Question 4, Exercise 2.3

Solutions of Question 4 of Exercise 2.3 of Unit 02: Matrices and Determinants. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Find the value of λ, so that the given matrix is singular [λ13218031].

Solution.

A=[λ13218031]|A|=λ(23)12+36=23λ2+18=23λ+16 A is singular, Then |A|=023λ+16=023λ=16λ=1623 Thus, λ=1623.

Find the value of λ, so that the given matrix is singular [λ20213λ21].

Solution.

A=[λ20213λ21]|A|=λ(5)2(23λ)=5λ2(23λ)=5λ4+6λ=λ4 We know that the matrix A is singular, then: |A|=0λ4=0λ=4

Find the value of λ, so that the given matrix is singular [λi1213312].

Solution.

Do yourself

Find the value of λ, so that the given matrix is singular [2+i162λ1302].

Solution.

A=[2+i162λ1302]|A|=(2+i)2λ11+6(3λ)=(2+i)2λ118λ=4λ+2iλ118λ=(4λ18λ+2iλ1)=14λ+2iλ1=(14+2i)λ1

A

(14+2i)λ1=0(14+2i)λ=1λ=114+2iλ=114+2i142i142i=142i(14)2(2i)2=142i196+4=142i200=142i200=7100i100 Hence λ=7100i100