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Determinants - Minors, co-factors and applications of determinants in finding the area of a triangle

Grade 12CBSE

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

🔑Concepts

Minor of an element aija_{ij}: The minor MijM_{ij} is the determinant of the sub-matrix obtained by deleting the ithi^{th} row and jthj^{th} column in which the element aija_{ij} lies. Visually, you can imagine crossing out the vertical and horizontal lines passing through aija_{ij} and looking at the remaining block of numbers.

Cofactor of an element aija_{ij}: The cofactor AijA_{ij} is defined as (1)i+j(-1)^{i+j} times the minor MijM_{ij}. Visually, this follows a checkerboard sign pattern starting with ++ at the top-left corner: (+++++)\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}.

Expansion of Determinant: The value of a determinant is the sum of products of elements of any row (or column) with their corresponding cofactors. For example, expanding along Row 1: Δ=a11A11+a12A12+a13A13\Delta = a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13}. Conversely, if elements of one row are multiplied with cofactors of a different row, the sum is always zero.

Area of a Triangle: The area of a triangle with vertices (x1,y1)(x_1, y_1), (x2,y2)(x_2, y_2), and (x3,y3)(x_3, y_3) is given by half the absolute value of the determinant formed by the coordinates. Because area is a scalar magnitude, we always take the positive value of the result.

Condition for Collinearity: If three points (x1,y1)(x_1, y_1), (x2,y2)(x_2, y_2), and (x3,y3)(x_3, y_3) are collinear, they lie on a single straight line and do not form a triangle. Geometrically, this means the area of the triangle formed by them must be zero, leading to the determinant of their coordinates being zero.

Equation of a Line: To find the equation of a line passing through two given points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2), we take an arbitrary point (x,y)(x, y) on the line and apply the condition of collinearity for these three points. This results in a linear equation in xx and yy.

📐Formulae

Minor of aij=Mija_{ij} = M_{ij}

Cofactor Aij=(1)i+jMijA_{ij} = (-1)^{i+j} M_{ij}

Value of determinant Δ=j=13aijAij\Delta = \sum_{j=1}^{3} a_{ij} A_{ij} (along any row ii)

Area of Triangle = 12x1y11x2y21x3y31\frac{1}{2} | \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} |

Collinearity Condition: x1y11x2y21x3y31=0\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0

Equation of a line through (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2): xy1x1y11x2y21=0\begin{vmatrix} x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end{vmatrix} = 0

💡Examples

Problem 1:

Find the minors and cofactors of all elements of the determinant Δ=1243\Delta = \begin{vmatrix} 1 & -2 \\ 4 & 3 \end{vmatrix}.

Solution:

  1. For element a11=1a_{11} = 1: Minor M11=3M_{11} = 3, Cofactor A11=(1)1+1(3)=3A_{11} = (-1)^{1+1}(3) = 3. \ 2. For element a12=2a_{12} = -2: Minor M12=4M_{12} = 4, Cofactor A12=(1)1+2(4)=4A_{12} = (-1)^{1+2}(4) = -4. \ 3. For element a21=4a_{21} = 4: Minor M21=2M_{21} = -2, Cofactor A21=(1)2+1(2)=2A_{21} = (-1)^{2+1}(-2) = 2. \ 4. For element a22=3a_{22} = 3: Minor M22=1M_{22} = 1, Cofactor A22=(1)2+2(1)=1A_{22} = (-1)^{2+2}(1) = 1.

Explanation:

To find the minor, we hide the row and column of the element. To find the cofactor, we multiply the minor by (1)i+j(-1)^{i+j} based on the element's position.

Problem 2:

Find the area of the triangle whose vertices are (3,8)(3, 8), (4,2)(-4, 2), and (5,1)(5, 1).

Solution:

The area Δ\Delta is given by: \ Δ=12381421511\Delta = \frac{1}{2} | \begin{vmatrix} 3 & 8 & 1 \\ -4 & 2 & 1 \\ 5 & 1 & 1 \end{vmatrix} | \ Expanding along R1R_1: \ Δ=123(21)8(45)+1(410)\Delta = \frac{1}{2} | 3(2 - 1) - 8(-4 - 5) + 1(-4 - 10) | \ Δ=123(1)8(9)+1(14)\Delta = \frac{1}{2} | 3(1) - 8(-9) + 1(-14) | \ Δ=123+7214\Delta = \frac{1}{2} | 3 + 72 - 14 | \ Δ=1261=30.5\Delta = \frac{1}{2} | 61 | = 30.5 sq. units.

Explanation:

We use the coordinate-based determinant formula for the area of a triangle. Expanding along the first row simplifies the calculation, and we take the absolute value of the final result.