Question

Using properties of determinants, prove that: ∣∣∣∣a2+2a2a+132a+1a+23111∣∣∣∣ = (a – 1)3.​

Answer 1

Answer:

∣

∣

∣

a

2

+2a

2a+1

3

2a+1

a+2

3

1

1

1

∣

∣

∣

∣

∣

∣

∣

∣

=(a−1)

3

Consider L.H.S=

∣

∣

∣

∣

∣

∣

∣

∣

a

2

+2a

2a+1

3

2a+1

a+2

3

1

1

1

∣

∣

∣

∣

∣

∣

∣

∣

apply R

1

→R

1

−R

3

and R

2

→R

2

−R

3

L.H.S=

∣

∣

∣

∣

∣

∣

∣

∣

a

2

+2a−3

2a−2

3

2a−2

a−1

3

0

0

1

∣

∣

∣

∣

∣

∣

∣

∣

take (a−1) common from R

1

and R

2

L.H.S=(a−1)

2

∣

∣

∣

∣

∣

∣

∣

∣

a+3

2

3

2

1

3

0

0

1

∣

∣

∣

∣

∣

∣

∣

∣

expand through C

3

and get

L.H.S=(a−1)

2

(a+3−4)

=(a−1)

3

= R.H.S

Hence proved

Hope it helps u