Questions: Calculate the 3 × 3 determinant: -5 7 8 6 0 1 -2 -4 -4 =

Solution Steps

To calculate the determinant of a \(3 \times 3\) matrix, we can use the rule of Sarrus or the cofactor expansion method. Here, we'll use the cofactor expansion along the first row. The determinant of a \(3 \times 3\) matrix \(\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}\) is calculated as:

\[ \text{det} = a(ei - fh) - b(di - fg) + c(dh - eg) \]

We are given the matrix

\[ A = \begin{bmatrix} -5 & 7 & 8 \\ 6 & 0 & 1 \\ -2 & -4 & -4 \end{bmatrix} \]

Using the formula for the determinant of a \(3 \times 3\) matrix, we have:

\[ \text{det}(A) = -5 \cdot (0 \cdot -4 - 1 \cdot -4) - 7 \cdot (6 \cdot -4 - 1 \cdot -2) + 8 \cdot (6 \cdot -4 - 0 \cdot -2) \]

Calculating each term:

1. \(0 \cdot -4 - 1 \cdot -4 = 0 + 4 = 4\)
2. \(6 \cdot -4 - 1 \cdot -2 = -24 + 2 = -22\)
3. \(6 \cdot -4 - 0 \cdot -2 = -24\)

Substituting these values back into the determinant formula:

\[ \text{det}(A) = -5 \cdot 4 - 7 \cdot (-22) + 8 \cdot (-24) \]

Calculating each term:

1. \(-5 \cdot 4 = -20\)
2. \(-7 \cdot -22 = 154\)
3. \(8 \cdot -24 = -192\)

Now, summing these results:

\[ \text{det}(A) = -20 + 154 - 192 = -58 \]

Final Answer