Mathematics 20 flashcards ~10 min

Vectors and Matrices

Twenty flashcards covering vector and matrix algebra. Includes vector magnitude, addition, the dot and cross products, unit vectors, matrix addition and multiplication, determinants, inverse matrices, and solving linear systems. Suited to A-Level Further Maths...

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Twenty flashcards covering vector and matrix algebra. Includes vector magnitude, addition, the dot and cross products, unit vectors, matrix addition and multiplication, determinants, inverse matrices, and solving linear systems. Suited to A-Level Further Maths, AP Precalculus/Calculus and introductory linear algebra.

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A quantity with both magnitude and direction (e.g. velocity, force). A scalar has magnitude only (e.g. speed, mass).
∣ 𝑣 ∣ = π‘₯ 2 + 𝑦 2 ∣v∣= x 2 +y 2 ​ β€” an application of the Pythagorean theorem.
Add or subtract corresponding components: ( π‘Ž , 𝑏 ) + ( 𝑐 , 𝑑 ) = ( π‘Ž + 𝑐 ,   𝑏 + 𝑑 ) (a,b)+(c,d)=(a+c,b+d), and similarly for subtraction.
A vector with magnitude 1, found by dividing a vector by its own magnitude: 𝑣 ^ = 𝑣 ∣ 𝑣 ∣ v ^ = ∣v∣ v ​ .
π‘Ž β‹… 𝑏 = π‘Ž 1 𝑏 1 + π‘Ž 2 𝑏 2 + π‘Ž 3 𝑏 3 = ∣ π‘Ž ∣ ∣ 𝑏 ∣ cos ⁑ πœƒ aβ‹…b=a 1 ​ b 1 ​ +a 2 ​ b 2 ​ +a 3 ​ b 3 ​ =∣a∣∣b∣cosΞΈ.
The two vectors are perpendicular (orthogonal) to each other, since cos ⁑ 90 ∘ = 0 cos90 ∘ =0.
∣aΓ—b∣=∣a∣∣b∣sinΞΈ, giving a vector perpendicular to both π‘Ž a and 𝑏 b (right-hand rule for direction).
Its magnitude equals the area of the parallelogram formed by the two vectors; the result is only defined in three dimensions.
π‘Ÿ = π‘Ž + 𝑑 𝑑 r=a+td, where π‘Ž a is the position vector of a known point, 𝑑 d is the direction vector, and 𝑑 t is a scalar parameter.
What is the order (dimensions) of a matrix?
When the number of columns in the first matrix equals the number of rows in the second matrix.
Generally no β€” 𝐴 𝐡 β‰  𝐡 𝐴 AB ξ€  =BA in most cases, even when both products are defined.
A square matrix with 1s on the diagonal and 0s elsewhere, denoted 𝐼 I, which leaves any compatible matrix unchanged under multiplication: 𝐴 𝐼 = 𝐼 𝐴 = 𝐴 AI=IA=A.
For ( π‘Ž 𝑏 𝑐 𝑑 ) ( a c ​ b d ​ ), det ⁑ = π‘Ž 𝑑 βˆ’ 𝑏 𝑐 det=adβˆ’bc.
The matrix is singular β€” it has no inverse, and its rows/columns are linearly dependent.
𝐴 βˆ’ 1 = 1 π‘Ž 𝑑 βˆ’ 𝑏 𝑐 ( 𝑑 βˆ’ 𝑏 βˆ’ 𝑐 π‘Ž ) A βˆ’1 = adβˆ’bc 1 ​ ( d βˆ’c ​ βˆ’b a ​ ), provided π‘Ž 𝑑 βˆ’ 𝑏 𝑐 β‰  0 adβˆ’bc ξ€  =0.
Write the system as 𝐴 π‘₯ = 𝑏 Ax=b, then solve π‘₯ = 𝐴 βˆ’ 1 𝑏 x=A βˆ’1 b, provided 𝐴 A is invertible.
The matrix obtained by swapping rows and columns, denoted 𝐴 𝑇 A T , so that element π‘Ž 𝑖 𝑗 a ij ​ becomes π‘Ž 𝑗 𝑖 a ji ​ .
One vector is a scalar multiple of the other: 𝑏 = π‘˜ π‘Ž b=ka for some scalar π‘˜ k, meaning they point in the same or exactly opposite directions.
cos ⁑ πœƒ = π‘Ž β‹… 𝑏 ∣ π‘Ž ∣ ∣ 𝑏 ∣ cosΞΈ= ∣a∣∣b∣ aβ‹…b ​ , rearranged from the dot product formula.