BCA Semester 1 – Mathematics I () Questions & Answers | Past TU Exam Papers

To prove: \[ \begin{vmatrix} x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1 \end{vmatrix} = (x - y)(y - z)(z - x) \] Proof: Given determinant: \[ =\begin{vmatrix} x & x^{2} & 1 \\ y & y^{2} & 1 \\ z

\[ \textbf{Definition (Singular Matrix):} \] \[ \text{A square matrix is called singular if its determinant is zero, i.e., }|A|=0. \] \[ \text{A singular matrix has no inverse.} \] --- \[ \textbf{Def

$\textbf{Given:}$ \[ A = \begin{pmatrix} 4 & 0 \\ 0 & 5 \end{pmatrix}, \quad AX = \begin{pmatrix} 12 \\ 24 \end{pmatrix} \] $\textbf{Solution:}$ $\text{Let } X = \begin{pmatrix} x1 \\ x2 \end{pmat

$\textbf{Definition of a matrix:}$ $\text{A matrix is a rectangular array of numbers arranged in rows and columns.}$ $\textbf{Given:}$ \[ A = \begin{pmatrix}2 & 0 \\ 1 & 3\end{pmatrix}, \quad B

$\textbf{Given determinant:}$ \[ D = \begin{vmatrix}a & b & c \\ a^2 & b^2 & c^2 \\ 1 & 1 & 1\end{vmatrix} \] $\textbf{Step 1:}$ Apply row operations. Subtract row 3 multiplied by $a$ from row 1 and

$\textbf{Given matrix:}$ \[ A = \begin{pmatrix} 1 & 4 & 1 \\ 3 & 3 & -2 \\ 0 & -4 & 1 \end{pmatrix} \] $\textbf{Step 1:}$ Find determinant of $A$ $$ \det(A) = 1 \cdot \begin{vmatrix} 3 & -2 \\ -4

$\textbf{Problem a:}$ Prove that \[ \begin{vmatrix} 1 & a & b & c \\ 1 & b & c & a \\ 1 & c & a & b \end{vmatrix} = (a - b)(b - c)(c - a) \] $\textbf{Step 1:}$ Expand along the first column or use p

$\textbf{Given determinant:}$ \[ D = \begin{vmatrix} 1 + x & 1 & 1 \\ 1 & 1 + y & 1 \\ 1 & 1 & 1 + z \end{vmatrix} \] $\textbf{Step 1:}$ Subtract the first column from the second and third columns: