Question:

Step 1: Identify the properties of matrix $A$ from the given condition. We are given that $a_{ij} + a_{ji} = 0$ for all $i, j$. This means $a_{ij} = -a_{ji}$, which is the defining property of a skew-symmetric matrix. Therefore, $A$ is a skew-symmetric matrix, i.e., $A^T = -A$. Step 2: Identify the properties of matrix $B$ from the given condition. We are given that $b_{ij} - b_{ji} = 0$ for all $i, j$. This means $b_{ij} = b_{ji}$, which is the defining property of a symmetric matrix. Therefore, $B$ is a symmetric matrix, i.e., $B^T = B$. Step 3: Determine the determinant of the skew-symmetric matrix $A$. For a skew-symmetric matrix $A$ of odd order (in this case $3 \times 3$), we can establish that $\det(A) = 0$. This is because: $$\det(A) = \det(A^T) = \det(-A) = (-1)^3 \det(A) = -\det(A)$$ The first equality holds since the determinant of a matrix equals the determinant of its transpose. The second equality uses the fact that $A^T = -A$ for a skew-symmetric matrix. The third equality applies the property that $\det(cM) = c^n \det(M)$ for an $n \times n$ matrix. From $\det(A) = -\det(A)$, we get $2\det(A) = 0$, which implies $\det(A) = 0$. Step 4: Calculate the determinant of $A^4 B^3$. Using the multiplicative property of determinants: $$\det(A^4 B^3) = \det(A^4) \cdot \det(B^3) = (\det A)^4 \cdot (\det B)^3$$ Since $\det(A) = 0$: $$\det(A^4 B^3) = (0)^4 \cdot (\det B)^3 = 0$$ Step 5: Conclude that $A^4 B^3$ is singular. A matrix is singular if and only if its determinant equals zero. Since $\det(A^4 B^3) = 0$, the matrix $A^4 B^3$ is singular. **Final Answer:** $A^4 B^3$ is **singular** (Option 1).