Question #1
\[\]
\[
\text{Given the matrix } A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}, \text{ compute the determinant using cofactor expansion.}
\]
Question #2
\[
\text{Given the matrix } B = \begin{pmatrix} 2 & 3 & 1 \\ 4 & 0 & 5 \\ 7 & 6 & 8 \end{pmatrix}, \text{ compute the determinant using cofactor expansion.}
\]
Question #3
\[
\text{Given the matrix } C = \begin{pmatrix} 3 & 2 & 4 \\ 1 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}, \text{ compute the determinant using cofactor expansion.}
\]
Question #4
\[
\text{Given the matrix } D = \begin{pmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \end{pmatrix}, \text{ compute the determinant using cofactor expansion.}
\]
Solution of Exercise
Question #1
\[
\text{Given the matrix } A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}, \text{ compute the determinant using cofactor expansion.}
\]
Solution:
\[
\text{We will expand along the first row. The determinant is:}
\]
\[
\text{det}(A) = 1 \cdot \begin{vmatrix} 5 & 6 \\ 8 & 9 \end{vmatrix} – 2 \cdot \begin{vmatrix} 4 & 6 \\ 7 & 9 \end{vmatrix} + 3 \cdot \begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix}
\]
\[
\text{Now, calculate the 2×2 determinants:}
\]
\[
\begin{vmatrix} 5 & 6 \\ 8 & 9 \end{vmatrix} = (5 \times 9) – (6 \times 8) = 45 – 48 = -3
\]
\[
\begin{vmatrix} 4 & 6 \\ 7 & 9 \end{vmatrix} = (4 \times 9) – (6 \times 7) = 36 – 42 = -6
\]
\[
\begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix} = (4 \times 8) – (5 \times 7) = 32 – 35 = -3
\]
\[
\text{Substitute these values back into the cofactor expansion:}
\]
\[
\text{det}(A) = 1 \cdot (-3) – 2 \cdot (-6) + 3 \cdot (-3)
\]
\[
\text{det}(A) = -3 + 12 – 9
\]
\[
\text{det}(A) = 0
\]
\[
\textbf{Answer: The determinant of matrix } A \text{ is } 0.
\]
Question #2
\[
\text{Given the matrix } B = \begin{pmatrix} 2 & 3 & 1 \\ 4 & 0 & 5 \\ 7 & 6 & 8 \end{pmatrix}, \text{ compute the determinant using cofactor expansion.}
\]
Solution:
\[
\text{We will expand along the first row. The determinant is:}
\]
\[
\text{det}(B) = 2 \cdot \begin{vmatrix} 0 & 5 \\ 6 & 8 \end{vmatrix} – 3 \cdot \begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix} + 1 \cdot \begin{vmatrix} 4 & 0 \\ 7 & 6 \end{vmatrix}
\]
\[
\text{Now, calculate the 2×2 determinants:}
\]
\[
\begin{vmatrix} 0 & 5 \\ 6 & 8 \end{vmatrix} = (0 \times 8) – (5 \times 6) = 0 – 30 = -30
\]
\[
\begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix} = (4 \times 8) – (5 \times 7) = 32 – 35 = -3
\]
\[
\begin{vmatrix} 4 & 0 \\ 7 & 6 \end{vmatrix} = (4 \times 6) – (0 \times 7) = 24 – 0 = 24
\]
\[
\text{Substitute these values back into the cofactor expansion:}
\]
\[
\text{det}(B) = 2 \cdot (-30) – 3 \cdot (-3) + 1 \cdot 24
\]
\[
\text{det}(B) = -60 + 9 + 24
\]
\[
\text{det}(B) = -27
\]
\[
\textbf{Answer: The determinant of matrix } B \text{ is } -27.
\]
Question #3
\[
\text{Given the matrix } C = \begin{pmatrix} 3 & 2 & 4 \\ 1 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}, \text{ compute the determinant using cofactor expansion.}
\]
Solution:
\[
\text{We will expand along the first row. The determinant is:}
\]
\[
\text{det}(C) = 3 \cdot \begin{vmatrix} 5 & 6 \\ 8 & 9 \end{vmatrix} – 2 \cdot \begin{vmatrix} 1 & 6 \\ 7 & 9 \end{vmatrix} + 4 \cdot \begin{vmatrix} 1 & 5 \\ 7 & 8 \end{vmatrix}
\]
\[
\text{Now, calculate the 2×2 determinants:}
\]
\[
\begin{vmatrix} 5 & 6 \\ 8 & 9 \end{vmatrix} = (5 \times 9) – (6 \times 8) = 45 – 48 = -3
\]
\[
\begin{vmatrix} 1 & 6 \\ 7 & 9 \end{vmatrix} = (1 \times 9) – (6 \times 7) = 9 – 42 = -33
\]
\[
\begin{vmatrix} 1 & 5 \\ 7 & 8 \end{vmatrix} = (1 \times 8) – (5 \times 7) = 8 – 35 = -27
\]
\[
\text{Substitute these values back into the cofactor expansion:}
\]
\[
\text{det}(C) = 3 \cdot (-3) – 2 \cdot (-33) + 4 \cdot (-27)
\]
\[
\text{det}(C) = -9 + 66 – 108
\]
\[
\text{det}(C) = -51
\]
\[
\textbf{Answer: The determinant of matrix } C \text{ is } -51.
\]
Question #4
\[
\text{Given the matrix } D = \begin{pmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \end{pmatrix}, \text{ compute the determinant using cofactor expansion.}
\]
Solution:
\[
\text{We will expand along the first row. The determinant is:}
\]
\[
\text{det}(D) = 1 \cdot \begin{vmatrix} 5 & 8 \\ 6 & 9 \end{vmatrix} – 4 \cdot \begin{vmatrix} 2 & 8 \\ 3 & 9 \end{vmatrix} + 7 \cdot \begin{vmatrix} 2 & 5 \\ 3 & 6 \end{vmatrix}
\]
\[
\text{Now, calculate the 2×2 determinants:}
\]
\[
\begin{vmatrix} 5 & 8 \\ 6 & 9 \end{vmatrix} = (5 \times 9) – (8 \times 6) = 45 – 48 = -3
\]
\[
\begin{vmatrix} 2 & 8 \\ 3 & 9 \end{vmatrix} = (2 \times 9) – (8 \times 3) = 18 – 24 = -6
\]
\[
\begin{vmatrix} 2 & 5 \\ 3 & 6 \end{vmatrix} = (2 \times 6) – (5 \times 3) = 12 – 15 = -3
\]
\[
\text{Substitute these values back into the cofactor expansion:}
\]
\[
\text{det}(D) = 1 \cdot (-3) – 4 \cdot (-6) + 7 \cdot (-3)
\]
\[
\text{det}(D) = -3 + 24 – 21
\]
\[
\text{det}(D) = 0
\]
\[
\textbf{Answer: The determinant of matrix } D \text{ is } 0.
\]
