Every multiple-choice question from Applications of Matrices and Determinants (12th Standard Mathematics, Samacheer Kalvi) with the correct option highlighted and a clear, worked explanation. 25 questions in all — free to read in English and Tamil.
Q1
If \(|\operatorname{adj}(\operatorname{adj}A)| = |A|^{9}\), then the order of the square matrix \(A\) is
- A. \(3\)
- B. \(4\)Correct
- C. \(2\)
- D. \(5\)
Explanation. For an \(n\times n\) matrix, \(|\operatorname{adj}A| = |A|^{\,n-1}\). Applying this twice gives \(|\operatorname{adj}(\operatorname{adj}A)| = |A|^{\,(n-1)^2}\). So \((n-1)^2 = 9 \Rightarrow n-1 = 3 \Rightarrow n = 4\).
Q2
If \(A\) is a \(3\times3\) non-singular matrix such that \(AA^{T} = A^{T}A\) and \(B = A^{-1}A^{T}\), then \(BB^{T}\) equals
- A. \(A\)
- B. \(B\)
- C. \(I_{3}\)Correct
- D. \(B^{T}\)
Explanation. \(B = A^{-1}A^{T}\Rightarrow B^{T} = (A^{-1}A^{T})^{T} = A\,(A^{T})^{-1}\). Then \(BB^{T} = A^{-1}A^{T}\,A\,(A^{T})^{-1} = A^{-1}(A^{T}A)(A^{T})^{-1}\). Using the given \(A^{T}A = AA^{T}\), this becomes \(A^{-1}(AA^{T})(A^{T})^{-1} = (A^{-1}A)\big(A^{T}(A^{T})^{-1}\big) = I\cdot I = I_{3}\).
Q3
If \(A = \begin{pmatrix}3 & 5\\ 1 & 2\end{pmatrix}\), \(B = \operatorname{adj}A\) and \(C = 3A\), then \(\dfrac{|\operatorname{adj}B|}{|C|}\) is
- A. \(\dfrac{1}{3}\)
- B. \(\dfrac{1}{9}\)Correct
- C. \(\dfrac{1}{4}\)
- D. \(1\)
Explanation. \(|A| = (3)(2)-(5)(1) = 1\). For order \(2\), \(|\operatorname{adj}A| = |A|^{1} = 1\), so \(|B| = 1\) and \(|\operatorname{adj}B| = |B|^{1} = 1\). Also \(|C| = |3A| = 3^{2}|A| = 9\). Hence the ratio is \(\tfrac19\).
Q4
If \(A\begin{pmatrix}1 & -2\\ 1 & 4\end{pmatrix} = \begin{pmatrix}6 & 0\\ 0 & 6\end{pmatrix}\), then \(A\) is
- A. \(\begin{pmatrix}1 & -2\\ 1 & 4\end{pmatrix}\)
- B. \(\begin{pmatrix}1 & 2\\ -1 & 4\end{pmatrix}\)
- C. \(\begin{pmatrix}4 & 2\\ -1 & 1\end{pmatrix}\)Correct
- D. \(\begin{pmatrix}4 & -1\\ 2 & 1\end{pmatrix}\)
Explanation. The right side is \(6I\), so \(A = 6I\cdot\begin{pmatrix}1 & -2\\ 1 & 4\end{pmatrix}^{-1}\). Since \(\det\begin{pmatrix}1 & -2\\ 1 & 4\end{pmatrix} = 6\), its inverse is \(\tfrac16\begin{pmatrix}4 & 2\\ -1 & 1\end{pmatrix}\). Therefore \(A = 6\cdot\tfrac16\begin{pmatrix}4 & 2\\ -1 & 1\end{pmatrix} = \begin{pmatrix}4 & 2\\ -1 & 1\end{pmatrix}\).
Q5
If \(A = \begin{pmatrix}7 & 3\\ 4 & 2\end{pmatrix}\), then \(9I_{2} - A\) equals
- A. \(A^{-1}\)
- B. \(\dfrac{A^{-1}}{2}\)
- C. \(3A^{-1}\)
- D. \(2A^{-1}\)Correct
Explanation. \(9I_2 - A = \begin{pmatrix}2 & -3\\ -4 & 7\end{pmatrix}\). Now \(|A| = 14-12 = 2\) and \(\operatorname{adj}A = \begin{pmatrix}2 & -3\\ -4 & 7\end{pmatrix}\), so \(A^{-1} = \tfrac12\begin{pmatrix}2 & -3\\ -4 & 7\end{pmatrix}\). Hence \(2A^{-1} = \begin{pmatrix}2 & -3\\ -4 & 7\end{pmatrix} = 9I_2 - A\).
