Evalute determinant of a 3x3 matrix
Evalute determinant of a 3x3 matrix
Last updated at April 16, 2024 by Teachoo
Misc 1 Prove that determinant |β 8(π₯&π ππβ‘π&πππ β‘π@βπ ππβ‘π&βπ₯&1@πππ β‘π&1&π₯)| is independent of ΞΈ. Let β = |β 8(π₯&π ππβ‘π&πππ β‘π@βπ ππβ‘π&βπ₯&1@πππ β‘π&1&π₯)| β = x |β 8(βπ₯&1@1&π₯)| β sin ΞΈ |β 8(βsinβ‘ΞΈ&1@cosβ‘ΞΈ&π₯)| + cos ΞΈ |β 8(βsinβ‘ΞΈ&βπ₯@cosβ‘ΞΈ&1)| = x ( βx2 β 1) β sin ΞΈ ( βxsin ΞΈ β cos ΞΈ) + cos ΞΈ (βsin ΞΈ + x cos ΞΈ) = βx3 β x + x sinβ‘γ2 ΞΈγ + π¬π’π§β‘π cos ΞΈ β sin ΞΈ cos ΞΈ + x cos2 ΞΈ = βx3 β x + x sin2 ΞΈ + x cos2 ΞΈ = βx3 β x + x (sin2 ΞΈ + cos2 ΞΈ) = βx3 β x + x (1) = βx3 Hence β = βx3 Which has no ΞΈ term Thus, the determinant is independent of ΞΈ Hence Proved