Memory Based GATE DA 2024 | Question: 4

Answer 1

By linear dependency (R1 + R2 → R3) or solving simply, we have |M| = 0

We know |A+B| <= |A| + |B|

$\therefore$ |M² + 12M| <= |M²| + |12M|

RHS of which is : 

|M|² + 12³|M| (By properties of determinants)

= 0 + 0 = 0

Answer 2

Det(M^2 + 12M) = Det(M) Det(M + 12I)(since Det(AB) = DET(A) DET(B)))

DET(M) = 0 { since rows are Linearly dependent}
hence Det(M^2 + 12M) = 0

Comments

It is Det(M$^2$ + 12M)

You can't apply |A+B| = |A|.|B| in 

Det(M^2 + 12M) = Det(M) Det(M + 12I)

directly

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at least see my explanation,
M^2 + 12M can be written as M(M + 12I)
Det(M(M + 12I)) = Det(M) * DET(M + 12 I)

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Got it now