TN State Board 12th Maths Question Paper 2026

Given $B=A^{-1}A^T$. Then $$BB^T=(A^{-1}A^T)(A^{-1}A^T)^T=(A^{-1}A^T)(A(A^{-1})^T).$$ Since $A$ is non-singular, $(A^{-1})^T=(A^T)^{-1}$. Hence $$BB^T=A^{-1}A^TA(A^T)^{-1}.$$ Using the given condition $AA^T=A^TA$, we get $A^TA=A A^T$, so $$BB^T=A^{-1}(AA^T)(A^T)^{-1}=I_3.$$ Therefore the correct answer is $I_3$. So, (a) is correct. The other options are not identity matrices and do not follow from the given relation.

For a system $AX=B$, if $\rho(A)=\rho([A\,|\,B])$, then the system is consistent. It may have one solution or infinitely many solutions depending on whether this common rank equals the number of unknowns. Since only consistency is guaranteed here, the correct option is (d) consistent.

If $\dfrac{1}{z}\in \mathbb{R}^+$, then $\dfrac{1}{z}$ is a positive real number. Hence $z$ must also be a positive real number. Since $z\in \mathbb{C}\setminus \mathbb{R}$ is given, this is inconsistent with the condition. Among the given choices, the positive real option is $2$. Therefore the intended correct choice is (a) $2$.

The four values are the fourth roots of unity obtained from the complex number $\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}$. The product of all the values is the constant term in the equation whose roots are these four values, and for the fourth roots of unity the product is $$1.$$ Hence the correct answer is (a) $1$.

If $f(x)$ has degree $m$ and $g(x)$ has degree $n$, then the composite function $$h(x)=(f\circ g)(x)=f(g(x))$$ has degree equal to the product of the degrees: $$\deg(h)=mn.$$ So the correct answer is (b) $mn$.