Question ID: #178
For a $3\times3$ matrix $M$, let trace($M$) denote the sum of all the diagonal elements of $M$. Let $A$ be a $3\times3$ matrix such that $|A|=\frac{1}{2}$ and trace($A$)=3. If $B=adj(adj(2A))$, then the value of $|B| + \text{trace}(B)$ equals:
- (1) 56
- (2) 132
- (3) 174
- (4) 280
Solution:
Let $C = 2A$. We know $B = adj(adj(C))$.
For a matrix of order $n=3$, $adj(adj(C)) = |C|^{n-2}C$.
$$ B = |2A|^{3-2}(2A) = |2A|(2A) $$
Calculating $|2A|$:
$$ |2A| = 2^3 |A| = 8 \left(\frac{1}{2}\right) = 4 $$
$$ \therefore B = 4(2A) = 8A $$
Now, find $|B|$ and trace($B$):
$$ |B| = |8A| = 8^3 |A| = 512 \left(\frac{1}{2}\right) = 256 $$
$$ \text{trace}(B) = \text{trace}(8A) = 8 \times \text{trace}(A) = 8 \times 3 = 24 $$
$$ |B| + \text{trace}(B) = 256 + 24 = 280 $$
Ans. (4)
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