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Question

Let M=01a1233b1 and adj(M)=-11-18-62-53-1 where a and b are real numbers. Which of the following options is/are correct?


A

a+b=3

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B

detadjM2=81

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C

adjM-1+adjM-1=-M

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D

If Mαβγ=123 then α-β+γ=3

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Solution

The correct option is D

If Mαβγ=123 then α-β+γ=3


Explanation for correct option(s):

Step 1: Given information

M=01a1233b1

And adjoint of matrix M is,

adj(M)=-11-18-62-53-1

Step 2: Calculating M

We also know that,

|adj(M)|=M2=-11-18-62-53-1M2=-1(6-6)+1(-10+8)-1(24-30)M2=4M=±2

Step 3: Calculating (adj(M))-1

We also know that,

M(adjM)=M

(adj(M))-1=1adj(M)0-2-6-2-4-2-4-6-2T

adj(M-1)=140-2-4-2-4-6-6-2-2

Step 4: Equating the two equations

Now

M=M(adj(M))-1

01a1233b1=|M|40-2-4-2-4-6-6-2-2

Comparing Left Hand Side and Right Hand Side we get,

M=-2

Therefore,

a=2,b=1

Step 5: Calculating a+b

Therefore,

a+b=2+1

a+b=3

Thus, the value a+b is equal 3 .

Hence, option (A) is correct.

Step 6: Calculating (adjM)-1+adj-1=-M

We know that,

(adjM)-1+4djM-1=2(adjM)-1

(adjM)-1+adjM-1=2M|M|adjM-1=adj-M

(adjM)-1+adjM-1=-2M2=-M(adjM)-1=MM

Therefore,

(adjM)-2+adjM-1=-M

Therefore,

adjM)-1+adjM-1=-M

Therefore, option (C) is correct.

Step 7: Checking for Option (D)

If, Mαβγ=123 then α-β+γ=3

M=012123311

Now if,

Mαβγ=123

012123311αβγ=123

β+2γ=1-11α+2β+3γ=2-(2)3α+β+γ=3-(3)

By Solving the above three Equation we get,

α=1,β=-1 and γ=1

Therefore,

α-β+γ=1-(-1)+1α-β+γ=1+1+1α-β+γ=3

Therefore, α-β+γ=3.

Hence, Option (D) is correct.

Explanation for incorrect option:

Step 8: Calculating detadjM2

We know that,

det(adjM2=M22

detadj˙M2=M4

detadjM2=16|m|2=4

Therefore,

detadjM2 is not equal to 81.

Thus, option (B) is incorrect.

Hence, Option (A), (C) and (D) are the correct answers.


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