Introduction to Determinants

Every square matrix A=[aij]A = [a_{ij}] of order nn can be associated with a number (real or complex) called the determinant of the square matrix AA.

Think of a system of linear equations:

a1x+b1y=c1a_1x + b_1y = c_1

a2x+b2y=c2a_2x + b_2y = c_2

This system can be represented in matrix form as: [a1b1a2b2][xy]=[c1c2]\begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}

This system of equations has a unique solution if and only if a1b2−a2b1≠0a_1b_2 - a_2b_1 \neq 0.

The expression (a1b2−a2b1)(a_1b_2 - a_2b_1) determines the uniqueness of the solution associated with the matrix A=[a1b1a2b2]A = \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix} and is called the determinant of AA.

Notation: The determinant of matrix AA is denoted by ∣A∣|A|, det(A)\text{det}(A), or Δ\Delta.

  • Important Note: For a matrix AA, ∣A∣|A| is read as the determinant of A, not the modulus of A. Only square matrices have determinants.

Evaluation of Determinants of Order 1 and 2

Determinant of a Matrix of Order 1

Let A=[a]A = [a] be the matrix of order 1×11 \times 1. The determinant of AA is defined to be equal to aa.

  • Example: If A=[−5]A = [-5], then ∣A∣=−5|A| = -5.

Determinant of a Matrix of Order 2

Let A=[a11a12a21a22]A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} be a matrix of order 2×22 \times 2.

The determinant of AA is defined as: ∣A∣=∣a11a12a21a22∣=a11a22−a21a12|A| = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} = a_{11}a_{22} - a_{21}a_{12}

Multiply the diagonal elements (top-left to bottom-right) and subtract the product of the off-diagonal elements (bottom-left to top-right).

  • Example: If A=[24−13]A = \begin{bmatrix} 2 & 4 \\ -1 & 3 \end{bmatrix}, then ∣A∣=(2)(3)−(4)(−1)=6+4=10|A| = (2)(3) - (4)(-1) = 6+4=10

Evaluation of Determinants of Order 3

A determinant of order 3 can be evaluated by expressing it in terms of second-order determinants. This process is called expansion of a determinant along a row or a column. There are 6 ways to expand a determinant of order 3 (along any of the 3 rows or 3 columns), and they all yield the same value.

LetA=[a11a12a13a21a22a23a31a32a33]Let A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}

Expansion along Row 1 (R1R_1)

Step 1: Multiply the first element a11a_{11} by (−1)1+1(-1)^{1+1} and the 2×22 \times 2 determinant obtained by deleting R1R_1 and C1C_1.

Step 2: Multiply the second element a12a_{12} by (−1)1+2(-1)^{1+2} and the 2×22 \times 2 determinant obtained by deleting R1R_1 and C2C_2.

Step 3: Multiply the third element a13a_{13} by (−1)1+3(-1)^{1+3} and the 2×22 \times 2 determinant obtained by deleting R1R_1 and C3C_3.

Δ=a11∣a22a23a32a33∣−a12∣a21a23a31a33∣+a13∣a21a22a31a32∣\Delta = a_{11}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} - a_{12}\begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} + a_{13}\begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix} Δ=a11(a22a33−a32a23)−a12(a21a33−a31a23)+a13(a21a32−a31a22)\Delta = a_{11}(a_{22}a_{33} - a_{32}a_{23}) - a_{12}(a_{21}a_{33} - a_{31}a_{23}) + a_{13}(a_{21}a_{32} - a_{31}a_{22})

Sign Convention: The sign (−1)i+j(-1)^{i+j} alternates: ∣+−+−+−+−+∣\begin{vmatrix} + & - & + \\ - & + & - \\ + & - & + \end{vmatrix}

Calculation Tip: For easier calculations, we shall expand the determinant along that row or column which contains maximum number of zeros.

