Let's evaluate the determinant of the given \( 4 \times 4 \) matrix \( A \) using cofactor expansion along the **first row** (since it contains zeros, which simplifies calculations). The matrix is: \[ A = \begin{pmatrix} 5 & 0 & 1 & 0 \\ 2 & 2 & -1 & 0 \\ 1 & 5 & 2 & 3 \\ 3 & 2 & 5 & 3 \\ \end{pmatrix} \] ### Step 1: Write the Cofactor Expansion for Row 1 \[ \det(A) = 5 \cdot C_{11} + 0 \cdot C_{12} + 1 \cdot C_{13} + 0 \cdot C_{14} \] Terms with 0 coefficients can be ignored: \[ \det(A) = 5 \cdot C_{11} + 1 \cdot C_{13} \] Where \( C_{ij} \) is the cofactor of the element in row \( i \), column \( j \). --- ### Step 2: Calculate \( C_{11} \) and \( C_{13} \) #### 2.1: \( C_{11} \) (Remove row 1 and column 1) \[ M_{11} = \begin{pmatrix} 2 & -1 & 0 \\ 5 & 2 & 3 \\ 2 & 5 & 3 \\ \end{pmatrix} \] \[ C_{11} = (-1)^{1+1} \cdot \det(M_{11}) = 1 \cdot \det(M_{11}) \] #### 2.2: \( C_{13} \) (Remove row 1 and column 3) \[ M_{13} = \begin{pmatrix} 2 & 2 & 0 \\ 1 & 5 & 3 \\ 3 & 2 & 3 \\ \end{pmatrix} \] \[ C_{13} = (-1)^{1+3} \cdot \det(M_{13}) = 1 \cdot \det(M_{13}) \] --- ### Step 3: Find the Determinants of the \( 3 \times 3 \) Matrices #### 3.1: Compute \( \det(M_{11}) \) \[ M_{11} = \begin{pmatrix} 2 & -1 & 0 \\ 5 & 2 & 3 \\ 2 & 5 & 3 \\ \end{pmatrix} \] Use the first row for expansion: \[ \det(M_{11}) = 2 \begin{vmatrix} 2 & 3 \\ 5 & 3 \end{vmatrix} - (-1) \begin{vmatrix} 5 & 3 \\ 2 & 3 \end{vmatrix} + 0 \] Calculate minors: - \( \begin{vmatrix} 2 & 3 \\ 5 & 3 \end{vmatrix} = (2)(3) - (3)(5) = 6 - 15 = -9 \) - \( \begin{vmatrix} 5 & 3 \\ 2 & 3 \end{vmatrix} = (5)(3) - (3)(2) = 15 - 6 = 9 \) So, \[ \det(M_{11}) = 2(-9) + 1(9) = -18 + 9 = -9 \] --- #### 3.2: Compute \( \det(M_{13}) \) \[ M_{13} = \begin{pmatrix} 2 & 2 & 0 \\ 1 & 5 & 3 \\ 3 & 2 & 3 \\ \end{pmatrix} \] Expand along the first row: \[ \det(M_{13}) = 2 \begin{vmatrix} 5 & 3 \\ 2 & 3 \end{vmatrix} - 2 \begin{vmatrix} 1 & 3 \\ 3 & 3 \end{vmatrix} + 0 \] Calculate minors: - \( \begin{vmatrix} 5 & 3 \\ 2 & 3 \end{vmatrix} = (5)(3) - (3)(2) = 15 - 6 = 9 \) - \( \begin{vmatrix} 1 & 3 \\ 3 & 3 \end{vmatrix} = (1)(3) - (3)(3) = 3 - 9 = -6 \) So, \[ \det(M_{13}) = 2(9) - 2(-6) = 18 + 12 = 30 \] --- ### Step 4: Substitute Back \[ \det(A) = 5 \cdot (-9) + 1 \cdot (30) = -45 + 30 = -15 \] --- ## **Final Answer** \[ \boxed{-15} \]

