krit.club logo

Determinants - Determinant of a matrix of order two

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

A determinant is a scalar value associated with a square matrix. For a matrix AA of order 22, the determinant is denoted by ∣A∣|A|, det⁡(A)\det(A), or Δ\Delta.

•

For a square matrix A=[a11a12a21a22]A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}, the determinant is calculated by taking the product of the elements in the principal diagonal and subtracting the product of the elements in the secondary diagonal.

•

Only square matrices have determinants. A matrix of order 2×22 \times 2 consists of 22 rows and 22 columns.

•

The determinant of a matrix of order 22 represents the signed area of the parallelogram formed by the column vectors of the matrix in a 2D plane.

📐Formulae

If A=\text{If } A = \begin{bmatrix} a & b \ c & d \end{bmatrix}, then ∣A∣=∣abcd∣=ad−bc, \text{ then } |A| = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc

Δ=a11a22−a21a12\Delta = a_{11}a_{22} - a_{21}a_{12}

det⁡(kA)=kndet⁡(A) (For order n=2,det⁡(kA)=k2∣A∣)\det(kA) = k^n \det(A) \text{ (For order } n=2, \det(kA) = k^2 |A|)

∣AB∣=∣A∣∣B∣|AB| = |A||B|

💡Examples

Problem 1:

Evaluate the determinant: Δ=∣24−12∣\Delta = \begin{vmatrix} 2 & 4 \\ -1 & 2 \end{vmatrix}

Solution:

Δ=(2)(2)−(4)(−1)\Delta = (2)(2) - (4)(-1) Δ=4−(−4)\Delta = 4 - (-4) Δ=4+4=8\Delta = 4 + 4 = 8

Explanation:

Apply the formula ad−bcad - bc where a=2,b=4,c=−1,d=2a=2, b=4, c=-1, d=2.

Problem 2:

Evaluate: ∣cos⁡θ−sin⁡θsin⁡θcos⁡θ∣\begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{vmatrix}

Solution:

Determinant=(cos⁡θ)(cos⁡θ)−(−sin⁡θ)(sin⁡θ)\text{Determinant} = (\cos \theta)(\cos \theta) - (-\sin \theta)(\sin \theta) =cos⁡2θ−(−sin⁡2θ)= \cos^2 \theta - (-\sin^2 \theta) =cos⁡2θ+sin⁡2θ= \cos^2 \theta + \sin^2 \theta =1= 1

Explanation:

The product of diagonal elements is cos⁡2θ\cos^2 \theta and off-diagonal elements is −sin⁡2θ-\sin^2 \theta. Subtracting them yields the trigonometric identity cos⁡2θ+sin⁡2θ=1\cos^2 \theta + \sin^2 \theta = 1.

Problem 3:

Find the value of xx for which ∣3xx1∣=∣3241∣\begin{vmatrix} 3 & x \\ x & 1 \end{vmatrix} = \begin{vmatrix} 3 & 2 \\ 4 & 1 \end{vmatrix}

Solution:

3(1)−x(x)=3(1)−4(2)3(1) - x(x) = 3(1) - 4(2) 3−x2=3−83 - x^2 = 3 - 8 3−x2=−53 - x^2 = -5 −x2=−5−3-x^2 = -5 - 3 −x2=−8-x^2 = -8 x2=8x^2 = 8 x=±8=±22x = \pm \sqrt{8} = \pm 2\sqrt{2}

Explanation:

Equate the determinants of both sides by expanding them using the ad−bcad-bc rule and solve the resulting quadratic equation for xx.