Given that \( A=\left(\begin{array}{lll}2 & 3 & 5 \\ 4 & 1 & 6 \\ 1 & 4 & 0\end{array}\right) \), find the inverse of \( A \)

We are given   A = [ 2 3 5 ]     [ 4 1 6 ]     [ 1 4 0 ] and we want to find A⁻¹. Step 1. Calculate the Determinant of A The determinant of A is:   det(A) = 2·det[[1, 6], [4, 0]] – 3·det[[4, 6], [1, 0]] + 5·det[[4, 1], [1, 4]] Calculate each 2×2 determinant:   det[[1, 6], [4, 0]] = (1·0 – 6·4) = –24   det[[4, 6], [1, 0]] = (4·0 – 6·1) = –6   det[[4, 1], [1, 4]] = (4·4 – 1·1) = 16 – 1 = 15 Thus:   det(A) = 2(–24) – 3(–6) + 5(15)        = –48 + 18 + 75        = 45 Since det(A) ≠ 0, the inverse exists, and we have det(A) = 45. Step 2. Find the Matrix of Cofactors We now compute the cofactor C₍ᵢⱼ₎ for each entry of A. For the first row:  • C₁₁: Remove row 1 and column 1 → submatrix [[1, 6], [4, 0]]    M₁₁ = 1·0 – 6·4 = –24  (sign: (+) because 1+1 is even)    C₁₁ = –24  • C₁₂: Remove row 1 and column 2 → submatrix [[4, 6], [1, 0]]    M₁₂ = 4·0 – 6·1 = –6  (sign: (–) because 1+2 is odd)    C₁₂ = –(–6) = 6  • C₁₃: Remove row 1 and column 3 → submatrix [[4, 1], [1, 4]]    M₁₃ = 4·4 – 1·1 = 16 – 1 = 15  (sign: (+) because 1+3 is even)    C₁₃ = 15 For the second row:  • C₂₁: Remove row 2 and column 1 → submatrix [[3, 5], [4, 0]]    M₂₁ = 3·0 – 5·4 = –20  (sign: (–) because 2+1 is odd)    C₂₁ = –(–20) = 20  • C₂₂: Remove row 2 and column 2 → submatrix [[2, 5], [1, 0]]    M₂₂ = 2·0 – 5·1 = –5  (sign: (+) because 2+2 is even)    C₂₂ = –5  • C₂₃: Remove row 2 and column 3 → submatrix [[2, 3], [1, 4]]    M₂₃ = 2·4 – 3·1 = 8 – 3 = 5  (sign: (–) because 2+3 is odd)    C₂₃ = –5 For the third row:  • C₃₁: Remove row 3 and column 1 → submatrix [[3, 5], [1, 6]]    M₃₁ = 3·6 – 5·1 = 18 – 5 = 13  (sign: (+) because 3+1 is even)    C₃₁ = 13  • C₃₂: Remove row 3 and column 2 → submatrix [[2, 5], [4, 6]]    M₃₂ = 2·6 – 5·4 = 12 – 20 = –8  (sign: (–) because 3+2 is odd)    C₃₂ = –(–8) = 8  • C₃₃: Remove row 3 and column 3 → submatrix [[2, 3], [4, 1]]    M₃₃ = 2·1 – 3·4 = 2 – 12 = –10  (sign: (+) because 3+3 is even)    C₃₃ = –10 Thus, the matrix of cofactors is:   [ –24  6  15 ]   [ 20 –5  –5 ]   [ 13  8 –10 ] Step 3. Form the Adjugate Matrix The adjugate (adj(A)) is the transpose of the matrix of cofactors. So, we have:   adj(A) =     [ –24 20 13 ]     [ 6  –5  8 ]     [ 15  –5 –10 ] Step 4. Write the Inverse Finally, the inverse is given by:   A⁻¹ = (1/det(A)) · adj(A) Since det(A) = 45, we obtain:   A⁻¹ = (1/45) ·     [ –24 20 13 ]     [ 6  –5  8 ]     [ 15  –5 –10 ] Thus, the inverse of A is:   A⁻¹ = (1/45) ·     [ –24 20 13 ]     [ 6  –5  8 ]     [ 15  –5 –10 ] You can leave your answer in this form or simplify further if needed.