Let's evaluate the determinant of the given \( 4 \times 4 \) matrix \( A \): \[ A = \begin{pmatrix} 5 & 0 & 1 & 0 \\ 2 & 2 & -1 & 0 \\ 1 & 5 & 2 & 3 \\ 3 & 2 & 5 & 3 \\ \end{pmatrix} \] ### Step 1: Expand along the first row The determinant of \( A \) using cofactor expansion along the first row: \[ \det(A) = a_{11} C_{11} + a_{12} C_{12} + a_{13} C_{13} + a_{14} C_{14} \] Given the first row: \( (5, 0, 1, 0) \), the terms with zero coefficients will vanish: \[ \det(A) = 5 \cdot C_{11} + 1 \cdot C_{13} \] where: \[ C_{ij} = (-1)^{i+j} \cdot \det(M_{ij}) \] and \( M_{ij} \) is the minor matrix obtained by deleting row \( i \) and column \( j \). --- ### Step 2: Compute minors \( M_{11} \) and \( M_{13} \) #### Minor \( M_{11} \): delete row 1, column 1 \[ M_{11} = \begin{pmatrix} 2 & -1 & 0 \\ 5 & 2 & 3 \\ 2 & 5 & 3 \\ \end{pmatrix} \] #### Minor \( M_{13} \): delete row 1, column 3 \[ M_{13} = \begin{pmatrix} 2 & 2 & 0 \\ 1 & 5 & 3 \\ 3 & 2 & 3 \\ \end{pmatrix} \] --- ### Step 3: Calculate \( \det(M_{11}) \) Using the first row for expansion: \[ \det(M_{11}) = 2 \cdot \det \begin{pmatrix} 2 & 3 \\ 5 & 3 \end{pmatrix} - (-1) \cdot \det \begin{pmatrix} 5 & 3 \\ 2 & 3 \end{pmatrix} + 0 \cdot (\text{something}) \] Calculate the minors: - \( \det \begin{pmatrix} 2 & 3 \\ 5 & 3 \end{pmatrix} = (2)(3) - (3)(5) = 6 - 15 = -9 \) - \( \det \begin{pmatrix} 5 & 3 \\ 2 & 3 \end{pmatrix} = (5)(3) - (3)(2) = 15 - 6 = 9 \) Now substitute back: \[ \det(M_{11}) = 2 \times (-9) + 1 \times 9 = -18 + 9 = -9 \] ### Step 4: Calculate \( \det(M_{13}) \) Using the first row: \[ \det(M_{13}) = 2 \cdot \det \begin{pmatrix} 5 & 3 \\ 2 & 3 \end{pmatrix} - 2 \cdot \det \begin{pmatrix} 1 & 3 \\ 3 & 3 \end{pmatrix} + 0 \cdot (\text{something}) \] Calculate minors: - \( \det \begin{pmatrix} 5 & 3 \\ 2 & 3 \end{pmatrix} = 15 - 6 = 9 \) - \( \det \begin{pmatrix} 1 & 3 \\ 3 & 3 \end{pmatrix} = (1)(3) - (3)(3) = 3 - 9 = -6 \) Substitute: \[ \det(M_{13}) = 2 \times 9 - 2 \times (-6) = 18 + 12 = 30 \] --- ### Step 5: Calculate cofactors and final determinant Recall: \[ C_{11} = (-1)^{1+1} \cdot \det(M_{11}) = 1 \times (-9) = -9 \] \[ C_{13} = (-1)^{1+3} \cdot \det(M_{13}) = 1 \times 30 = 30 \] Therefore: \[ \det(A) = 5 \times C_{11} + 1 \times C_{13} = 5 \times (-9) + 1 \times 30 = -45 + 30 = -15 \] --- ## **Final Answer:** \[ \boxed{-15} \] The determinant of the matrix \( A \) is **\(-15\)**.