Inverse, Rank & Linear Systems — JEE Main Maths PYQ MCQs with Solutions

Solution: For a 2×2 matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the inverse is $A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$. Here, $\det(A) = 1(4) - 2(3) = -2$. Therefore, $A^{-1} = \frac{1}{-2} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ 3/2 & -1/2 \end{bmatrix}$.

Solution: Notice that $R_3 = R_1 + R_2$ (component-wise: $3=2+1$, $3=3+0$, $3=1+2$). Since one row is a linear combination of others, the matrix is singular. Alternatively, expanding: $\det(A) = 2(0-6) - 3(3-6) + 1(3-0) = -12 + 9 + 3 = 0$.

Solution: $\det(A) = 1(3) - (-1)(2) = 3 + 2 = 5 \neq 0$. Using the 2×2 inverse formula: $A^{-1} = \frac{1}{5} \begin{bmatrix} 3 & 1 \\ -2 & 1 \end{bmatrix}$. Comparing, $k = 5$.

Solution: Every row is a multiple of the first row: $R_2 = 2R_1$, $R_3 = 3R_1$. Thus only one row is linearly independent, giving rank = 1.

Solution: Row reducing: $R_2 - 2R_1 \to R_2$ and $R_3 - 3R_1 \to R_3$ gives $\begin{bmatrix} 1 & 2 & 3 \\ 0 & -1 & -2 \\ 0 & -1 & -2 \end{bmatrix}$. Then $R_3 - R_2 \to R_3$ gives $\begin{bmatrix} 1 & 2 & 3 \\ 0 & -1 & -2 \\ 0 & 0 & 0 \end{bmatrix}$. Two non-zero rows remain, so rank = 2.

Solution: The second equation is exactly 2 times the first: $2(x+2y) = 2(3)$. Both represent the same line in the $xy$-plane, so there are infinitely many solutions (all points on the line $x + 2y = 3$).

Solution: All three equations represent the same plane (equation 2 = $2 \times$ equation 1, equation 3 = $3 \times$ equation 1). The rank of the coefficient matrix equals the rank of the augmented matrix (both 1), and both are less than 3 (number of variables). This indicates a dependent system with infinitely many solutions.

Solution: Since $\det(A) = 2(5) - 3(4) = -2 \neq 0$, $A$ is invertible. $X = A^{-1}B$. Computing $A^{-1} = \frac{1}{-2}\begin{bmatrix} 5 & -3 \\ -4 & 2 \end{bmatrix}$. Thus $X = \frac{1}{-2}\begin{bmatrix} 5 & -3 \\ -4 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \frac{1}{-2}\begin{bmatrix} -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1/2 \\ 0 \end{bmatrix}$.

Solution: For a triangular matrix (upper or lower), the determinant is the product of diagonal elements: $\det(A) = 1 \times 2 \times 4 = 8$.

Solution: A matrix is singular when $\det = 0$. Here, $\det = 2(6) - k(3) = 12 - 3k$. Setting this to zero: $12 - 3k = 0 \Rightarrow k = 4$.

Solution: The cofactor $C_{22} = (-1)^{2+2} M_{22}$, where $M_{22}$ is the minor. Deleting row 2 and column 2 gives $M_{22} = \begin{vmatrix} 2 & 0 \\ 3 & -3 \end{vmatrix} = 2(-3) - 0(3) = -6$. Since $(-1)^4 = 1$, we have $C_{22} = 1 \times (-6) = -6$.

Solution: For an $n \times n$ matrix, $\det(kA) = k^n \det(A)$. Here, $n = 3$ and $k = 2$, so $\det(2A) = 2^3 \cdot 5 = 8 \times 5 = 40$.

Solution: By the multiplicative property of determinants: $\det(AB) = \det(A) \times \det(B) = 3 \times 4 = 12$.

Solution: By definition, $(A^{-1})^{-1} = A$. This property holds for any invertible matrix.

Solution: For a homogeneous system $AX = 0$, there is always at least the trivial solution $X = 0$. The number of free variables = (number of columns) - (rank of $A$) = $3 - 2 = 1$. Since there is at least one free variable, there are infinitely many solutions.

Solution: For $AX = B$ where $A$ is an $m \times n$ matrix: a unique solution exists if and only if $A$ is square (i.e., $m = n$) and non-singular (i.e., $\det(A) \neq 0$). In this case, $X = A^{-1}B$ is the unique solution.

Solution: $\det(A) = \cos^2\theta + \sin^2\theta = 1 \neq 0$. Using the 2×2 formula: $A^{-1} = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}^T = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} = A^T$. (This is because $A$ is a rotation matrix, which is orthogonal.)

Solution: $R_2 = 2R_1$ and $R_3 = R_1$, so all rows are scalar multiples of the first row. Only one row is linearly independent, hence rank = 1.

Solution: $\det(A) = 3 - 0 = 3$. $A^{-1} = \frac{1}{3}\begin{bmatrix} 3 & 0 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -2/3 & 1/3 \end{bmatrix}$. So $a = 1$, $d = 1/3$, giving $a + d = 4/3$.

Solution: Equations 1 and 3 are identical on the left ($x + 2y + 3z$) but have different constants (1 vs 5). This is a contradiction. The system is inconsistent and has no solution.

Solution: This is a standard property: $(A^T)^{-1} = (A^{-1})^T$. To verify: $A^T (A^{-1})^T = (A^{-1} A)^T = I^T = I$, confirming the result.

Solution: This gives the system $x + 2y = 5$ and $3x + 4y = 11$. Multiply the first equation by 3: $3x + 6y = 15$. Subtract the second equation: $(3x + 6y) - (3x + 4y) = 15 - 11 \Rightarrow 2y = 4 \Rightarrow y = 2$. Substituting back: $x + 2(2) = 5 \Rightarrow x = 1$. Therefore, $x + y = 1 + 2 = 3$.