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Determinants - Determinant of a matrix of order 3 × 3

Grade 12CBSE

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

🔑Concepts

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A determinant of order 3 is a scalar value associated with a square matrix of order 3×33 \times 3.

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It is evaluated by expanding along any one of the three rows or three columns.

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The sign of each term in the expansion is determined by the position (i,j)(i, j) using the formula (−1)i+j(-1)^{i+j}.

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The sign convention follows a 'chessboard' pattern: [+−+−+−+−+]\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}

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To simplify calculations, it is often best to expand along the row or column that contains the maximum number of zeros.

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The determinant of a 3×33 \times 3 matrix is the sum of the products of elements of any row (or column) with their corresponding cofactors.

📐Formulae

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

det(A)=∣A∣=∣a11a12a13a21a22a23a31a32a33∣\text{det}(A) = |A| = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}

Expansion along R1:∣A∣=a11∣a22a23a32a33∣−a12∣a21a23a31a33∣+a13∣a21a22a31a32∣\text{Expansion along } R_1: |A| = 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}

∣A∣=a11(a22a33−a32a23)−a12(a21a33−a31a23)+a13(a21a32−a31a22)|A| = 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})

💡Examples

Problem 1:

Evaluate the determinant: Δ=∣3−4511−2231∣\Delta = \begin{vmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{vmatrix}

Solution:

Expanding along the first row (R1R_1): Δ=3∣1−231∣−(−4)∣1−221∣+5∣1123∣\Delta = 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} Δ=3(1(1)−3(−2))+4(1(1)−2(−2))+5(1(3)−2(1))\Delta = 3(1(1) - 3(-2)) + 4(1(1) - 2(-2)) + 5(1(3) - 2(1)) Δ=3(1+6)+4(1+4)+5(3−2)\Delta = 3(1 + 6) + 4(1 + 4) + 5(3 - 2) Δ=3(7)+4(5)+5(1)\Delta = 3(7) + 4(5) + 5(1) Δ=21+20+5\Delta = 21 + 20 + 5 2120+546\begin{array}{r} 21 \\ 20 \\ + 5 \\ \hline 46 \end{array} Δ=46\Delta = 46

Explanation:

We expanded the 3×33 \times 3 determinant along the first row. Each element of the row was multiplied by its corresponding 2×22 \times 2 minor, applying the alternating signs (+,−,+)(+, -, +). The 2×22 \times 2 determinants were then solved using cross-multiplication (ad−bc)(ad - bc).

Problem 2:

Find the value of xx if ∣x218x∣=∣62186∣\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}

Solution:

Equating the determinants on both sides: LHS: x(x)−18(2)=x2−36x(x) - 18(2) = x^2 - 36 RHS: 6(6)−18(2)=36−36=06(6) - 18(2) = 36 - 36 = 0 Therefore: x2−36=0x^2 - 36 = 0 x2=36x^2 = 36 x=±6x = \pm 6

Explanation:

Determinants are evaluated as single values. Unlike matrices, we do not equate corresponding elements; we must calculate the numerical value of both determinants and then solve the resulting equation.