NCERT Class 12 Maths · Chapter 4 Determinants · Exercise 4.5 Question 6

✅ Step-by-Step Solution

1. Expand the determinant along row 1, using cofactors Cᵢⱼ=(−1)ⁱ⁺ʲMᵢⱼ.
2. The cofactors in this row are \(\left[\begin{matrix}28 & 13 & -19\end{matrix}\right]\).
3. The expansion terms are \(140\), \(-13\), \(-76\).
4. Adding and simplifying gives det A = \(51\).
5. First check whether this determinant is zero.
6. The cofactor matrix is \(\left[\begin{matrix}28 & 13 & -19\\-2 & 10 & 5\\-17 & -17 & 17\end{matrix}\right]\).
7. Transpose the cofactor matrix: adj A=\(\left[\begin{matrix}28 & -2 & -17\\13 & 10 & -17\\-19 & 5 & 17\end{matrix}\right]\).
8. Divide by det A: A⁻¹=\(\left[\begin{matrix}\frac{28}{51} & - \frac{2}{51} & - \frac{1}{3}\\\frac{13}{51} & \frac{10}{51} & - \frac{1}{3}\\- \frac{19}{51} & \frac{5}{51} & \frac{1}{3}\end{matrix}\right]\).
9. Multiplication verifies AA⁻¹=A⁻¹A=I.
10. Multiply X=A⁻¹B=\(\left[\begin{matrix}3\\2\\-2\end{matrix}\right]\).
11. Substituting this vector into the original equations gives AX=\(\left[\begin{matrix}5\\2\\-1\end{matrix}\right]\)=B. Hence the system has a unique solution.