Let A and B be two square matrices of order 3 | vedclass

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(A) Given $|A|=3$ and $|B|=2$ for matrices of order $n=3$. We know that $|A^T| = |A| = 3$,$|A^T A| = |A^T||A| = |A|^2 = 3^2 = 9$. Also,$|\operatorname

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$64$

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(A) Given $|A|=3$ and $|B|=2$ for matrices of order $n=3$. We know that $|A^T| = |A| = 3$,$|A^T A| = |A^T||A| = |A|^2 = 3^2 = 9$. Also,$|\operatorname{adj}(kA)| = |kA|^{n-1} = (k^n |A|)^{n-1} = (k^3 \cdot 3)^2 = k^6 \cdot 9$. For $k=2$,$|\operatorname{adj}(2A)| = 2^6 \cdot 3^2 = 64 \cdot 9 = 576$. For $k=4$,$|\operatorname{adj}(4B)| = (4^3 |B|)^2 = (64 \cdot 2)^2 = 128^2 = 16384$. $|\operatorname{adj}(AB)| = |AB|^{n-1} = (|A||B|)^{3-1} = (3 \cdot 2)^2 = 6^2 = 36$. Now,the expression is: $|A^T A| \cdot |(\operatorname{adj}(2A))^{-1}| \cdot |\operatorname{adj}(4B)| \cdot |(\operatorname{adj}(AB))^{-1}| \cdot |AA^T|$ $= 9 \cdot \frac{1}{576} \cdot 16384 \cdot \frac{1}{36} \cdot 9$ $= \frac{81 \cdot 16384}{576 \cdot 36} = \frac{81 \cdot 16384}{20736} = \frac{1327104}{20736} = 64$.

Column $I$	Column $II$
$(A)$ $a+b+c \neq 0$ and $a^2+b^2+c^2=ab+bc+ca$	$(p)$ The equations represent planes meeting only at a single point.
$(B)$ $a+b+c=0$ and $a^2+b^2+c^2 \neq ab+bc+ca$	$(q)$ The equations represent the line $x=y=z$.
$(C)$ $a+b+c \neq 0$ and $a^2+b^2+c^2 \neq ab+bc+ca$	$(r)$ The equations represent identical planes.
$(D)$ $a+b+c=0$ and $a^2+b^2+c^2=ab+bc+ca$	$(s)$ The equations represent the whole of the three-dimensional space.

Let $A$ and $B$ be two square matrices of order $3$ such that $|A|=3$ and $|B|=2$. Then $\left|A^{T} A(\operatorname{adj}(2A))^{-1}(\operatorname{adj}(4B))(\operatorname{adj}(AB))^{-1} AA^{T}\right|$ is equal to:

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