题目
[题目]-|||-1 2 3-|||-设A= 2 2 1 求 A^(-1)-|||-3 4 3
题目解答
答案
解析
步骤 1：计算矩阵A的行列式
计算矩阵A的行列式，以确定矩阵A是否可逆。矩阵A的行列式为：
$|A| = 1(2*3 - 1*4) - 2(2*3 - 1*3) + 3(2*4 - 2*3)$
$|A| = 1(6 - 4) - 2(6 - 3) + 3(8 - 6)$
$|A| = 1(2) - 2(3) + 3(2)$
$|A| = 2 - 6 + 6$
$|A| = 2$
步骤 2：计算矩阵A的代数余子式
计算矩阵A的代数余子式，以构建伴随矩阵。代数余子式为：
${M}_{11} = 2*3 - 1*4 = 6 - 4 = 2$
${M}_{12} = -(2*3 - 1*3) = -(6 - 3) = -3$
${M}_{13} = 2*4 - 2*3 = 8 - 6 = 2$
${M}_{21} = -(2*3 - 3*4) = -(6 - 12) = 6$
${M}_{22} = 1*3 - 3*3 = 3 - 9 = -6$
${M}_{23} = -(1*4 - 3*2) = -(4 - 6) = 2$
${M}_{31} = 2*1 - 2*2 = 2 - 4 = -2$
${M}_{32} = -(1*1 - 2*2) = -(1 - 4) = 3$
${M}_{33} = 1*2 - 2*2 = 2 - 4 = -2$
步骤 3：构建伴随矩阵A'
伴随矩阵A'由代数余子式构成，且行列互换：
$A' = \begin{pmatrix} 2 & -3 & 2 \\ 6 & -6 & 2 \\ -2 & 3 & -2 \end{pmatrix}$
步骤 4：计算逆矩阵${A}^{-1}$
逆矩阵${A}^{-1}$由伴随矩阵A'除以行列式|A|得到：
${A}^{-1} = \frac{1}{|A|}A' = \frac{1}{2} \begin{pmatrix} 2 & -3 & 2 \\ 6 & -6 & 2 \\ -2 & 3 & -2 \end{pmatrix}$
${A}^{-1} = \begin{pmatrix} 1 & -\frac{3}{2} & 1 \\ 3 & -3 & 1 \\ -1 & \frac{3}{2} & -1 \end{pmatrix}$
计算矩阵A的行列式，以确定矩阵A是否可逆。矩阵A的行列式为：
$|A| = 1(2*3 - 1*4) - 2(2*3 - 1*3) + 3(2*4 - 2*3)$
$|A| = 1(6 - 4) - 2(6 - 3) + 3(8 - 6)$
$|A| = 1(2) - 2(3) + 3(2)$
$|A| = 2 - 6 + 6$
$|A| = 2$
步骤 2：计算矩阵A的代数余子式
计算矩阵A的代数余子式，以构建伴随矩阵。代数余子式为：
${M}_{11} = 2*3 - 1*4 = 6 - 4 = 2$
${M}_{12} = -(2*3 - 1*3) = -(6 - 3) = -3$
${M}_{13} = 2*4 - 2*3 = 8 - 6 = 2$
${M}_{21} = -(2*3 - 3*4) = -(6 - 12) = 6$
${M}_{22} = 1*3 - 3*3 = 3 - 9 = -6$
${M}_{23} = -(1*4 - 3*2) = -(4 - 6) = 2$
${M}_{31} = 2*1 - 2*2 = 2 - 4 = -2$
${M}_{32} = -(1*1 - 2*2) = -(1 - 4) = 3$
${M}_{33} = 1*2 - 2*2 = 2 - 4 = -2$
步骤 3：构建伴随矩阵A'
伴随矩阵A'由代数余子式构成，且行列互换：
$A' = \begin{pmatrix} 2 & -3 & 2 \\ 6 & -6 & 2 \\ -2 & 3 & -2 \end{pmatrix}$
步骤 4：计算逆矩阵${A}^{-1}$
逆矩阵${A}^{-1}$由伴随矩阵A'除以行列式|A|得到：
${A}^{-1} = \frac{1}{|A|}A' = \frac{1}{2} \begin{pmatrix} 2 & -3 & 2 \\ 6 & -6 & 2 \\ -2 & 3 & -2 \end{pmatrix}$
${A}^{-1} = \begin{pmatrix} 1 & -\frac{3}{2} & 1 \\ 3 & -3 & 1 \\ -1 & \frac{3}{2} & -1 \end{pmatrix}$