Given $A = \left[ \begin{array} { c c c }
0 & 2 & 4 \
- 6 & - 4 & 2 \
0 & 8 & 2
\end{array} \right]$ , find $| A |$ .
A)-168
B)-80
C)-88
D)40
E)120

Question

Given $A = \left[ \begin{array} { c c c } 
0 & 2 & 4 \
- 6 & - 4 & 2 \
0 & 8 & 2
\end{array} \right]$ , find $| A |$ .
A)-168
B)-80
C)-88
D)40
E)120

Quizplus · Accepted Answer

The determinant $|A|$ of matrix $A$ can be calculated using the formula for a 3x3 matrix: $|A| = a(ei - fh) - b(di - fg) + c(dh - eg)$, where $a, b, c, ...$ are the elements of the matrix. Substituting the values from matrix $A$, we get $|A| = 0(2 - 64) - 2(-12 - 0) + 4(-48 - 0) = 0(-62) + 2(-12) + 4(-48) = -24 - 192 = -216$. However, there seems to be a mistake in my calculation process as none of the options match the result. Let's correctly calculate the determinant:Using the first row for expansion, we get:$$|A| = 0\cdot\left|\begin{array}{cc}-4 & 2 \ 8 & 2\end{array}\right| - 2\cdot\left|\begin{array}{cc}-6 & 2 \ 0 & 2\end{array}\right| + 4\cdot\left|\begin{array}{cc}-6 & -4 \ 0 & 8\end{array}\right|$$$$= 0 - 2(-12 - 0) + 4(-48 + 0)$$$$= 24 - 192$$$$= -168$$Therefore, the correct calculation shows that the determinant of matrix $A$ is $-168$.

Given A=[024−6−42082]A = \left[ \begin{array} { c c c } 0 & 2 & 4 \\- 6 & - 4 & 2 \\0 & 8 & 2\end{array} \right]A=​0−60​2−48​422​​

Given $A = \left[ \begin{array} { c c c } 0 & 2 & 4 \\- 6 & - 4 & 2 \\0 & 8 & 2\end{array} \right]$