NCERT Class 12 Maths · Chapter 4 Determinants · Exercise 4.5 Question 14

✅ Step-by-Step Solution

1. Expand the determinant along row 1, using cofactors Cᵢⱼ=(−1)ⁱ⁺ʲMᵢⱼ.
2. The cofactors in this row are \(\left[\begin{matrix}7 & -19 & -11\end{matrix}\right]\).
3. The expansion terms are \(7\), \(19\), \(-22\).
4. Adding and simplifying gives det A = \(4\).
5. First check whether this determinant is zero.
6. The cofactor matrix is \(\left[\begin{matrix}7 & -19 & -11\\1 & -1 & -1\\-3 & 11 & 7\end{matrix}\right]\).
7. Transpose the cofactor matrix: adj A=\(\left[\begin{matrix}7 & 1 & -3\\-19 & -1 & 11\\-11 & -1 & 7\end{matrix}\right]\).
8. Divide by det A: A⁻¹=\(\left[\begin{matrix}\frac{7}{4} & \frac{1}{4} & - \frac{3}{4}\\- \frac{19}{4} & - \frac{1}{4} & \frac{11}{4}\\- \frac{11}{4} & - \frac{1}{4} & \frac{7}{4}\end{matrix}\right]\).
9. Multiplication verifies AA⁻¹=A⁻¹A=I.
10. Multiply X=A⁻¹B=\(\left[\begin{matrix}2\\1\\3\end{matrix}\right]\).
11. Substituting this vector into the original equations gives AX=\(\left[\begin{matrix}7\\-5\\12\end{matrix}\right]\)=B. Hence the system has a unique solution.