Evaluate the determinant in each case : A =[(2,-1,0,1)(-3,0,1,-2)(1,1,-1,1)(2,-1,5,0)] ​

Question

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 = \(\begin{bmatrix} 2 &-1 & 0 &1 \\[0.3em] -3 & 0 & 1&-2 \\[0.3em] 1 & 1 &-1&1\\[0.3em]2 & -1 &5&0 \end{bmatrix}\)

Answer 1

Let M_ij and C_ij 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, 

C_ij = (–1)^i+j × M_ij

A = \(\begin{bmatrix} 2 &-1 & 0 &1 \\[0.3em] -3 & 0 & 1&-2 \\[0.3em] 1 & 1 &-1&1\\[0.3em]2 & -1 &5&0 \end{bmatrix}\) 

⇒ M₁₁ = \(\begin{bmatrix} 0 &1 & -2 \\[0.3em] 1 & -1 & 1 \\[0.3em] -1 & 5 &0\\[0.3em] \end{bmatrix}\) 

M₁₁ = 0(–1×0 – 5×1) – 1(1×0 – (–1)×1) + (–2)(1×5 – (–1)×(–1)) 

M₁₁ = – 9

⇒ M₂₁ = \(\begin{bmatrix} -1 &0 & 1 \\[0.3em] 1 & -1 & 1 \\[0.3em] -1 & 5 &0\\[0.3em] \end{bmatrix}\) 

M₂₁ = –1(–1×0 – 5×1) – 0(1×0 – (–1)×1) + 1(1×5 – (–1)×(–1)) 

M₂₁ = 9

⇒ M₃₁ = \(\begin{bmatrix} -1 &0 & 1 \\[0.3em] 0 & 1 & -2 \\[0.3em] -1 & 5 &0\\[0.3em] \end{bmatrix}\)

M₃₁ = –1(1×0 – 5×(–2)) – 0(0×0 – (–1)×(–2)) + 1(0×5 – (–1)×1) 

M₃₁ = – 9

⇒ M₄₁ = \(\begin{bmatrix} -1 &0 & 1 \\[0.3em] 0 & 1 & -2 \\[0.3em] -1 & -1 &1\\[0.3em] \end{bmatrix}\) 

M₄₁ = –1(1×1 – (–1)×(–2)) – 0(0×1 – 1×(–2)) + 1(0×(–1) – 1×1)

M₄₁ = 0

C₁₁ = (–1)¹⁺¹ × M₁₁ 

= 1 × (–9) 

= –9

C₂₁ = (–1)²⁺¹ × M₂₁ 

= –1 × 9 

= –9

C₃₁ = (–1)³⁺¹ × M₃₁ 

= 1 × –9 

= –9

C₄₁ = (–1)⁴⁺¹ × M₄₁ 

= –1 × 0 

= 0

Now expanding along the first column we get 

|A| = a₁₁ × C₁₁ + a₂₁× C₂₁+ a₃₁× C₃₁ + a₄₁× C₄₁ 

= 2× (–9) + (–3) × –9 + 1× (–9) + 2× 0 

= – 18 + 27 – 9 

= 0