Let tr(X) and adj(X) denote the trace and adjoint of a square matrix X. If M is a non-singular square matrix of order 3 such that M−1=⎡⎢⎣345453534⎤⎥⎦, then the value of |15tr(adj(M))| is
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Solution
M−1=⎡⎢⎣345453534⎤⎥⎦ ⇒|M−1|=3(20−9)−4(16−15)+5(12−25) ⇒|M−1|=−36
and tr(M−1)=3+5+4=12
We know that M−1=adj(M)|M| ⇒adj(M)=|M|⋅M−1 ⇒tr(adj(M))=|M|tr(M−1) ⇒tr(adj(M))=−136×12=−13 |15tr(adj(M))|=∣∣∣15×(−13)∣∣∣=5