Ravi Kant Kumar

Construct 2×3 Matrix using aij = i + j Constructing a Matrix using aij = i + j Question: Construct a \( 2 \times 3 \) matrix \( A = [a_{ij}] \) whose elements are given by \( a_{ij} = i + j \). Step 1: Matrix Order A \( 2 \times 3 \) matrix […]

Construct 2×3 Matrix using aij = 2i − j Constructing a Matrix using aij = 2i − j Question: Construct a \( 2 \times 3 \) matrix \( A = [a_{ij}] \) whose elements are given by \( a_{ij} = 2i – j \). Step 1: Matrix Order A \( 2 \times 3 \) matrix

Construct 2×3 Matrix using aij = i×j Constructing a Matrix using aij = i × j Question: Construct a \( 2 \times 3 \) matrix \( A = [a_{ij}] \) whose elements are given by \( a_{ij} = i \times j \). Step 1: Matrix Order A \( 2 \times 3 \) matrix has: 2

Order of R1 and C2 of a 3×4 Matrix Order of First Row and Second Column of a Matrix Question: Let \( A \) be a matrix of order \( 3 \times 4 \). If \( R_1 \) denotes the first row and \( C_2 \) denotes the second column, find their orders. Concept Used

Find a22 + b21 and a11b11 + a22b22 Matrix Element Problem Solution Question: If \( A = [a_{ij}] = \begin{bmatrix} 2 & 3 & -5 \\ 1 & 4 & 9 \\ 0 & 7 & -2 \end{bmatrix} \) and \( B = [b_{ij}] = \begin{bmatrix} 2 & -1 \\ -3 & 4 \\ 1

Possible Orders of a Matrix with Given Number of Elements Possible Orders of a Matrix with Given Number of Elements Question: If a matrix has 8 elements, what are the possible orders it can have? What if it has 5 elements? Concept Used If a matrix has order \( m \times n \), then the

Value of tan(cos⁻¹(3/5) + tan⁻¹(1/4)) Question Evaluate: \[ \tan\left(\cos^{-1}\left(\frac{3}{5}\right) + \tan^{-1}\left(\frac{1}{4}\right)\right) \] Solution Let \[ A = \cos^{-1}\left(\frac{3}{5}\right), \quad B = \tan^{-1}\left(\frac{1}{4}\right) \] Find tan A: \[ \cos A = \frac{3}{5} \Rightarrow \sin A = \frac{4}{5} \] \[ \tan A = \frac{4}{3} \] Find tan B: \[ \tan B = \frac{1}{4} \] Use identity: \[

Domain of cos⁻¹(x² − 4) Question Find the domain of: \[ \cos^{-1}(x^2 – 4) \] Solution For \( \cos^{-1}(t) \) to be defined: \[ -1 \le t \le 1 \] So, \[ -1 \le x^2 – 4 \le 1 \] Add 4: \[ 3 \le x^2 \le 5 \] Thus, \[ \sqrt{3} \le |x| \le

Value of 2tan⁻¹x + sin⁻¹(2x/(1+x²)) Question If \( x > 1 \), evaluate: \[ 2\tan^{-1}x + \sin^{-1}\left(\frac{2x}{1+x^2}\right) \] Solution Use identity: \[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}x \quad \text{(for } x \le 1\text{)} \] But for \( x > 1 \), principal value adjustment gives: \[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \pi – 2\tan^{-1}x \] Thus, \[ 2\tan^{-1}x + \left(\pi

Value of sin(2 tan⁻¹(0.75)) Question Evaluate: \[ \sin\left(2\tan^{-1}(0.75)\right) \] Solution Let \[ \theta = \tan^{-1}(0.75) = \tan^{-1}\left(\frac{3}{4}\right) \] Then, \[ \tan\theta = \frac{3}{4} \Rightarrow \text{Opposite} = 3,\ \text{Adjacent} = 4 \Rightarrow \text{Hypotenuse} = 5 \] \[ \sin\theta = \frac{3}{5}, \quad \cos\theta = \frac{4}{5} \] Now use identity: \[ \sin 2\theta = 2\sin\theta \cos\theta \] \[