – c2b)
= c2b – a2b
C31 = (–1)3+1 × M31
= 1 × (a2c – b2c)
= a2c – b2c
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