Answer to Question #87347 - Math – Linear Algebra
Question:
5. Given that A=(123321132)A = \begin{pmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \\ 1 & 3 & 2 \end{pmatrix}A=⎝⎛131223312⎠⎞. Find the determinant of AAA.
a. 2
b. 3
c. 1
d. zero
6. A matrix is said to be singular if the determinant is equal to
a. 3
b. 1
c. zero
d. 2
Solution:
5. ∣A∣=∣123321132∣=1×(2×2−1×3)−2×(3×2−1×1)+3×(3×3−2×1)=1−10+21=12|A| = \begin{vmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \\ 1 & 3 & 2 \end{vmatrix} = 1 \times (2 \times 2 - 1 \times 3) - 2 \times (3 \times 2 - 1 \times 1) + 3 \times (3 \times 3 - 2 \times 1) = 1 - 10 + 21 = 12∣A∣=∣∣131223312∣∣=1×(2×2−1×3)−2×(3×2−1×1)+3×(3×3−2×1)=1−10+21=12.
6. A matrix is said to be singular if the determinant is equal to zero.
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