Question

Даны матрицы А и В. Найти det А, АВ, В+2А-3Е, В^-1 А: 6; -6; -1
5; -1; -3
8; -2; -3
В: -5; -7; -4
5; 5; -7
2; 8; 1

Answer 1

$\displaystyle\tt A=\left(\begin{array}{ccc}6&-6&-1\\5&-1&-3\\8&-2&-3\end{array}\right)$

$\displaystyle\tt B=\left(\begin{array}{ccc}-5&-7&-4\\5&5&-7\\2&8&1\end{array}\right)$

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$\displaystyle\tt det\ A=6\cdot(-1)\cdot(-3) + (-6)\cdot(-3)\cdot8 + (-1)\cdot5\cdot(-2) - (-1)\cdot(-1)\cdot8 - 6\cdot(-3)\cdot(-2) - (-6)\cdot5\cdot(-3) = 18 + 144 + 10 - 8 - 36 -90=38$

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$\displaystyle\tt A\cdot B=\left(\begin{array}{ccc}6&-6&-1\\5&-1&-3\\8&-2&-3\end{array}\right)\cdot\left(\begin{array}{ccc}-5&-7&-4\\5&5&-7\\2&8&1\end{array}\right)=\left(\begin{array}{ccc}-62&-80&17\\-36&-64&-16\\-56&-90&-21\end{array}\right)$

c₁₁ = a₁₁ · b₁₁ + a₁₂ · b₂₁ + a₁₃ · b₃₁ = 6 · (-5) + (-6) · 5 + (-1) · 2 = (-30) - 30 - 2 = -62

c₁₂ = a₁₁ · b₁₂ + a₁₂ · b₂₂ + a₁₃ · b₃₂ = 6 · (-7) + (-6) · 5 + (-1) · 8 = (-42) - 30 - 8 = -80  

c₁₃ = a₁₁ · b₁₃ + a₁₂ · b₂₃ + a₁₃ · b₃₃ = 6 · (-4) + (-6) · (-7) + (-1) · 1 = (-24) + 42 - 1 = 17  

c₂₁ = a₂₁ · b₁₁ + a₂₂ · b₂₁ + a₂₃ · b₃₁ = 5 · (-5) + (-1) · 5 + (-3) · 2 = (-25) - 5 - 6 = -36

c₂₂ = a₂₁ · b₁₂ + a₂₂ · b₂₂ + a₂₃ · b₃₂ = 5 · (-7) + (-1) · 5 + (-3) · 8 = (-35) - 5 - 24 = -64  

c₂₃ = a₂₁ · b₁₃ + a₂₂ · b₂₃ + a₂₃ · b₃₃ = 5 · (-4) + (-1) · (-7) + (-3) · 1 = (-20) + 7 - 3 = -16  

c₃₁ = a₃₁ · b₁₁ + a₃₂ · b₂₁ + a₃₃ · b₃₁ = 8 · (-5) + (-2) · 5 + (-3) · 2 = (-40) - 10 - 6 = -56  

c₃₂ = a₃₁ · b₁₂ + a₃₂ · b₂₂ + a₃₃ · b₃₂ = 8 · (-7) + (-2) · 5 + (-3) · 8 = (-56) - 10 - 24 = -90  

c₃₃ = a₃₁ · b₁₃ + a₃₂ · b₂₃ + a₃₃ · b₃₃ = 8 · (-4) + (-2) · (-7) + (-3) · 1 = (-32) + 14 - 3 = -21

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$\displaystyle\tt B+2A-3E=\left(\begin{array}{ccc}-5&-7&-4\\5&5&-7\\2&8&1\end{array}\right)+2\cdot\left(\begin{array}{ccc}6&-6&-1\\5&-1&-3\\8&-2&-3\end{array}\right)-3\cdot\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right)=\\\\\\\displaystyle\tt=\left(\begin{array}{ccc}-5&-7&-4\\5&5&-7\\2&8&1\end{array}\right)+\left(\begin{array}{ccc}12&-12&-2\\10&-2&-6\\16&-4&-6\end{array}\right)-\left(\begin{array}{ccc}3&0&0\\0&3&0\\0&0&3\end{array}\right)=$

$\displaystyle\tt=\left(\begin{array}{ccc}1&-13&-5\\10&4&-10\\10&6&-2\end{array}\right)-\left(\begin{array}{ccc}3&0&0\\0&3&0\\0&0&3\end{array}\right)=\\\\\\\displaystyle\tt=\left(\begin{array}{ccc}-2&-13&-5\\10&1&-10\\10&6&-5\end{array}\right)$

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$\displaystyle\tt B^{-1}=\left(\begin{array}{ccc}-5&-7&-4\\5&5&-7\\2&8&1\end{array}\Bigg|\left\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right)=$

$=\displaystyle\tt \left(\begin{array}{ccc}1&1.4&0.8\\5&5&-7\\2&8&1\end{array}\Bigg|\left\begin{array}{ccc}-0.2&0&0\\0&1&0\\0&0&1\end{array}\right)=$

$=\displaystyle\tt \left(\begin{array}{ccc}1&1.4&0.8\\0&-2&-11\\0&5.2&-0.6\end{array}\Bigg|\left\begin{array}{ccc}-0.2&0&0\\1&1&0\\0.4&0&1\end{array}\right)=$

$=\displaystyle\tt \left(\begin{array}{ccc}1&1.4&0.8\\0&1&5.5\\0&5.2&-0.6\end{array}\Bigg|\left\begin{array}{ccc}-0.2&0&0\\-0.5&-0.5&0\\0.4&0&1\end{array}\right)=$

$=\displaystyle\tt \left(\begin{array}{ccc}1&0&-6.9\\0&1&5.5\\0&0&-29.2\end{array}\Bigg|\left\begin{array}{ccc}0.5&0.7&0\\-0.5&-0.5&0\\3&2.6&1\end{array}\right)=$

$=\displaystyle\tt \left(\begin{array}{ccc}1&0&-6.9\\0&1&5.5\\0&0&1\end{array}\Bigg|\left\begin{array}{ccc}0.5&0.7&0\\-0.5&-0.5&0\\-\dfrac{15}{146}&-\dfrac{13}{146}&-\dfrac{5}{146}\end{array}\right)=$

$=\displaystyle\tt \left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\Bigg|\left\begin{array}{ccc}-\dfrac{61}{292}&\dfrac{25}{292}&-\dfrac{69}{292}\\\dfrac{19}{292}&-\dfrac{3}{292}&\dfrac{55}{292}\\-\dfrac{15}{146}&-\dfrac{13}{146}&-\dfrac{5}{146}\end{array}\right)$

$\displaystyle\tt B^{-1}=\left(\begin{array}{ccc}-\dfrac{61}{292}&\dfrac{25}{292}&-\dfrac{69}{292}\\\dfrac{19}{292}&-\dfrac{3}{292}&\dfrac{55}{292}\\-\dfrac{15}{146}&-\dfrac{13}{146}&-\dfrac{5}{146}\end{array}\right)$