1. Chapter 3 Class 12 Matrices (Term 1)
2. Concept wise
3. Solving Equation

Transcript

Ex 3.2, 18 If A =[β 8(0&βtan πΌ/2 " " @tan πΌ/2 " " &0)] and I is the identity matrix of order 2, Show that I + A = ( I β A)[β 8(cosβ‘πΌ&βsinβ‘πΌ@sinβ‘πΌ&cosβ‘πΌ )] Given I the identity matrix of order 2 i.e. I = [β 8(1&0@0&1)] Taking L.H.S. I + A = [β 8(1&0@0&1)] + [β 8(0&βtan πΌ/2 " " @tan πΌ/2 " " &0)] = [β 8(1+0&0βtan πΌ/2 " " @0+tan πΌ/2 " " &1+0)] = [β 8(1&βtan πΌ/2 " " @tan πΌ/2 " " &1)] Taking R.H.S (I β A) [β 8(cosβ‘πΌ&βsinβ‘πΌ@sinβ‘πΌ&cosβ‘πΌ )] = (" " [β 8(1&0@0&1)]β[β 8(0&βtan πΌ/2 " " @tan πΌ/2 " " &0)]) [β 8(cosβ‘πΌ&βsinβ‘πΌ@sinβ‘πΌ&cosβ‘πΌ )] = [β 8(1β0&0+tan πΌ/2 " " @0βtan πΌ/2 " " &1)][β 8(cosβ‘πΌ&βsinβ‘πΌ@sinβ‘πΌ&cosβ‘πΌ )] = [β 8(1&tan Ξ±/2 " " @β tan Ξ±/2 " " &1)][β 8(cosβ‘πΌ&βsinβ‘πΌ@sinβ‘πΌ&cosβ‘πΌ )] = [β 8(1(πππ β‘πΌ )+π‘ππ πΌ/2 (π ππβ‘πΌ)&1(γβπ ππγβ‘πΌ )+π‘ππ πΌ/2 (π ππβ‘πΌ)@βπ‘ππ πΌ/2 (πππ β‘πΌ )+1(π ππβ‘πΌ) &βπ‘ππ πΌ/2 (γβπ ππγβ‘πΌ )+1(π ππβ‘πΌ))] = [β 8(1((1 β π‘ππ2 πΌ/2)/(1 + π‘ππ2 πΌ/2))"+ " π‘ππ πΌ/2 ((2 tanβ‘γ πΌ/2γ)/(1 + π‘ππ2 πΌ/2))" " &1((β2 tanβ‘γ πΌ/2γ)/(1 + π‘ππ2 ( πΌ)/2))"+(" π‘ππ πΌ/2) ((1 β π‘ππ2 ( πΌ)/2)/(1 + π‘ππ2 ( πΌ)/2))" " @βπ‘ππ πΌ/2 ((1 β π‘ππ2 ( πΌ)/2)/(1 + π‘ππ2 ( πΌ)/2))" +1" ((2 tanβ‘γ πΌ/2γ)/(1 + π‘ππ2 πΌ/2))&βπ‘ππ πΌ/2 ((β2 γtan γβ‘γπΌ/2γ)/(1 + π‘ππ2 πΌ/2))" +1" ((1 β π‘ππ2 ( πΌ)/2)/(1 + π‘ππ2 πΌ/2)) )] We know that cos 2ΞΈ = (1 β π‘ππ2π)/(1 + π‘ππ2π) & sin 2ΞΈ = (2 tanβ‘π)/(1 + π‘ππ2π) Replacing ΞΈ with π/2 So, cos ΞΈ = (1 β π‘ππ2 π/2)/(1 + π‘ππ2 π/2) & sin ΞΈ = (2 π‘ππβ‘γ π/2γ)/(1 + π‘ππ2 π/2) = [β 8((1 β π‘ππ2 πΌ/2)/(1 + π‘ππ2 πΌ/2) " +" (2π‘ππ2 πΌ/2)/(1 + π‘ππ2 πΌ/2) " " &(β2 tanβ‘γ πΌ/2γ " +" tanβ‘γ πΌ/2γ β π‘ππ3 πΌ/2)/(1 + π‘ππ2 πΌ/2) " " @(βγtan γβ‘γπΌ/2γ (1 β π‘ππ2 πΌ/2))/(1 + π‘ππ2 πΌ/2) " +" (2 tanβ‘γ πΌ/2γ)/(1 + π‘ππ2 πΌ/2)&(2π‘ππ2 πΌ/2)/(1+π‘ππ2 πΌ/2)+ " " (1 β π‘ππ2 πΌ/2)/(1 + π‘ππ2 πΌ/2))] = [β 8((1 β π‘ππ2 πΌ/2 + 2π‘ππ2 πΌ/2)/(1 + π‘ππ2 πΌ/2) " " &(β2 γtan γβ‘γ πΌ/2γ " + " tanβ‘γ πΌ/2γ β π‘ππ3 πΌ/2)/(1 + π‘ππ2 πΌ/2) " " @(βγtan γβ‘γπΌ/2γ + π‘ππ3 πΌ/2 +2 tanβ‘γ πΌ/2γ)/(1 + π‘ππ2 πΌ/2) " " &" " (2π‘ππ2 πΌ/2 + 1 β π‘ππ2 πΌ/2)/(1 + π‘ππ2 πΌ/2))] = [β 8((1 + π‘ππ2 πΌ/2 )/(1 + π‘ππ2 πΌ/2) " " &(βγtan γβ‘γπΌ/2γ " " (1 + π‘ππ2 πΌ/2) )/(1 + π‘ππ2 πΌ/2) " " @(γtan γβ‘γπΌ/2γ (1 + π‘ππ2 πΌ/2))/(1 + π‘ππ2 πΌ/2) " " &" " (1 + π‘ππ2 πΌ/2 )/(1 + π‘ππ2 ( πΌ)/2))] = [β 8(1&βtan πΌ/2 " " @tan πΌ/2 " " &1)] = R.H.S. Hence proved

Solving Equation

Chapter 3 Class 12 Matrices (Term 1)
Concept wise