Question:

I) Let $A,B$ be square matrices of the same order, $O$ be the null matrix, and $I$ be the identity matrix. The statement is: If $AB=O$, then $(\det A)^2+(\det B)^2=0$. Consider the matrices $A = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. First, verify the premise $AB=O$: $$AB = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = O$$ The premise is satisfied. Next, evaluate the conclusion $(\det A)^2+(\det B)^2=0$: The determinant of $A$ is $\det A = 0$. The determinant of $B$ is $\det B = 1$. Then, $(\det A)^2 + (\det B)^2 = (0)^2 + (1)^2 = 0 + 1 = 1$. Since $1 \ne 0$, the conclusion $(\det A)^2+(\det B)^2=0$ is false for this choice of $A$ and $B$. Therefore, statement I is false. II) The statement is: If $l_i,m_i,n_i$ $(i=1,2,3)$ are direction cosines of 3 mutually perpendicular vectors, then $A=\begin{pmatrix}l_1&m_1&n_1\\l_2&m_2&n_2\\l_3&m_3&n_3\end{pmatrix}$ is orthogonal. Let the three vectors be $\vec{v_1} = (l_1, m_1, n_1)$, $\vec{v_2} = (l_2, m_2, n_2)$, and $\vec{v_3} = (l_3, m_3, n_3)$. These vectors form the rows of matrix $A$. Since $l_i, m_i, n_i$ are direction cosines, each vector is a unit vector. This means the magnitude squared of each vector is 1: $$\vec{v_i} \cdot \vec{v_i} = l_i^2 + m_i^2 + n_i^2 = 1 \quad \text{for } i=1,2,3$$ The vectors are mutually perpendicular, which means their dot products are zero for distinct vectors: $$\vec{v_i} \cdot \vec{v_j} = 0 \quad \text{for } i \ne j$$ A matrix $A$ is orthogonal if $A A^T = I$. Let's compute $A A^T$: $$A A^T = \begin{pmatrix}l_1&m_1&n_1\\l_2&m_2&n_2\\l_3&m_3&n_3\end{pmatrix} \begin{pmatrix}l_1&l_2&l_3\\m_1&m_2&m_3\\n_1&n_2&n_3\end{pmatrix}$$ The element in the $i$-th row and $j$-th column of $A A^T$ is the dot product of the $i$-th row vector of $A$ and the $j$-th row vector of $A$. That is, $(A A^T)_{ij} = \vec{v_i} \cdot \vec{v_j}$. Using the properties derived from the direction cosines and mutual perpendicularity: For the diagonal elements ($i=j$): $$(A A^T)_{ii} = \vec{v_i} \cdot \vec{v_i} = l_i^2 + m_i^2 + n_i^2 = 1$$ For the off-diagonal elements ($i \ne j$): $$(A A^T)_{ij} = \vec{v_i} \cdot \vec{v_j} = 0$$ Therefore, $A A^T = \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix} = I$. Hence, $A$ is an orthogonal matrix. Statement II is true. III) The statement is: For two distinct lines with shortest distance $d\ne 0$, a point $A$ on line 1 and $B$ on line 2 with $AB=d$: $A$ and $B$ are unique. Consider two distinct parallel lines. For example, let line 1 be the x-axis, $L_1: \vec{r_1}(t) = (t,0,0)$, and line 2 be the line $L_2: \vec{r_2}(s) = (s,1,0)$. These lines are distinct and parallel. The shortest distance between them is $d=1$, which is non-zero. Let $A$ be a point on $L_1$ and $B$ be a point on $L_2$. For the distance $AB$ to be the shortest distance $d=1$, the vector $\vec{AB}$ must be perpendicular to the direction vector of the lines, which is $(1,0,0)$. Let $A = (t_0, 0, 0)$ for some $t_0 \in \mathbb{R}$. Let $B = (s_0, 1, 0)$ for some $s_0 \in \mathbb{R}$. The vector $\vec{AB} = (s_0 - t_0, 1, 0)$. For $\vec{AB}$ to be perpendicular to the direction vector $(1,0,0)$, their dot product must be zero: $$(s_0 - t_0)(1) + (1)(0) + (0)(0) = 0 \implies s_0 - t_0 = 0 \implies s_0 = t_0$$ Thus, for any real number $t_0$, the points $A = (t_0, 0, 0)$ and $B = (t_0, 1, 0)$ satisfy the condition that $AB=1$ (the shortest distance) and $\vec{AB}$ is perpendicular to both lines. Since $t_0$ can be any real number, there are infinitely many such pairs of points $(A,B)$. For instance, $A=(0,0,0), B=(0,1,0)$ is one pair, and $A=(1,0,0), B=(1,1,0)$ is another. Therefore, $A$ and $B$ are not unique. The uniqueness only holds for skew lines, which are not specified here. Statement III is false.