DeterminantHard
Question
Let A, B, C be three 3 × 3 matrices with real entries. If BA + BC + AC = I and det(A + B) = 0 then value of det(A + B + C - BAC) equals
Options
A.1
B.2
C.- 1
D.0
Solution
BA = I - AC - BC
⇒ BAC = C - AC2 - BC2
⇒ C - BAC = AC2 + BC2
⇒ C - BAC = (A + B)C2
⇒ A + B + C - BAC = A + B + (A + B) C2 = (A + B) (I + C2)
∴ det(A + B + C - BAC) = det((A + B) (I + C2)) = det(A + B) · det(I + C2) = 0
⇒ BAC = C - AC2 - BC2
⇒ C - BAC = AC2 + BC2
⇒ C - BAC = (A + B)C2
⇒ A + B + C - BAC = A + B + (A + B) C2 = (A + B) (I + C2)
∴ det(A + B + C - BAC) = det((A + B) (I + C2)) = det(A + B) · det(I + C2) = 0
Create a free account to view solution
View Solution FreeMore Determinant Questions
If DP = , then D1 + D2 + D3 + D4 + D5 is equal to -...Let A and B be two 2 × 2 matrix with real entries. If AB = O and tr(A) = tr(B) = 0 then...The value of is -...If A, B and C are the angles of a triangle and then the triangle ABC is-...For a non-zero real a, b and c = α abc, then the values of α is -...