If A is a 3 times 3 matrix such that |A|=27 a | vedclass

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(A) Given $\operatorname{Adj}(A)=k A^T$. Taking the determinant on both sides, we get $|\operatorname{Adj}(A)|=|k A^T|$. We know that for a square mat

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$5$

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(A) Given $\operatorname{Adj}(A)=k A^T$. Taking the determinant on both sides, we get $|\operatorname{Adj}(A)|=|k A^T|$. We know that for a square matrix $A$ of order $n$, $|\operatorname{Adj}(A)|=|A|^{n-1}$, $|k A|=k^n|A|$, and $|A^T|=|A|$. Here, $n=3$, so $|\operatorname{Adj}(A)|=|A|^{3-1}=|A|^2$. Also, $|k A^T|=k^3|A^T|=k^3|A|$. Equating these, we have $|A|^2=k^3|A|$. Since $|A|=27 
eq 0$, we can divide by $|A|$ to get $k^3=|A|=27$. Thus, $k=3$. Finally, substituting $k=3$ into the expression, $k^2-3 k+5 = 3^2-3(3)+5 = 9-9+5 = 5$.

If $A$ is a $3 \times 3$ matrix such that $|A|=27$ and $\operatorname{Adj}(A)=k A^T$, then find the value of $k^2-3 k+5$.

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