Example 1: Basic 2x2 Determinant

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

Solution: Step 1: Identify the elements. a11=2a_{11} = 2, a12=4a_{12} = 4, a21=−1a_{21} = -1, a22=2a_{22} = 2.

Step 2: Apply the 2×22 \times 2 determinant formula: Δ=(a11×a22)−(a21×a12)\Delta = (a_{11} \times a_{22}) - (a_{21} \times a_{12}) Step 3: Substitute the values: Δ=(2×2)−(−1×4)\Delta = (2 \times 2) - (-1 \times 4) Δ=4−(−4)=4+4=8\Delta = 4 - (-4) = 4 + 4 = 8

Answer: 88

Example 2: Algebraic 2x2 Determinant

Evaluate: Δ=∣xx+1x−1x∣\Delta = \begin{vmatrix} x & x+1 \\ x-1 & x \end{vmatrix}

Solution: Step 1: Apply the formula for a 2×22 \times 2 determinant. Δ=(x×x)−((x+1)×(x−1))\Delta = (x \times x) - ((x+1) \times (x-1)) Step 2: Use the algebraic identity (a+b)(a−b)=a2−b2(a+b)(a-b) = a^2 - b^2. Δ=x2−(x2−12)\Delta = x^2 - (x^2 - 1^2) Step 3: Simplify the expression. Δ=x2−(x2−1)=x2−x2+1=1\Delta = x^2 - (x^2 - 1) = x^2 - x^2 + 1 = 1

Answer: 11

Example 3: Solving a Determinant Equation

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

Solution: Step 1: Evaluate the determinant on the Left Hand Side (LHS). LHS=(3×1)−(x×x)=3−x2LHS = (3 \times 1) - (x \times x) = 3 - x^2 Step 2: Evaluate the determinant on the Right Hand Side (RHS). RHS=(3×1)−(4×2)=3−8=−5RHS = (3 \times 1) - (4 \times 2) = 3 - 8 = -5 Step 3: Equate the LHS and RHS. 3−x2=−53 - x^2 = -5 Step 4: Solve for xx. 3+5=x23 + 5 = x^2 x2=8x^2 = 8 x=±8=±22x = \pm \sqrt{8} = \pm 2\sqrt{2}

Answer: x=±22x = \pm 2\sqrt{2}

Example 4: Trigonometric Determinant

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

Solution: Step 1: Apply the 2×22 \times 2 determinant formula. Δ=(cos⁡θ×cos⁡θ)−(sin⁡θ×(−sin⁡θ))\Delta = (\cos \theta \times \cos \theta) - (\sin \theta \times (-\sin \theta)) Step 2: Simplify the terms. Δ=cos⁡2θ−(−sin⁡2θ)\Delta = \cos^2 \theta - (-\sin^2 \theta) Δ=cos⁡2θ+sin⁡2θ\Delta = \cos^2 \theta + \sin^2 \theta Step 3: Use the fundamental trigonometric identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 Δ=1\Delta = 1

Answer: 11

Example 5: Basic 3x3 Determinant Expansion

Evaluate: Δ=∣124−130410∣\Delta = \begin{vmatrix} 1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0 \end{vmatrix}

Solution: Step 1: Observe the determinant. Column 3 (C3C_3) has two zeros. Expanding along C3C_3 will make the calculation much easier.

Step 2: Expand along C3C_3. Remember the sign convention for the third column: +,−,++,-,+. Δ=4∣−1341∣−0∣1241∣+0∣12−13∣\Delta = 4 \begin{vmatrix} -1 & 3 \\ 4 & 1 \end{vmatrix} - 0 \begin{vmatrix} 1 & 2 \\ 4 & 1 \end{vmatrix} + 0 \begin{vmatrix} 1 & 2 \\ -1 & 3 \end{vmatrix} Step 3: The terms multiplied by zero vanish. Δ=4[(−1×1)−(4×3)]−0+0\Delta = 4 [(-1 \times 1) - (4 \times 3)] - 0 + 0 Step 4: Simplify the remaining term. Δ=4[−1−12]=4[−13]=−52\Delta = 4 [-1 - 12] = 4 [-13] = -52