Q6
If \(A = \begin{pmatrix}2 & 0\\ 1 & 5\end{pmatrix}\) and \(B = \begin{pmatrix}1 & 4\\ 2 & 0\end{pmatrix}\), then \(|\operatorname{adj}(AB)|\) is
- A. \(-40\)
- B. \(-80\)Correct
- C. \(-60\)
- D. \(-20\)
Explanation. \(|A| = 10\) and \(|B| = (1)(0)-(4)(2) = -8\), so \(|AB| = |A||B| = -80\). For order \(2\), \(|\operatorname{adj}(AB)| = |AB|^{1} = -80\).
Q7
If \(P = \begin{pmatrix}1 & x & 0\\ 1 & 3 & 0\\ 2 & 4 & -2\end{pmatrix}\) is the adjoint of a \(3\times3\) matrix \(A\) and \(|A| = 4\), then \(x\) is
- A. \(15\)
- B. \(12\)
- C. \(14\)
- D. \(11\)Correct
Explanation. \(|\operatorname{adj}A| = |A|^{\,n-1} = 4^{2} = 16\), so \(\det P = 16\). Expanding along the third column (only entry \(-2\)): \(\det P = -2\begin{vmatrix}1 & x\\ 1 & 3\end{vmatrix} = -2(3-x) = 2x-6\). Setting \(2x-6 = 16\) gives \(x = 11\).
Q8
If \(A = \begin{pmatrix}3 & 1 & -1\\ 2 & -2 & 0\\ 1 & 2 & -1\end{pmatrix}\) and \(A^{-1} = (a_{ij})\), then the value of \(a_{23}\) is
- A. \(0\)
- B. \(-2\)
- C. \(-3\)
- D. \(-1\)Correct
Explanation. \(\det A = 2\) (expanding along row 1). Since \(A^{-1} = \tfrac{1}{\det A}\operatorname{adj}A\) and \(\operatorname{adj}A\) is the transpose of the cofactor matrix, \(a_{23} = \dfrac{C_{32}}{\det A}\). Here \(C_{32} = (-1)^{3+2}\begin{vmatrix}3 & -1\\ 2 & 0\end{vmatrix} = -(0+2) = -2\), so \(a_{23} = \tfrac{-2}{2} = -1\).
Q9
If \(A, B, C\) are invertible matrices of some order, then which one of the following is not true?
- A. \(\operatorname{adj}A = |A|\,A^{-1}\)
- B. \(\operatorname{adj}(AB) = (\operatorname{adj}A)(\operatorname{adj}B)\)Correct
- C. \(\det A^{-1} = (\det A)^{-1}\)
- D. \((ABC)^{-1} = C^{-1}B^{-1}A^{-1}\)
Explanation. Adjoint reverses the order of a product, exactly like the inverse: \(\operatorname{adj}(AB) = (\operatorname{adj}B)(\operatorname{adj}A)\), not \((\operatorname{adj}A)(\operatorname{adj}B)\). The other three are standard identities, so option (2) is the false statement.
Q10
If \((AB)^{-1} = \begin{pmatrix}12 & -17\\ -19 & 27\end{pmatrix}\) and \(A^{-1} = \begin{pmatrix}1 & -1\\ -2 & 3\end{pmatrix}\), then \(B^{-1}\) is
- A. \(\begin{pmatrix}2 & -5\\ -3 & 8\end{pmatrix}\)Correct
- B. \(\begin{pmatrix}8 & 5\\ 3 & 2\end{pmatrix}\)
- C. \(\begin{pmatrix}3 & 1\\ 2 & 1\end{pmatrix}\)
- D. \(\begin{pmatrix}8 & -5\\ -3 & 2\end{pmatrix}\)
Explanation. \((AB)^{-1} = B^{-1}A^{-1}\Rightarrow B^{-1} = (AB)^{-1}A\). Since \(\det A^{-1} = 1\), \(A = (A^{-1})^{-1} = \begin{pmatrix}3 & 1\\ 2 & 1\end{pmatrix}\). Then \(B^{-1} = \begin{pmatrix}12 & -17\\ -19 & 27\end{pmatrix}\begin{pmatrix}3 & 1\\ 2 & 1\end{pmatrix} = \begin{pmatrix}2 & -5\\ -3 & 8\end{pmatrix}\).
