Applications of Matrices and Determinants — Book Back Answers, Exercise 1.8 (Class 12 Maths)

Quick reference for the 25 book-back MCQs of Exercise 1.8. Each line gives the correct option and a one-line reason. For full worked solutions, open the solved MCQs page for this chapter.

1. Q1. Answer: (2) \(4\). For an \(n\times n\) matrix, \(|\operatorname{adj}A| = |A|^{\,n-1}\).
2. Q2. Answer: (3) \(I_{3}\). \(B = A^{-1}A^{T}\Rightarrow B^{T} = (A^{-1}A^{T})^{T} = A\,(A^{T})^{-1}\).
3. Q3. Answer: (2) \(\dfrac{1}{9}\). \(|A| = (3)(2)-(5)(1) = 1\).
4. Q4. Answer: (3) \(\begin{pmatrix}4 & 2\\ -1 & 1\end{pmatrix}\). The right side is \(6I\), so \(A = 6I\cdot\begin{pmatrix}1 & -2\\ 1 & 4\end{pmatrix}^{-1}\).
5. Q5. Answer: (4) \(2A^{-1}\). \(9I_2 - A = \begin{pmatrix}2 & -3\\ -4 & 7\end{pmatrix}\).
6. Q6. Answer: (2) \(-80\). \(|A| = 10\) and \(|B| = (1)(0)-(4)(2) = -8\), so \(|AB| = |A||B| = -80\).
7. Q7. Answer: (4) \(11\). \(|\operatorname{adj}A| = |A|^{\,n-1} = 4^{2} = 16\), so \(\det P = 16\).
8. Q8. Answer: (4) \(-1\). \(\det A = 2\) (expanding along row 1).
9. Q9. Answer: (2) \(\operatorname{adj}(AB) = (\operatorname{adj}A)(\operatorname{adj}B)\). Adjoint reverses the order of a product, exactly like the inverse: \(\operatorname{adj}(AB) = (\operatorname{adj}B)(\operatorname{adj}A)\), not \((\operatorname{adj}A)(\operatorname{adj}B)\).
10. Q10. Answer: (1) \(\begin{pmatrix}2 & -5\\ -3 & 8\end{pmatrix}\). \((AB)^{-1} = B^{-1}A^{-1}\Rightarrow B^{-1} = (AB)^{-1}A\).
11. Q11. Answer: (2) \((A^{T})^{2}\). If \(A^{T}A^{-1}\) is symmetric, then \(A^{T}A^{-1} = (A^{T}A^{-1})^{T} = (A^{-1})^{T}A = (A^{T})^{-1}A\).
12. Q12. Answer: (4) \(\begin{pmatrix}5 & -2\\ 3 & -1\end{pmatrix}\). Transpose and inverse commute: \((A^{T})^{-1} = (A^{-1})^{T}\).
13. Q13. Answer: (1) \(-\dfrac45\). \(A^{T} = A^{-1}\) means \(A\) is orthogonal, so its rows are orthonormal.
14. Q14. Answer: (2) \(\left(\cos^{2}\tfrac{\theta}{2}\right)A^{T}\). \(AB = I\Rightarrow B = A^{-1}\).
15. Q15. Answer: (4) \(1\). For any square matrix, \(A(\operatorname{adj}A) = |A|\,I\).
16. Q16. Answer: (3) \(19\). From \(\lambda A^{-1} = A\), multiply both sides by \(A\): \(\lambda I = A^{2}\).
17. Q17. Answer: (2) \(\begin{pmatrix}-6 & 5\\ -2 & -10\end{pmatrix}\). \(\operatorname{adj}(AB) = (\operatorname{adj}B)(\operatorname{adj}A)\).
18. Q18. Answer: (1) \(1\). Row \(2 = 2\times\)Row 1 and Row \(3 = -1\times\)Row 1, so after row reduction only one non-zero row survives.
19. Q19. Answer: (4) \(e^{\Delta_1/\Delta_3},\ e^{\Delta_2/\Delta_3}\). Taking natural logarithms gives the linear system \(a\ln x + b\ln y = m\) and \(c\ln x + d\ln y = n\).
20. Q20. Answer: (4) (i), (ii) and (iv). Statements (i), (ii) and (iv) are standard true properties.
21. Q21. Answer: (2) consistent. When the rank of the coefficient matrix equals the rank of the augmented matrix, a solution exists, i.e.
22. Q22. Answer: (4) \(\dfrac{\pi}{4}\). A homogeneous system has a non-trivial solution only when its coefficient determinant is zero.
23. Q23. Answer: (4) \(\lambda = 7,\ \mu = -5\). What matters is the last row \([\,0\ \ 0\ \ \lambda-7\ |\ \mu+5\,]\).
24. Q24. Answer: (4) \(1\). \(B = A^{-1}\Rightarrow 4B = 4A^{-1} = \tfrac{4}{|A|}\operatorname{adj}A\).
25. Q25. Answer: (1) \(\begin{pmatrix}3 & -3 & 4\\ 2 & -3 & 4\\ 0 & -1 & 1\end{pmatrix}\). For order \(n\), \(\operatorname{adj}(\operatorname{adj}A) = |A|^{\,n-2}A\).