Answer: −52-52

Example 6: 3x3 Determinant Expansion along Row 1

Evaluate: Δ=∣3−1−200−13−50∣\Delta = \begin{vmatrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{vmatrix}

Solution: Step 1: Let's expand along the first row (R1R_1). Sign convention: +,−,++,-,+. Δ=3∣0−1−50∣−(−1)∣0−130∣+(−2)∣003−5∣\Delta = 3 \begin{vmatrix} 0 & -1 \\ -5 & 0 \end{vmatrix} - (-1) \begin{vmatrix} 0 & -1 \\ 3 & 0 \end{vmatrix} + (-2) \begin{vmatrix} 0 & 0 \\ 3 & -5 \end{vmatrix} Step 2: Evaluate the 2×22 \times 2 determinants.

For the 1st matrix: (0×0)−(−5×−1)=0−5=−5(0 \times 0) - (-5 \times -1) = 0 - 5 = -5

For the 2nd matrix: (0×0)−(3×−1)=0−(−3)=3(0 \times 0) - (3 \times -1) = 0 - (-3) = 3

For the 3rd matrix: (0×−5)−(3×0)=0−0=0(0 \times -5) - (3 \times 0) = 0 - 0 = 0

Step 3: Substitute back into the expansion. Δ=3(−5)+1(3)−2(0)\Delta = 3(-5) + 1(3) - 2(0) Δ=−15+3−0=−12\Delta = -15 + 3 - 0 = -12

(Note: Expanding along R2R_2 would have been faster since it has two zeros! Try it): Δ=−(−1)∣3−13−5∣=1(−15−(−3))=−12\Delta = -(-1) \begin{vmatrix} 3 & -1 \\ 3 & -5 \end{vmatrix} = 1(-15 - (-3)) = -12

Answer: −12-12

Example 7: Proving a Matrix Determinant Property

Let A=[101012004]A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix}. Show that ∣3A∣=27∣A∣|3A| = 27|A|.

Solution: Step 1: First, evaluate the determinant of matrix AA (∣A∣|A|). Expanding along C1C_1 (since it has two zeros): ∣A∣=1∣1204∣−0+0=1(4−0)=4|A| = 1 \begin{vmatrix} 1 & 2 \\ 0 & 4 \end{vmatrix} - 0 + 0 = 1(4 - 0) = 4 Step 2: Find matrix 3A3A by multiplying every element of AA by 33. 3A=[3030360012]3A = \begin{bmatrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{bmatrix} Step 3: Evaluate ∣3A∣|3A|. Expanding along C1C_1: ∣3A∣=3∣36012∣−0+0=3(36−0)=108|3A| = 3 \begin{vmatrix} 3 & 6 \\ 0 & 12 \end{vmatrix} - 0 + 0 = 3(36 - 0) = 108 Step 4: Check the relationship. 27∣A∣=27×4=10827|A| = 27 \times 4 = 108.

Since ∣3A∣=108|3A| = 108 and 27∣A∣=10827|A| = 108, we have proved that ∣3A∣=27∣A∣|3A| = 27|A|.

(Property check: For a square matrix of order n, ∣kA∣=kn∣A∣|kA| = k^n|A| Here k=3,n=3k=3, n=3, so ∣3A∣=33∣A∣=27∣A∣|3A| = 3^3|A| = 27|A|).

Answer: Proved.

