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Elite lesson· 40 min Hinglish Unlock every lessonFormulas At a glance 11 formulas
1 2×2 determinant ∣ a b c d ∣ = a d − b c \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc a c b d = a d − b c 2 3×3 expansion Expand along the row or column with the most zeros: each entry times the 2×2 left when its row and column are deleted, with checkerboard signs + − + +\,-\,+ + − + from the top left. 3 Triangular matrix When every entry below (or above) the diagonal is 0, the determinant is the product of the diagonal entries, not their sum. 4 Determinant is zero Two equal or proportional rows (or columns), or a row of zeros, make the determinant 0. Look for them before expanding. 5 Row operations R i ↔ R j : sign flips R_i \leftrightarrow R_j:\ \text{sign flips} R i ↔ R j : sign flips R i → k R i : × k ; R i → R i + k R j : no change R_i \to kR_i:\ \times k;\ R_i \to R_i + kR_j:\ \text{no change} R i → k R i : × k ; R i → R i + k R j : no change 6 Scalar multiple det ( k A ) = k n det A \det(kA) = k^n \det A det ( k A ) = k n det A n = order of A n = \text{order of } A n = order of A 7 Product, transpose, inverse det ( A B ) = det A ⋅ det B \det(AB) = \det A \cdot \det B det ( A B ) = det A ⋅ det B det A T = det A , det A − 1 = 1 det A \det A^T = \det A,\ \det A^{-1} = \dfrac{1}{\det A} det A T = det A , det A − 1 = det A 1 8 Adjoint det ( adj A ) = ( det A ) n − 1 \det(\text{adj}\,A) = (\det A)^{n-1} det ( adj A ) = ( det A ) n − 1 A ( adj A ) = ( det A ) I , n = order , I = identity A\,(\text{adj}\,A) = (\det A)\,I,\ n = \text{order},\ I = \text{identity} A ( adj A ) = ( det A ) I , n = order , I = identity 9 Singular matrix det A = 0 ⟺ A − 1 does not exist \det A = 0 \iff A^{-1} \text{ does not exist} det A = 0 ⟺ A − 1 does not exist det ( A 2 + k A ) = det A ⋅ det ( A + k I ) , I = identity \det(A^2 + kA) = \det A \cdot \det(A + kI),\ I = \text{identity} det ( A 2 + k A ) = det A ⋅ det ( A + k I ) , I = identity 10 Rows 1, a, a² ∣ 1 1 1 a b c a 2 b 2 c 2 ∣ = ( b − a ) ( c − a ) ( c − b ) \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} = (b - a)(c - a)(c - b) 1 a a 2 1 b b 2 1 c c 2 = ( b − a ) ( c − a ) ( c − b ) 11 Cyclic rows ∣ a b c b c a c a b ∣ = 3 a b c − a 3 − b 3 − c 3 \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 3abc - a^3 - b^3 - c^3 a b c b c a c a b = 3 ab c − a 3 − b 3 − c 3 = − ( a + b + c ) ( a 2 + b 2 + c 2 − a b − b c − c a ) = -(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = − ( a + b + c ) ( a 2 + b 2 + c 2 − ab − b c − c a ) Test yourself Q1 Easy
If ∣ k 6 4 3 ∣ = 0 \begin{vmatrix} k & 6 \\ 4 & 3 \end{vmatrix} = 0 k 4 6 3 = 0 , what is k k k ?
A − 8 -8 − 8 B 2 2 2 C 4.5 4.5 4.5 D 8 8 8
Correct answer· 8 8 8
Why
A 2 × 2 2 \times 2 2 × 2 determinant is a d − b c ad - bc a d − b c .
Here that is 3 k − 6 × 4 = 3 k − 24 3k - 6 \times 4 = 3k - 24 3 k − 6 × 4 = 3 k − 24 , which is 0 when 3 k = 24 3k = 24 3 k = 24 .
So k = 8 k = 8 k = 8 .
Q2 Easy
What is the value of ∣ 3 7 − 2 0 − 2 9 0 0 5 ∣ \begin{vmatrix} 3 & 7 & -2 \\ 0 & -2 & 9 \\ 0 & 0 & 5 \end{vmatrix} 3 0 0 7 − 2 0 − 2 9 5 ?
A − 30 -30 − 30 B − 6 -6 − 6 C 6 6 6 D 30 30 30
Correct answer· − 30 -30 − 30
Why
Everything below the diagonal is 0, so the determinant is the product of the diagonal.
3 × ( − 2 ) × 5 = − 30 3 \times (-2) \times 5 = -30 3 × ( − 2 ) × 5 = − 30 (adding the diagonal gives the trap 6).
So the value is − 30 -30 − 30 .
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