Let A=[[∣−11−1101001]]∣ satisfy
A2+α(adj(adj(A)))+β(adj(A)(adj(adj(A))))=[[∣2−22−20−100−1]]∣ for some α,β∈R.
Then (α−β)2 is equal to ____
Answer: 4
Solutions · 1
Official solution
We need to find α and β from
A2+α(adj(adj(A)))+β(adj(A)adj(adj(A)))=[[∣2−22−20−100−1]]∣.
Let us solve step by step.
1. Given matrix
A=[[∣−11−1101001]]∣
2. Find det(A)
Since the third row is (0,0,1), expand along the third row:
det(A)=1⋅∣−1110∣=(−1)(0)−(1)(1)=−1.
So,
det(A)=−1=0.
Hence A is invertible.
3. Use NCERT property of adjoint of adjoint
For an n×n invertible matrix,
adj(adj(A))=(detA)n−2A.
Here n=3, so
adj(adj(A))=(detA)1A=(−1)A=−A.
Thus,
adj(adj(A))=−A.
Also, for any invertible matrix,
Aadj(A)=det(A)I.
So,
Aadj(A)=−I.
Multiplying both sides by A−1 is not necessary; instead use
adj(A)=det(A)A−1=−A−1 if needed.
Now,
adj(A)adj(adj(A))=adj(A)(−A)=−(adj(A)A).
And since
adj(A)A=det(A)I=−I,
we get
adj(A)adj(adj(A))=−(−I)=I.
So the given equation becomes
A2+α(−A)+βI=[[∣2−22−20−100−1]]∣.
That is,
A2−αA+βI=[[∣2−22−20−100−1]]∣.
4. Compute A2A=[[∣−11−1101001]]∣
Now multiply A by itself:
A2=[[∣−11−1101001]]∣[[∣−11−1101001]]∣.
Entrywise calculation:
First row:
(1,1):(−1)(−1)+(1)(1)+(−1)(0)=2(1,2):(−1)(1)+(1)(0)+(−1)(0)=−1(1,3):(−1)(−1)+(1)(1)+(−1)(1)=1
Second row:
(2,1):(1)(−1)+(0)(1)+(1)(0)=−1(2,2):(1)(1)+(0)(0)+(1)(0)=1(2,3):(1)(−1)+(0)(1)+(1)(1)=0
Third row:
(3,1)=0,(3,2)=0,(3,3)=1
Hence
A2=[[∣2−11−110001]]∣.
5. Form A2−αA+βI
First,
−αA=[[∣α−αα−α0−α00−α]]∣
and
βI=[[∣β000β000β]]∣.
So,
A2−αA+βI=[[∣2+α+β−1−α1+α−1−α1+β−α001−α+β]]∣.
This must equal
[[∣2−22−20−100−1]]∣.
6. Compare corresponding entries
From the (1,2) entry:
−1−α=−2⇒α=1.
From the (2,2) entry:
1+β=0⇒β=−1.
These also satisfy the other entries:
(1,1):
2+α+β=2+1−1=2(1,3):
1+α=2(2,1):
−1−α=−2(2,3):
−α=−1(3,3):
1−α+β=1−1−1=−1
All correct.
Thus,
α=1,β=−1.
7. Find (α−β)2α−β=1−(−1)=2
Therefore,
(α−β)2=22=4.
So the required answer is
4.