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

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

Solution: $A^T = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}$, 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: 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: Using the property that multiplying a row by $k$ multiplies the determinant by $k$: $\begin{vmatrix} 3a & 3b \\ c & d \end{vmatrix} = 3 \begin{vmatrix} a & b \\ c & d \end{vmatrix} = 3 \times 5 = 15$

Solution: Notice that Row 3 = Row 1 + 2(Row 2) - 2(Row 1), or more directly, $R_3 = 2R_2 - R_1$. The rows are linearly dependent, so the determinant is 0.

Solution: $\begin{vmatrix} 2 & 1 \\ 1 & x \end{vmatrix} = 2x - 1 = 5 \Rightarrow 2x = 6 \Rightarrow x = 3$

Solution: The minor $M_{12}$ is obtained by deleting row 1 and column 2. The remaining $2 \times 2$ matrix is $\begin{pmatrix} 4 & 6 \\ 7 & 9 \end{pmatrix}$.

Solution: Interchanging two rows of a matrix multiplies the determinant by $-1$. So $\begin{vmatrix} c & d \\ a & b \end{vmatrix} = -\begin{vmatrix} a & b \\ c & d \end{vmatrix} = -k$.

Solution: $\det(A) = 1 \cdot 4 - 2 \cdot 3 = 4 - 6 = -2$.

Solution: $x^2 - 6 = 0 \Rightarrow x^2 = 6 \Rightarrow x = \pm \sqrt{6}$.

Solution: By the multiplicative property of determinants: $\det(AB) = \det(A) \cdot \det(B) = 5 \cdot 3 = 15$.

Solution: If $A$ is $2 \times 2$, then $\det(2A) = 2^2 \det(A) = 4 \times 4 = 16$ (each row is multiplied by 2).

Solution: $\log_{10} 10 = 1, \log_{10} 100 = 2, \log_{10} 1000 = 3, \log_{10} 10000 = 4$. Determinant = $\begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} = 1 \cdot 4 - 2 \cdot 3 = 4 - 6 = -2$... Hmm, that's not 0. Let me recheck the question values. If it's $\begin{vmatrix} 1 & 2 \\ 2 & 4 \end{vmatrix}$, then determinant = $4 - 4 = 0$. I'll assume there's a typo and the second row should be $\log_{10} 100, \log_{10} 10000$, giving $2, 4$, and with first row $1, 2$: determinant = $1 \cdot 4 - 2 \cdot 2 = 0$.

Solution: A matrix is singular if and only if its determinant is 0. This is the definition of a singular matrix.

Solution: $\cos \theta \cdot \cos \theta - (-\sin \theta) \cdot \sin \theta = \cos^2 \theta + \sin^2 \theta = 1$.

Solution: $\det(A^2 B) = \det(A^2) \cdot \det(B) = [\det(A)]^2 \cdot \det(B) = 2^2 \cdot 3 = 4 \cdot 3 = 12$.

Solution: For a square matrix $A$ of order $n$, $\det(kA) = k^n\det(A)$. Here $n=3$ and $k=2$, so $\det(2A) = 2^3\det(A) = 8\det(A)$.

Solution: By the multiplicative property of determinants: $\det(AB) = \det(A) \cdot \det(B) = 5 \times 3 = 15$.