Calculate $| \mathbf { A } |$
-$$A = \left| \begin{array} { r r r }
- 1 & 4 & 2 \
- 2 & - 2 & 5 \
- 3 & - 3 & - 3
\end{array} \right|$$
A)15
B)105
C)-3
D)-105

Question

Calculate $| \mathbf { A } |$
-$$A = \left| \begin{array} { r r r } 
- 1 & 4 & 2 \
- 2 & - 2 & 5 \
- 3 & - 3 & - 3
\end{array} \right|$$
A)15
B)105
C)-3
D)-105

Quizplus · Accepted Answer

The determinant $| \mathbf { A } |$ of matrix A can be calculated using the formula for a 3x3 matrix: $$|A| = a(ei - fh) - b(di - fg) + c(dh - eg)$$ Substituting the values from matrix A, we get: $$|A| = (-1)((-2)(-3) - (5)(-3)) - (4)((-2)(-3) - (5)(-3)) + (2)((-2)(-3) - (-2)(-3))$$ Simplifying, we find: $$|A| = -1(-6 + 15) - 4(-6 + 15) + 2(6 - 6)$$ $$|A| = -1(9) - 4(9) + 2(0)$$ $$|A| = -9 - 36$$ $$|A| = -45$$ However, it seems there was a mistake in my calculation as none of the options match -45. Let's correct the calculation error: $$|A| = (-1)(-2*-3 - 5*-3) - 4(-2*-3 - 5*-3) + 2(-2*-3 - -2*-3)$$ Correctly calculating, we should find: $$|A| = (-1)(-6 + 15) - 4(-6 + 15) + 2(0)$$ $$|A| = -9 - 4 * 9$$ $$|A| = -9 - 36$$ $$|A| = -45$$ Given the mistake in the recalculation and the options provided, let's correctly calculate the determinant directly: $$|A| = (-1)(-2*(-3) - 5*(-3)) - 4((-2)*(-3) - 5*(-3)) + 2((-2)*(-3) - (-2)*(-3))$$ $$|A| = (-1)(6 + 15) - 4(6 + 15) + 2(0)$$ $$|A| = -21 - 4(21)$$ $$|A| = -21 - 84$$ $$|A| = -105$$ Therefore, the correct calculation leads to the determinant being -105.

Calculate ∣A∣| \mathbf { A } |∣A∣ - A=∣−142−2−25−3−3−3∣A = \left| \begin{array} { r r r } - 1 & 4 & 2 \\- 2 & - 2 & 5 \\- 3 & - 3 & - 3\end{array} \right|A=​−1−2−3​4−2−3​25−3​​

Calculate $| \mathbf { A } |$ - $A = \left| \begin{array} { r r r } - 1 & 4 & 2 \\- 2 & - 2 & 5 \\- 3 & - 3 & - 3\end{array} \right|$