設 $\overset{\large\rightharpoonup}{a}, \overset{\large\rightharpoonup}{b}, \overset{\large\rightharpoonup}{c}$ 三向量排成的矩陣行列式絕對值為 $5$： $$\left| \det \begin{bmatrix} \overset{\large\rightharpoonup}{a} \\ \overset{\large\rightharpoonup}{b} \\ \overset{\large\rightharpoonup}{c} \end{bmatrix} \right| = 5$$ 我們需要求 $\overset{\large\rightharpoonup}{x} = 2\overset{\large\rightharpoonup}{a}+3\overset{\large\rightharpoonup}{b}$，$\overset{\large\rightharpoonup}{y} = \overset{\large\rightharpoonup}{b}$，$\overset{\large\rightharpoonup}{z} = \overset{\large\rightharpoonup}{c}$ 這三向量所張平行六面體的體積，即求下列行列式的絕對值： $$\det \begin{bmatrix} 2\overset{\large\rightharpoonup}{a}+3\overset{\large\rightharpoonup}{b} \\ \overset{\large\rightharpoonup}{b} \\ \overset{\large\rightharpoonup}{c} \end{bmatrix}$$ 根據行列式的線性與列運算基本性質，將第一列減去第二列的 $3$ 倍（此運算不改變行列式的值）： $$\det \begin{bmatrix} 2\overset{\large\rightharpoonup}{a}+3\overset{\large\rightharpoonup}{b} \\ \overset{\large\rightharpoonup}{b} \\ \overset{\large\rightharpoonup}{c} \end{bmatrix} = \det \begin{bmatrix} 2\overset{\large\rightharpoonup}{a} \\ \overset{\large\rightharpoonup}{b} \\ \overset{\large\rightharpoonup}{c} \end{bmatrix}$$ 再將第一列常數倍數 $2$ 提出： $$\det \begin{bmatrix} 2\overset{\large\rightharpoonup}{a} \\ \overset{\large\rightharpoonup}{b} \\ \overset{\large\rightharpoonup}{c} \end{bmatrix} = 2 \det \begin{bmatrix} \overset{\large\rightharpoonup}{a} \\ \overset{\large\rightharpoonup}{b} \\ \overset{\large\rightharpoonup}{c} \end{bmatrix}$$ 取絕對值後，可得其體積為： $$\text{體積} = 2 \times \left| \det \begin{bmatrix} \overset{\large\rightharpoonup}{a} \\ \overset{\large\rightharpoonup}{b} \\ \overset{\large\rightharpoonup}{c} \end{bmatrix} \right| = 2 \times 5 = 10$$ 因此答案為 $10$。