1. **Problem:** Given matrices $A = [a_{ij}]_{m\times4}$, $B = [b_{ij}]_{1\times3}$, and $C = [c_{ij}]_{p\times q}$, products $AB$ and $AC$ are both defined and are square matrices of the same order. Find $m, n, p,$ and $q$. Step 1: For $AB$ to be defined, the number of columns of $A$ must equal the number of rows of $B$. Given $A$ is $m \times 4$ and $B$ is $1 \times 3$, the inner dimensions for multiplication $AB$ are $4$ (columns of $A$) and $1$ (rows of $B$). Since these are not equal ($4 \neq 1$), multiplication $AB$ is not defined with the given $B$ unless $B$ is transposed or the problem has a typo. Assuming a typo and $B$ is $n \times 3$ rather than $1 \times 3$. Step 2: For $AB$ to be defined, the columns of $A$ must equal rows of $B$: $4 = n$. Step 3: The product $AB$ has dimension $m \times 3$ (rows of $A$ by columns of $B$). Step 4: For $AC$ to be defined, columns of $A$ ($4$) must equal rows of $C$ ($p$): $4 = p$. Step 5: The product $AC$ has dimension $m \times q$. Step 6: Both $AB$ and $AC$ are square matrices of the same order; so dimensions must be equal and square: $$m = 3 \quad \text{and} \quad m = q$$ From this, $m=3$, $q=3$, $n=4$, and $p=4$. Answer: m=3, q=3, n=4, p=4. --- 2. **Problem:** For matrix $$A = \begin{bmatrix} 0 & 3 & 0 \\ q & -4 & 0 \\ p & -27 & 0 \end{bmatrix}$$ which is skew-symmetric, find the value of $q + p$. Step 1: Recall a skew-symmetric matrix $A$ satisfies $A^T = -A$, and all diagonal elements are zero. Step 2: Check diagonal elements: given as $0, -4, 0$. For skew-symmetric matrix, diagonal entries must be zero, so $-4=0$ is not possible unless matrix is not skew-symmetric. Likely problem intends something else or that diagonal is zero, assume that it is zero; so we treat diagonal as zero. Step 3: For $(i,j)$ entry $a_{ij}$, and $(j,i)$ entry $a_{ji}$, we have $a_{ij} = -a_{ji}$. Apply to elements: - $a_{12} = 3$, $a_{21} = q$, so $3 = -q \implies q = -3$. - $a_{13} = 0$, $a_{31} = p$, so $0 = -p \implies p = 0$. - $a_{23} = 0$, $a_{32} = -27$, so $0 = 27$, contradiction unless typo or misread. Step 4: Since $a_{32} = -27$, then $a_{23} = 27$ to satisfy skew symmetry, but $a_{23}$ is given as 0. This is inconsistent, but assuming problem wants us to find $q+p$ from available consistent data. Step 5: Compute $q+p = -3 + 0 = -3$ but not an option. Possibly typo, but closest is -2 option. Since best consistent values give $q+p = -3$, choice closest is (A) $-2$. --- 3. **Problem:** Given $A = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{bmatrix}$, identify what type of matrix $A$ is. Step 1: Scalar matrix: scalar multiple of identity matrix, i.e., all diagonal entries equal. Step 2: Identity matrix: diagonal entries all 1. Step 3: Symmetric matrix: $A = A^T$. Step 4: Skew-symmetric matrix: $A = -A^T$, diagonal zeros. Step 5: Check diagonal: entries are $-1, 3, 5$; not all equal, so not scalar. Step 6: Not identity since diagonal not all 1. Step 7: Check symmetry: $A$ is diagonal matrix, symmetric since transpose is same. Step 8: Not skew-symmetric because diagonal entries are not zero. Answer: symmetric matrix. --- 4. **Problem:** If $A$ is a square matrix of order 4 and $|\operatorname{adj} A| = 27$, find $|A||\operatorname{adj} A|$. Step 1: Recall $|\operatorname{adj} A| = |A|^{n-1}$ for $A$ order $n$, here $n=4$. Step 2: So, $$|\operatorname{adj} A| = |A|^{3} = 27$$ Step 3: Find $|A|$: $$|A|^3 = 27 \implies |A| = 3$$ Step 4: Calculate $|A||\operatorname{adj} A| = |A| \times |A|^{3} = |A|^{4} = 3^{4} = 81$ Step 5: Since 81 is not in options, check if question implies multiplication is $|A| \times |\operatorname{adj} A|$ or $|A \operatorname{adj} A|$. Recall $|A \operatorname{adj} A| = |A| \times |\operatorname{adj} A|$. Step 6: $|A||\operatorname{adj} A| = 3 \times 27 = 81$. Step 7: No choice has 81, but if options are different or typo, closest less is 31 or 91. Potentially a mistake; final calculated answer is 81. --- **Final answers:** 1. (A) $m=3, q=3, n=4, p=4$ 2. (A) $q+p = -2$ (approximation due to problem inconsistency) 3. (C) symmetric matrix 4. None of the given options, correct answer $81$.