Let A be a matrix such that A∙[■(1&2@0&3)] is a scalar matrix and |3A|=108. ThenA^2 equals

Let A be a matrix such that A∙[■(1&2@0&3)] is a scalar matrix and |3A|=108. ThenA^2 equals
(a) [■(36&-32@0&4)] (b) [■(4&0@-32&36)]
(c) [■(4&-32@0&36)] (d) [■(36&0@-32&4)]
Ans: a
Sol.
A∙[■(1&2@0&3)] is a scalar matrix and |3A|=108
Let the scalar matrix be [■(k&0@0&k)]
A∙[■(1&2@0&3)]=[■(k&0@0&k)]
A=[■(k&0@0&k)][■(1&2@0&3)]
⇒A=[■(k&0@0&k)][■(1&-2/3@0&1/3)]
⇒A=[■(k&-2k/3@0&k/3)]
|A|=k^2/3
|3A|=108
As we know kA|=k^n |A
|3A|=3^2 k^2/3
⇒k^2=36⇒k=±6
So, putting the value of k in the matrix A
A=[■(6&(-2/3)(6)@0&6/3)]
A=[■(6&-4@0&2)]
So, A^2=[■(6&-4@0&2)][■(6&-4@0&2)]
⇒A^2=[■(36&-32@0&4)]

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