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05.05 · Modern Maths

Matrices

Elite lesson10 formulas10 questions

On the syllabus of

Elite lesson1 h 27 minHinglish

Formulas

At a glance 10 formulas

  1. 1Order of a product(m×n)(n×p)→m×p(m \times n)(n \times p) \to m \times pAB needs columns of A=rows of BAB \text{ needs columns of } A = \text{rows of } B
  2. 2Entry of a product(AB)ij=ai1b1j+ai2b2j+⋯(AB)_{ij} = a_{i1}b_{1j} + a_{i2}b_{2j} + \cdotsrow i of A times column j of B\text{row } i \text{ of } A \text{ times column } j \text{ of } B
  3. 3Order mattersAB≠BA in generalAB \ne BA \text{ in general}(A+B)2=A2+AB+BA+B2(A + B)^2 = A^2 + AB + BA + B^2
  4. 4Transpose of a product(AB)T=BTAT(AB)^T = B^TA^TAT turns rows into columns; (AT)T=AA^T \text{ turns rows into columns};\ (A^T)^T = A
  5. 5Symmetric and skewAT=A (symmetric), AT=−A (skew-symmetric)A^T = A \text{ (symmetric)},\ A^T = -A \text{ (skew-symmetric)}skew-symmetric: every diagonal entry is 0\text{skew-symmetric: every diagonal entry is } 0
  6. 6Inverse of a 2×2[abcd]−1=1ad−bc[d−b−ca]\begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} = \dfrac{1}{ad - bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}none if ad−bc=0; (AB)−1=B−1A−1\text{none if } ad - bc = 0;\ (AB)^{-1} = B^{-1}A^{-1}
  7. 7Cayley-Hamilton (2×2)A2=(a+d)A−(ad−bc)IA^2 = (a + d)A - (ad - bc)IA=[abcd], I=identity matrixA = \begin{bmatrix} a & b \\ c & d \end{bmatrix},\ I = \text{identity matrix}
  8. 8Special squaresA2=I⇒Aeven=I, Aodd=AA^2 = I \Rightarrow A^{\text{even}} = I,\ A^{\text{odd}} = AA2=A⇒An=A for every n; I=identityA^2 = A \Rightarrow A^n = A \text{ for every } n;\ I = \text{identity}
  9. 9Square is zeroA2=O⇒(I+A)n=I+nAA^2 = O \Rightarrow (I + A)^n = I + nAO=zero matrix, I=identityO = \text{zero matrix},\ I = \text{identity}
  10. 10Entry sum of ABsum of entries of AB=c1r1+c2r2+⋯\text{sum of entries of } AB = c_1r_1 + c_2r_2 + \cdotsck=column k sum of A, rk=row k sum of Bc_k = \text{column } k \text{ sum of } A,\ r_k = \text{row } k \text{ sum of } B

Test yourself

10 questions · 3 easy · 4 medium · 3 hard

Tap an option to answer. The first tap counts.

  1. Q1Easy

    Matrix AA has order 3×23 \times 2 and matrix BB has order 2×42 \times 4. How many entries does the product ABAB have?

    Correct answer12

    Why

    Use the order rule: (m×n)(n×p)(m \times n)(n \times p) gives m×pm \times p.

    The inner 2s match and drop out, leaving a 3×43 \times 4 matrix.

    So ABAB has 3×4=123 \times 4 = 12 entries.

  2. Q2Easy

    For a campus analytics dashboard, two matrices are defined as A=[2−134]A=\begin{bmatrix}2&-1\\3&4\end{bmatrix} and B=[152−3]B=\begin{bmatrix}1&5\\2&-3\end{bmatrix}. If C=ABC=AB, what is the sum of all the elements of matrix CC?

    Correct answer27

    Why

    Skip the full product: the entries of ABAB total (column 1 sum of AA)(row 1 sum of BB) plus (column 2 sum of AA)(row 2 sum of BB).

    That is (2+3)(1+5)+(−1+4)(2−3)=30−3(2 + 3)(1 + 5) + (-1 + 4)(2 - 3) = 30 - 3.

    So the entries of CC add up to 27.

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