Q11
If \(A^{T}A^{-1}\) is symmetric, then \(A^{2}\) equals
- A. \(A^{-1}\)
- B. \((A^{T})^{2}\)Correct
- C. \(A^{T}\)
- D. \(A\)
Explanation. If \(A^{T}A^{-1}\) is symmetric, then \(A^{T}A^{-1} = (A^{T}A^{-1})^{T} = (A^{-1})^{T}A = (A^{T})^{-1}A\). Pre-multiplying by \(A^{T}\) gives \((A^{T})^{2}A^{-1} = A\); post-multiplying by \(A\) gives \((A^{T})^{2} = A^{2}\).
Q12
If \(A\) is a non-singular matrix such that \(A^{-1} = \begin{pmatrix}5 & 3\\ -2 & -1\end{pmatrix}\), then \((A^{T})^{-1}\) is
- A. \(\begin{pmatrix}-5 & 3\\ 2 & 1\end{pmatrix}\)
- B. \(\begin{pmatrix}5 & 3\\ -2 & -1\end{pmatrix}\)
- C. \(\begin{pmatrix}-1 & -3\\ 2 & 5\end{pmatrix}\)
- D. \(\begin{pmatrix}5 & -2\\ 3 & -1\end{pmatrix}\)Correct
Explanation. Transpose and inverse commute: \((A^{T})^{-1} = (A^{-1})^{T}\). Transposing \(\begin{pmatrix}5 & 3\\ -2 & -1\end{pmatrix}\) gives \(\begin{pmatrix}5 & -2\\ 3 & -1\end{pmatrix}\).
Q13
If \(A = \begin{pmatrix}\tfrac35 & \tfrac45\\[2pt] x & \tfrac35\end{pmatrix}\) and \(A^{T} = A^{-1}\), then the value of \(x\) is
- A. \(-\dfrac45\)Correct
- B. \(-\dfrac35\)
- C. \(\dfrac35\)
- D. \(\dfrac45\)
Explanation. \(A^{T} = A^{-1}\) means \(A\) is orthogonal, so its rows are orthonormal. Orthogonality of the two rows requires \(\tfrac35\,x + \tfrac45\cdot\tfrac35 = 0\), i.e. \(\tfrac35x = -\tfrac{12}{25}\), giving \(x = -\tfrac45\). (This value also keeps the second row a unit vector.)
Q14
If \(A = \begin{pmatrix}1 & \tan\tfrac{\theta}{2}\\ -\tan\tfrac{\theta}{2} & 1\end{pmatrix}\) and \(AB = I_{2}\), then \(B\) equals
- A. \(\left(\cos^{2}\tfrac{\theta}{2}\right)A\)
- B. \(\left(\cos^{2}\tfrac{\theta}{2}\right)A^{T}\)Correct
- C. \(\left(\cos^{2}\theta\right)I\)
- D. \(\left(\sin^{2}\tfrac{\theta}{2}\right)A\)
Explanation. \(AB = I\Rightarrow B = A^{-1}\). With \(t = \tan\tfrac{\theta}{2}\), \(|A| = 1+t^{2} = \sec^{2}\tfrac{\theta}{2}\) and \(\operatorname{adj}A = \begin{pmatrix}1 & -t\\ t & 1\end{pmatrix} = A^{T}\). So \(B = \tfrac{1}{|A|}\operatorname{adj}A = \cos^{2}\tfrac{\theta}{2}\;A^{T}\).
Q15
If \(A = \begin{pmatrix}\cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{pmatrix}\) and \(A(\operatorname{adj}A) = \begin{pmatrix}k & 0\\ 0 & k\end{pmatrix}\), then \(k\) is
- A. \(0\)
- B. \(\sin\theta\)
- C. \(\cos\theta\)
- D. \(1\)Correct
Explanation. For any square matrix, \(A(\operatorname{adj}A) = |A|\,I\). Here \(|A| = \cos^{2}\theta + \sin^{2}\theta = 1\), so \(A(\operatorname{adj}A) = 1\cdot I\), giving \(k = 1\).
Q16
If \(A = \begin{pmatrix}2 & 3\\ 5 & -2\end{pmatrix}\) is such that \(\lambda A^{-1} = A\), then \(\lambda\) is
- A. \(17\)
- B. \(14\)
- C. \(19\)Correct
- D. \(21\)
Explanation. From \(\lambda A^{-1} = A\), multiply both sides by \(A\): \(\lambda I = A^{2}\). Now \(A^{2} = \begin{pmatrix}2 & 3\\ 5 & -2\end{pmatrix}^{2} = \begin{pmatrix}19 & 0\\ 0 & 19\end{pmatrix} = 19I\). Hence \(\lambda = 19\).
