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05.06 · Modern Maths

Determinants

Elite lesson11 formulas10 questions

On the syllabus of

Elite lesson40 minHinglish

Formulas

At a glance 11 formulas

  1. 12×2 determinant∣abcd∣=ad−bc\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc
  2. 23×3 expansionExpand 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. 3Triangular matrixWhen every entry below (or above) the diagonal is 0, the determinant is the product of the diagonal entries, not their sum.
  4. 4Determinant is zeroTwo equal or proportional rows (or columns), or a row of zeros, make the determinant 0. Look for them before expanding.
  5. 5Row operationsRi↔Rj: sign flipsR_i \leftrightarrow R_j:\ \text{sign flips}Ri→kRi: ×k; Ri→Ri+kRj: no changeR_i \to kR_i:\ \times k;\ R_i \to R_i + kR_j:\ \text{no change}
  6. 6Scalar multipledet⁡(kA)=kndet⁡A\det(kA) = k^n \det An=order of An = \text{order of } A
  7. 7Product, transpose, inversedet⁡(AB)=det⁡A⋅det⁡B\det(AB) = \det A \cdot \det Bdet⁡AT=det⁡A, det⁡A−1=1det⁡A\det A^T = \det A,\ \det A^{-1} = \dfrac{1}{\det A}
  8. 8Adjointdet⁡(adj A)=(det⁡A)n−1\det(\text{adj}\,A) = (\det A)^{n-1}A (adj A)=(det⁡A) I, n=order, I=identityA\,(\text{adj}\,A) = (\det A)\,I,\ n = \text{order},\ I = \text{identity}
  9. 9Singular matrixdet⁡A=0  ⟺  A−1 does not exist\det A = 0 \iff A^{-1} \text{ does not exist}det⁡(A2+kA)=det⁡A⋅det⁡(A+kI), I=identity\det(A^2 + kA) = \det A \cdot \det(A + kI),\ I = \text{identity}
  10. 10Rows 1, a, a²∣111abca2b2c2∣=(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)
  11. 11Cyclic rows∣abcbcacab∣=3abc−a3−b3−c3\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 3abc - a^3 - b^3 - c^3=−(a+b+c)(a2+b2+c2−ab−bc−ca)= -(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)

Test yourself

10 questions · 3 easy · 4 medium · 3 hard

Tap an option to answer. The first tap counts.

  1. Q1Easy

    If ∣k643∣=0\begin{vmatrix} k & 6 \\ 4 & 3 \end{vmatrix} = 0, what is kk?

    Correct answer88

    Why

    A 2×22 \times 2 determinant is ad−bcad - bc.

    Here that is 3k−6×4=3k−243k - 6 \times 4 = 3k - 24, which is 0 when 3k=243k = 24.

    So k=8k = 8.

  2. Q2Easy

    What is the value of ∣37−20−29005∣\begin{vmatrix} 3 & 7 & -2 \\ 0 & -2 & 9 \\ 0 & 0 & 5 \end{vmatrix}?

    Correct answer−30-30

    Why

    Everything below the diagonal is 0, so the determinant is the product of the diagonal.

    3×(−2)×5=−303 \times (-2) \times 5 = -30 (adding the diagonal gives the trap 6).

    So the value is −30-30.

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