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Determinants - Determinant of a square matrix (up to 3 x 3 matrices)

Grade 12CBSE

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

🔑Concepts

Determinant Definition: Every square matrix A=[aij]A = [a_{ij}] is associated with a scalar value called its determinant, denoted by A|A| or det(A)det(A). For a 1×11 \times 1 matrix A=[a]A = [a], the determinant is simply A=a|A| = a. It is important to remember that only square matrices have determinants.

Determinant of a 2×22 \times 2 Matrix: For a matrix A=[abcd]A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, the determinant is calculated by the 'cross-multiplication' method. Visually, imagine drawing two diagonal arrows: one from aa to dd (principal diagonal) and one from bb to cc (off-diagonal). The value is the product of the first diagonal minus the product of the second: adbcad - bc.

Expansion of a 3×33 \times 3 Matrix: The determinant of a 3×33 \times 3 matrix is found by breaking it down into smaller 2×22 \times 2 determinants (minors). You can expand along any row or column. For example, expanding along the first row involves taking each element a1ja_{1j}, multiplying it by its 2×22 \times 2 minor (the determinant remaining after deleting the row and column of a1ja_{1j}), and applying a specific sign.

Sign Convention (Checkerboard Pattern): When expanding a 3×33 \times 3 determinant, the sign of each term is determined by its position (i,j)(i, j) using the formula (1)i+j(-1)^{i+j}. Visually, this creates a checkerboard pattern of signs: [+++++]\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}. This pattern ensures that adjacent elements always have opposite signs.

Singular vs. Non-Singular Matrices: A square matrix AA is called 'Singular' if its determinant A=0|A| = 0. Visually, this means the matrix maps a space into a lower dimension (e.g., a 2D plane into a 1D line). If A0|A| \neq 0, the matrix is 'Non-singular' and possesses an inverse.

Area of a Triangle: Determinants provide a geometric tool to find 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). If the calculated determinant value is 00, it visually confirms that the three points are collinear, meaning they lie on the same straight line and do not form a triangle.

Minors and Cofactors: The minor MijM_{ij} of an element aija_{ij} is the determinant of the sub-matrix obtained by deleting the ithi^{th} row and jthj^{th} column. The cofactor AijA_{ij} is the minor with the positional sign applied, calculated as Aij=(1)i+jMijA_{ij} = (-1)^{i+j} M_{ij}.

Properties of Determinants: (1) The determinant of a matrix and its transpose are equal: A=AT|A| = |A^T|. (2) If any two rows (or columns) of a determinant are identical, the value of the determinant is zero. (3) If all elements of a row or column are zero, the determinant is zero.

📐Formulae

For A=[a11a12a21a22],A=a11a22a12a21A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}, |A| = a_{11}a_{22} - a_{12}a_{21}

For A=[a11a12a13a21a22a23a31a32a33],A=a11(a22a33a23a32)a12(a21a33a23a31)+a13(a21a32a22a31)A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}, |A| = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})

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

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} |

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

AB=AB|AB| = |A| |B| (where AA and BB are square matrices of same order)

kA=knA|kA| = k^n |A|, where nn is the order of the square matrix AA

💡Examples

Problem 1:

Evaluate the determinant of the matrix A=[2412]A = \begin{bmatrix} 2 & 4 \\ -1 & 2 \end{bmatrix}.

Solution:

Step 1: Use the formula for a 2×22 \times 2 determinant A=adbc|A| = ad - bc. \ Step 2: Identify a=2,b=4,c=1,d=2a=2, b=4, c=-1, d=2. \ Step 3: A=(2×2)(4×1)|A| = (2 \times 2) - (4 \times -1) \ Step 4: A=4(4)=4+4=8|A| = 4 - (-4) = 4 + 4 = 8.

Explanation:

We apply the cross-multiplication rule for 2×22 \times 2 matrices, ensuring the negative sign in the formula is correctly handled when multiplying by the negative element 1-1.

Problem 2:

Evaluate the determinant of A=[345112231]A = \begin{bmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{bmatrix} by expanding along the first row.

Solution:

Step 1: Apply the expansion along R1R_1: A=31231(4)1221+51123|A| = 3 \begin{vmatrix} 1 & -2 \\ 3 & 1 \end{vmatrix} - (-4) \begin{vmatrix} 1 & -2 \\ 2 & 1 \end{vmatrix} + 5 \begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix} \ Step 2: Solve the 2×22 \times 2 determinants: \ 1231=(1)(1)(2)(3)=1+6=7\begin{vmatrix} 1 & -2 \\ 3 & 1 \end{vmatrix} = (1)(1) - (-2)(3) = 1 + 6 = 7 \ 1221=(1)(1)(2)(2)=1+4=5\begin{vmatrix} 1 & -2 \\ 2 & 1 \end{vmatrix} = (1)(1) - (-2)(2) = 1 + 4 = 5 \ 1123=(1)(3)(1)(2)=32=1\begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix} = (1)(3) - (1)(2) = 3 - 2 = 1 \ Step 3: Substitute back into the expansion: A=3(7)+4(5)+5(1)|A| = 3(7) + 4(5) + 5(1) \ Step 4: A=21+20+5=46|A| = 21 + 20 + 5 = 46.

Explanation:

The 3×33 \times 3 determinant is solved by decomposing it into 2×22 \times 2 minors. Note the sign change for the middle term (4)-(-4) due to the checkerboard sign pattern.