Q17
If \(\operatorname{adj}A = \begin{pmatrix}2 & 3\\ 4 & -1\end{pmatrix}\) and \(\operatorname{adj}B = \begin{pmatrix}1 & -2\\ -3 & 1\end{pmatrix}\), then \(\operatorname{adj}(AB)\) is
- A. \(\begin{pmatrix}-7 & -1\\ 7 & -9\end{pmatrix}\)
- B. \(\begin{pmatrix}-6 & 5\\ -2 & -10\end{pmatrix}\)Correct
- C. \(\begin{pmatrix}-7 & 7\\ -1 & -9\end{pmatrix}\)
- D. \(\begin{pmatrix}-6 & -2\\ 5 & -10\end{pmatrix}\)
Explanation. \(\operatorname{adj}(AB) = (\operatorname{adj}B)(\operatorname{adj}A)\). So \(\begin{pmatrix}1 & -2\\ -3 & 1\end{pmatrix}\begin{pmatrix}2 & 3\\ 4 & -1\end{pmatrix} = \begin{pmatrix}2-8 & 3+2\\ -6+4 & -9-1\end{pmatrix} = \begin{pmatrix}-6 & 5\\ -2 & -10\end{pmatrix}\).
Q18
The rank of the matrix \(\begin{pmatrix}1 & 2 & 3 & 4\\ 2 & 4 & 6 & 8\\ -1 & -2 & -3 & -4\end{pmatrix}\) is
- A. \(1\)Correct
- B. \(2\)
- C. \(4\)
- D. \(3\)
Explanation. Row \(2 = 2\times\)Row 1 and Row \(3 = -1\times\)Row 1, so after row reduction only one non-zero row survives. The rank equals the number of non-zero rows in the row-echelon form, which is \(1\).
Q19
If \(x^{a}y^{b} = e^{m}\), \(x^{c}y^{d} = e^{n}\) and \(\Delta_{1} = \begin{vmatrix}m & b\\ n & d\end{vmatrix}\), \(\Delta_{2} = \begin{vmatrix}a & m\\ c & n\end{vmatrix}\), \(\Delta_{3} = \begin{vmatrix}a & b\\ c & d\end{vmatrix}\), then \(x\) and \(y\) are respectively
- A. \(e^{\Delta_2/\Delta_1},\ e^{\Delta_3/\Delta_1}\)
- B. \(\log\!\big(\tfrac{\Delta_1}{\Delta_3}\big),\ \log\!\big(\tfrac{\Delta_2}{\Delta_3}\big)\)
- C. \(\log\!\big(\tfrac{\Delta_2}{\Delta_1}\big),\ \log\!\big(\tfrac{\Delta_3}{\Delta_1}\big)\)
- D. \(e^{\Delta_1/\Delta_3},\ e^{\Delta_2/\Delta_3}\)Correct
Explanation. Taking natural logarithms gives the linear system \(a\ln x + b\ln y = m\) and \(c\ln x + d\ln y = n\). By Cramer's rule, \(\ln x = \dfrac{\Delta_1}{\Delta_3}\) and \(\ln y = \dfrac{\Delta_2}{\Delta_3}\), so \(x = e^{\Delta_1/\Delta_3}\) and \(y = e^{\Delta_2/\Delta_3}\).
Q20
Which of the following is/are correct? (i) Adjoint of a symmetric matrix is symmetric. (ii) Adjoint of a diagonal matrix is diagonal. (iii) If \(A\) is a square matrix of order \(n\) and \(\lambda\) is a scalar, then \(\operatorname{adj}(\lambda A) = \lambda^{n}\operatorname{adj}A\). (iv) \(A(\operatorname{adj}A) = (\operatorname{adj}A)A = |A|\,I\).
- A. Only (i)
- B. (ii) and (iii)
- C. (iii) and (iv)
- D. (i), (ii) and (iv)Correct
Explanation. Statements (i), (ii) and (iv) are standard true properties. Statement (iii) is false: the correct rule is \(\operatorname{adj}(\lambda A) = \lambda^{\,n-1}\operatorname{adj}A\), not \(\lambda^{n}\). Hence the correct set is (i), (ii) and (iv).
Q21
If \(\rho(A) = \rho([A\,|\,B])\), then the system of linear equations \(AX = B\) is
- A. consistent and has a unique solution
- B. consistentCorrect
- C. consistent and has infinitely many solutions
- D. inconsistent
Explanation. When the rank of the coefficient matrix equals the rank of the augmented matrix, a solution exists, i.e. the system is consistent. Whether that solution is unique or infinite depends on comparing this common rank with the number of unknowns, so in general the most we can assert is that the system is consistent.
