Questions: Given: Matrix A : [[-3, 4, 2], [-2, -6, 1], [-1, 4, -2]] Find det A

Solution Steps

To find the determinant of a 3x3 matrix, we can use the rule of Sarrus or the general formula for the determinant of a 3x3 matrix. The determinant of matrix \( A \) can be calculated as follows:

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

Where the matrix \( A \) is:

\[ A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]

For the given matrix:

\[ A = \begin{pmatrix} -3 & 4 & 2 \\ -2 & -6 & 1 \\ -1 & 4 & -2 \end{pmatrix} \]

We can substitute the values into the formula to find the determinant.

Given the matrix \( A \):

\[ A = \begin{pmatrix} -3 & 4 & 2 \\ -2 & -6 & 1 \\ -1 & 4 & -2 \end{pmatrix} \]

The determinant of a 3x3 matrix \( A \) is given by:

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

For the given matrix \( A \):

\[ a = -3, \quad b = 4, \quad c = 2 \] \[ d = -2, \quad e = -6, \quad f = 1 \] \[ g = -1, \quad h = 4, \quad i = -2 \]

Substitute the values into the determinant formula:

\[ \operatorname{det} A = -3((-6 \cdot -2) - (1 \cdot 4)) - 4((-2 \cdot -2) - (1 \cdot -1)) + 2((-2 \cdot 4) - (-6 \cdot -1)) \]

Simplify each term inside the parentheses:

\[ \operatorname{det} A = -3((12) - (4)) - 4((4) - (-1)) + 2((-8) - (6)) \]

\[ \operatorname{det} A = -3(8) - 4(5) + 2(-14) \]

\[ \operatorname{det} A = -24 - 20 - 28 \]

\[ \operatorname{det} A = -72 \]

Final Answer