NCERT Class 12 Maths · Chapter 4 Determinants · Exercise 4.5 Question 13

✅ Step-by-Step Solution

1. Expand the determinant along row 1, using cofactors Cᵢⱼ=(−1)ⁱ⁺ʲMᵢⱼ.
2. The cofactors in this row are \(\left[\begin{matrix}5 & 5 & 5\end{matrix}\right]\).
3. The expansion terms are \(10\), \(15\), \(15\).
4. Adding and simplifying gives det A = \(40\).
5. First check whether this determinant is zero.
6. The cofactor matrix is \(\left[\begin{matrix}5 & 5 & 5\\3 & -13 & 11\\9 & 1 & -7\end{matrix}\right]\).
7. Transpose the cofactor matrix: adj A=\(\left[\begin{matrix}5 & 3 & 9\\5 & -13 & 1\\5 & 11 & -7\end{matrix}\right]\).
8. Divide by det A: A⁻¹=\(\left[\begin{matrix}\frac{1}{8} & \frac{3}{40} & \frac{9}{40}\\\frac{1}{8} & - \frac{13}{40} & \frac{1}{40}\\\frac{1}{8} & \frac{11}{40} & - \frac{7}{40}\end{matrix}\right]\).
9. Multiplication verifies AA⁻¹=A⁻¹A=I.
10. Multiply X=A⁻¹B=\(\left[\begin{matrix}1\\2\\-1\end{matrix}\right]\).
11. Substituting this vector into the original equations gives AX=\(\left[\begin{matrix}5\\-4\\3\end{matrix}\right]\)=B. Hence the system has a unique solution.