Example 8: Quadratic Determinant Equation

Find values of xx, if ∣2451∣=∣2x46x∣\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}

Solution: Step 1: Evaluate LHS. LHS=(2×1)−(5×4)=2−20=−18LHS = (2 \times 1) - (5 \times 4) = 2 - 20 = -18 Step 2: Evaluate RHS. RHS=(2x×x)−(6×4)=2x2−24RHS = (2x \times x) - (6 \times 4) = 2x^2 - 24 Step 3: Equate LHS and RHS. −18=2x2−24-18 = 2x^2 - 24 Step 4: Solve for xx. 2x2=24−182x^2 = 24 - 18   ⟹  2x2=6\implies 2x^2 = 6   ⟹  x2=3  ⟹  x=±3\implies x^2 = 3 \implies x = \pm \sqrt{3}

Answer: x=3,−3x = \sqrt{3}, -\sqrt{3}

Example 9: Complex 3x3 Determinant

Evaluate: Δ=∣0sin⁡α−cos⁡α−sin⁡α0sin⁡βcos⁡α−sin⁡β0∣\Delta = \begin{vmatrix} 0 & \sin \alpha & -\cos \alpha \\ -\sin \alpha & 0 & \sin \beta \\ \cos \alpha & -\sin \beta & 0 \end{vmatrix}

Solution: Step 1: Expand along R1R_1. Δ=0∣0sin⁡β−sin⁡β0∣−sin⁡α∣−sin⁡αsin⁡βcos⁡α0∣+(−cos⁡α)∣−sin⁡α0cos⁡α−sin⁡β∣\Delta = 0 \begin{vmatrix} 0 & \sin \beta \\ -\sin \beta & 0 \end{vmatrix} - \sin \alpha \begin{vmatrix} -\sin \alpha & \sin \beta \\ \cos \alpha & 0 \end{vmatrix} + (-\cos \alpha) \begin{vmatrix} -\sin \alpha & 0 \\ \cos \alpha & -\sin \beta \end{vmatrix} Step 2: Calculate the 2×22 \times 2 determinants. First term =0= 0

Second term =−sin⁡α[(−sin⁡α×0)−(cos⁡α×sin⁡β)]= -\sin \alpha [(-\sin \alpha \times 0) - (\cos \alpha \times \sin \beta)] =−sin⁡α[0−sin⁡βcos⁡α]=sin⁡αsin⁡βcos⁡α= -\sin \alpha [0 - \sin \beta \cos \alpha] = \sin \alpha \sin \beta \cos \alpha Third term =−cos⁡α[(−sin⁡α×−sin⁡β)−(cos⁡α×0)]= -\cos \alpha [(-\sin \alpha \times -\sin \beta) - (\cos \alpha \times 0)] =−cos⁡α[sin⁡αsin⁡β−0]=−sin⁡αsin⁡βcos⁡α= -\cos \alpha [\sin \alpha \sin \beta - 0] = -\sin \alpha \sin \beta \cos \alpha Step 3: Combine all terms. Δ=0+sin⁡αsin⁡βcos⁡α−sin⁡αsin⁡βcos⁡α=0\Delta = 0 + \sin \alpha \sin \beta \cos \alpha - \sin \alpha \sin \beta \cos \alpha = 0 (Note: This is a skew-symmetric matrix of odd order. The determinant of any skew-symmetric matrix of odd order is always 0).

Answer: 00

Example 10: Evaluating using polynomial elements

Evaluate: Δ=∣x2−x+1x−1x+11∣\Delta = \begin{vmatrix} x^2 - x + 1 & x-1 \\ x+1 & 1 \end{vmatrix}

Solution: Step 1: Apply the 2×22 \times 2 determinant formula. Δ=(x2−x+1)×1−(x+1)×(x−1)\Delta = (x^2 - x + 1) \times 1 - (x+1) \times (x-1) Step 2: Use the identity (a+b)(a−b)=a2−b2(a+b)(a-b) = a^2 - b^2 Δ=(x2−x+1)−(x2−12)\Delta = (x^2 - x + 1) - (x^2 - 1^2) Step 3: Expand and simplify the expression. Δ=x2−x+1−x2+1\Delta = x^2 - x + 1 - x^2 + 1 Δ=−x+2\Delta = -x + 2

Answer: 2−x2 - x