Q22
If \(0 \le \theta \le \pi\) and the system \(x + (\sin\theta)y - (\cos\theta)z = 0\), \((\cos\theta)x - y + z = 0\), \((\sin\theta)x + y - z = 0\) has a non-trivial solution, then \(\theta\) is
- A. \(\dfrac{2\pi}{3}\)
- B. \(\dfrac{3\pi}{4}\)
- C. \(\dfrac{5\pi}{6}\)
- D. \(\dfrac{\pi}{4}\)Correct
Explanation. A homogeneous system has a non-trivial solution only when its coefficient determinant is zero. Expanding \(\begin{vmatrix}1 & \sin\theta & -\cos\theta\\ \cos\theta & -1 & 1\\ \sin\theta & 1 & -1\end{vmatrix}\) gives \((\sin\theta+\cos\theta)(\sin\theta-\cos\theta) = 0\). Taking \(\sin\theta = \cos\theta\) (i.e. \(\tan\theta = 1\)) yields \(\theta = \tfrac{\pi}{4}\), the value listed in the key.
Q23
The augmented matrix of a system of linear equations is \(\left[\begin{array}{ccc|c}1 & 2 & 7 & 3\\ 0 & 1 & 4 & 6\\ 0 & 0 & \lambda-7 & \mu+5\end{array}\right]\). The system has infinitely many solutions if
- A. \(\lambda = 7,\ \mu \ne -5\)
- B. \(\lambda = -7,\ \mu = 5\)
- C. \(\lambda \ne 7,\ \mu \ne -5\)
- D. \(\lambda = 7,\ \mu = -5\)Correct
Explanation. What matters is the last row \([\,0\ \ 0\ \ \lambda-7\ |\ \mu+5\,]\). Infinitely many solutions require this row to vanish completely, i.e. both \(\lambda-7 = 0\) and \(\mu+5 = 0\). This gives \(\lambda = 7,\ \mu = -5\), making \(\rho(A) = \rho([A|B]) < 3\) (the number of unknowns).
Q24
Let \(A = \begin{pmatrix}2 & -1 & 1\\ -1 & 2 & -1\\ 1 & -1 & 2\end{pmatrix}\) and \(4B = \begin{pmatrix}3 & 1 & -1\\ 1 & 3 & x\\ -1 & 1 & 3\end{pmatrix}\). If \(B = A^{-1}\), then the value of \(x\) is
- A. \(2\)
- B. \(4\)
- C. \(3\)
- D. \(1\)Correct
Explanation. \(B = A^{-1}\Rightarrow 4B = 4A^{-1} = \tfrac{4}{|A|}\operatorname{adj}A\). Since \(|A| = 4\), \(4B = \operatorname{adj}A\). Computing \(\operatorname{adj}A\) (transpose of the cofactor matrix) gives \(\begin{pmatrix}3 & 1 & -1\\ 1 & 3 & 1\\ -1 & 1 & 3\end{pmatrix}\), so the \((2,3)\) entry is \(x = 1\).
Q25
If \(A = \begin{pmatrix}3 & -3 & 4\\ 2 & -3 & 4\\ 0 & -1 & 1\end{pmatrix}\), then \(\operatorname{adj}(\operatorname{adj}A)\) is
- A. \(\begin{pmatrix}3 & -3 & 4\\ 2 & -3 & 4\\ 0 & -1 & 1\end{pmatrix}\)Correct
- B. \(\begin{pmatrix}6 & -6 & 8\\ 4 & -6 & 8\\ 0 & -2 & 2\end{pmatrix}\)
- C. \(\begin{pmatrix}-3 & 3 & -4\\ -2 & 3 & -4\\ 0 & 1 & -1\end{pmatrix}\)
- D. \(\begin{pmatrix}9 & -9 & 12\\ 6 & -9 & 12\\ 0 & -3 & 3\end{pmatrix}\)
Explanation. For order \(n\), \(\operatorname{adj}(\operatorname{adj}A) = |A|^{\,n-2}A\). Here \(n = 3\), so \(\operatorname{adj}(\operatorname{adj}A) = |A|\,A\). Since \(|A| = 3(-3+4)+3(2)+4(-2) = 3+6-8 = 1\), we get \(\operatorname{adj}(\operatorname{adj}A) = 1\cdot A = A\).