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
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