Matrices & Operations — JEE Main Maths PYQ MCQs with Solutions

Solution: Matrix addition is performed element-wise. $A + B$ adds corresponding elements: $(1+2, 2+0; 3+1, 4+5) = \begin{pmatrix} 3 & 2 \\ 4 & 9 \end{pmatrix}$

Solution: The transpose $A^T$ swaps rows and columns. The first row $(1, 2, 3)$ becomes the first column, and the second row $(4, 5, 6)$ becomes the second column, yielding a $3 \times 2$ matrix.

Solution: Matrix multiplication: $(AB)_{11} = 1 \cdot 2 + 2 \cdot 0 = 2$, $(AB)_{12} = 1 \cdot 1 + 2 \cdot 1 = 3$, $(AB)_{21} = 3 \cdot 2 + 4 \cdot 0 = 6$, $(AB)_{22} = 3 \cdot 1 + 4 \cdot 1 = 7$.

Solution: A matrix is symmetric if $A = A^T$. For a $2 \times 2$ matrix, this means $\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a & c \\ b & d \end{pmatrix}$, so $b = c$.

Solution: The order of a matrix is rows × columns. This matrix has 2 rows and 4 columns, so the order is $2 \times 4$.

Solution: The transpose of a transpose returns the original matrix: $(A^T)^T = A$. This is a fundamental property of matrix operations.

Solution: For the identity matrix $I$, $I^2 = I \cdot I = I$. The identity matrix is idempotent.

Solution: $(BA)_{11} = 2 \cdot 1 + 1 \cdot 0 = 2$, $(BA)_{12} = 2 \cdot 2 + 1 \cdot 3 = 7$, $(BA)_{21} = 1 \cdot 1 + 0 \cdot 0 = 1$, $(BA)_{22} = 1 \cdot 2 + 0 \cdot 3 = 2$.

Solution: Since $B$ is the identity matrix, $AB = A \cdot I = A$. Multiplying any matrix by the identity matrix leaves it unchanged.

Solution: $(A^2)_{11} = 0 \cdot 0 + 1 \cdot (-1) = -1$, $(A^2)_{12} = 0 \cdot 1 + 1 \cdot 0 = 0$, $(A^2)_{21} = (-1) \cdot 0 + 0 \cdot (-1) = 0$, $(A^2)_{22} = (-1) \cdot 1 + 0 \cdot 0 = -1$, so $A^2 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = -I$.

Solution: For $A = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$, $A^T = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = -A$. The diagonal elements of a skew-symmetric matrix must be 0.

Solution: For matrix multiplication $AB$, if $A$ is $m \times n$ and $B$ is $n \times p$, then $AB$ is $m \times p$. Here, $A$ is $3 \times 2$ and $B$ is $2 \times 4$, so $AB$ is $3 \times 4$.

Solution: $(AB)_1 = 1 \cdot 1 + 2 \cdot 2 + 3 \cdot 3 = 1 + 4 + 9 = 14$, $(AB)_2 = 4 \cdot 1 + 5 \cdot 2 + 6 \cdot 3 = 4 + 10 + 18 = 32$. The result is a $2 \times 1$ matrix.

Solution: Matrix multiplication is generally NOT commutative: $AB \neq BA$ in most cases. The transpose of a product reverses the order: $(AB)^T = B^T A^T$, not $A^T B^T$. Addition and transpose are commutative in their respective senses.

Solution: $(A^2)_{11} = 1 \cdot 1 + 1 \cdot 1 = 2$, $(A^2)_{12} = 1 \cdot 1 + 1 \cdot 1 = 2$, $(A^2)_{21} = 1 \cdot 1 + 1 \cdot 1 = 2$, $(A^2)_{22} = 1 \cdot 1 + 1 \cdot 1 = 2$.

Solution: Adding the two equations: $(A + B) + (A - B) = 2A = \begin{pmatrix} 4 & 6 \\ 2 & 6 \end{pmatrix}$, so $A = \begin{pmatrix} 2 & 3 \\ 1 & 3 \end{pmatrix}$.

Solution: $A^T = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}$, so $A + A^T = \begin{pmatrix} 2+2 & 3+1 \\ 1+3 & 2+2 \end{pmatrix} = \begin{pmatrix} 4 & 4 \\ 4 & 4 \end{pmatrix}$.

Solution: Since $A$ is the identity matrix, $BA = B \cdot I = B = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$.

Solution: The transpose of a $1 \times 3$ matrix (row vector) is a $3 \times 1$ matrix (column vector). $A^T = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$.

Solution: Scalar multiplication multiplies each element by the scalar: $3A = 3 \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 3 & 6 \\ 9 & 12 \end{pmatrix}$.

Solution: The identity matrix commutes with any matrix of the same order: $IA = AI = A$. Other matrices generally do not commute.

Solution: Counterexample: $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$ give $AB = 0$ but neither matrix is zero. Matrices do not form an integral domain.

Solution: For a symmetric matrix ($A^T = A$), we require $\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a & c \\ b & d \end{pmatrix}$. This means $b = c$. The diagonal elements $a$ and $d$ can be any values.

Solution: $A^T = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$. Then $(A^T A)_{11} = \cos^2\theta + \sin^2\theta = 1$, $(A^T A)_{12} = -\cos\theta \sin\theta + \sin\theta \cos\theta = 0$, $(A^T A)_{21} = \sin\theta \cos\theta - \cos\theta \sin\theta = 0$, $(A^T A)_{22} = \sin^2\theta + \cos^2\theta = 1$.

Solution: $A^T = \begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix}$. $2A = \begin{pmatrix} 4 & 2 \\ 0 & 6 \end{pmatrix}$, $3A^T = \begin{pmatrix} 6 & 0 \\ 3 & 9 \end{pmatrix}$. $2A + 3A^T = \begin{pmatrix} 10 & 2 \\ 3 & 15 \end{pmatrix}$. The $(2,1)$ entry is $3$.