Question

Dadas as matrizes a=
Oba: questão 60

Answer 1

Explicação passo-a-passo:

a) det B

   $B=\left[\begin{array}{ccc}1&2&1\\2&-1&-2\\3&0&-1\end{array}\right]\left[\begin{array}{ccc}1&2\\2&-1\\3&0\end{array}\right]$

   Multiplique a diagonal principal, somando seus termos.

   Multiplique a diagonal secundária, subtraindo seus termos.

   Depois some o resultado das duas diagonais.

   diagonal principal

   1 × (-1) × (-1) + 2 × (-2) × 3 + 1 × 2 × 0

   1 - 12 + 0 = -11

   diagonal secundária (comece com o sinal negativo)

   -1 × (-1) × 3 - 1 × (-2) × 0 - 2 × 2 × (-1)

   3 + 0 + 4 = 7

   det B = -11 + 7 → det B = -4

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b) det $A^{t}$

   A transposta de A é

   $A=\left[\begin{array}{ccc}3&-2&1\\2&4&-3\\4&1&2\end{array}\right]$  →  $A^{t}=\left[\begin{array}{ccc}3&2&4\\-2&4&1\\1&-3&2\end{array}\right]$

   $\left[\begin{array}{ccc}3&2&4\\-2&4&1\\1&-3&2\end{array}\right]\left[\begin{array}{ccc}3&2\\-2&4\\1&-3\end{array}\right]$

   diagonal principal

   3 × 4 × 2 + 2 × 1 × 1 + 4 × (-2) × (-3)

   24 + 2 + 24 = 50

   diagonal secundária (comece com o sinal negativo)

   -4 × 4 × 1 - 3 × 1 × (-3) - 2 × (-2) × 2

   -16 + 9 + 8 = 1

   det $A^{t}$ = 50 + 1  →  det $A^{t}$ = 51

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c) det (A . C)

   $A=\left[\begin{array}{ccc}3&-2&1\\2&4&-3\\4&1&2\end{array}\right]$ × $C=\left[\begin{array}{ccc}1&2&-3\\3&3&1\\2&1&-2\end{array}\right]$

   $A.C=\left[\begin{array}{ccc}3.1+(-2).3+1.2&3.2+(-2).3+1.1&3.(-3)+(-2).1+1.(-2)\\2.1+4.3+(-3).2&2.2+4.3+(-3).1&2.(-3)+4.1+(-3).(-2)\\4.1+1.3+2.2&4.2+1.3+2.1&4.(-3)+1.1+2.(-2)\end{array}\right]$

   $A.C=\left[\begin{array}{ccc}3-6+2&6-6+1&-9-2-2\\2+12-6&4+12-3&-6+4+6\\4+3+4&8+3+2&-12+1-4\end{array}\right]$

   $A.C=\left[\begin{array}{ccc}-1&1&-13\\8&13&4\\11&13&-15\end{array}\right]$

   det (A.C)

   $\left[\begin{array}{ccc}-1&1&-13\\8&13&4\\11&13&-15\end{array}\right]\left[\begin{array}{ccc}-1&1\\8&13\\11&13\end{array}\right]$

   diagonal principal

   -1 × 13 × (-15) + 1 × 4 × 11 + (-13) × 8 × 13

   195 + 44 - 1352 = -1113

   diagonal secundária (comece com o sinal negativo)

   -(-13) × 13 × 11 - (-1) × 4 × 13 - 1 × 8 × (-15)

   1859 + 52 + 120 = 2031

   det (A . C) = -1113 + 2031  →  det (A . C) = 918

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d) det (C . B)

   $C=\left[\begin{array}{ccc}1&2&-3\\3&3&1\\2&1&-2\end{array}\right]$ × $B=\left[\begin{array}{ccc}1&2&1\\2&-1&-2\\3&0&-1\end{array}\right]$

   $C.B=\left[\begin{array}{ccc}1.1+2.2+(-3).3&1.2+2.(-1)+(-3).0&1.1+2.(-2)+(-3).(-1)\\3.1+3.2+1.3&3.2+3.(-1)+1.0&3.1+3.(-2)+1.(-1)\\2.1+1.2+(-2).3&2.2+1.(-1)+(-2).0&2.1+1.(-2)+(-2).(-1)\end{array}\right]$

   $C.B=\left[\begin{array}{ccc}1+4-9&2-2-0&1-4+3\\3+6+3&6-3+0&3-6-1\\2+2-6&4-1-0&2-2+2\end{array}\right]$

   $C.B=\left[\begin{array}{ccc}-4&0&0\\12&3&-4\\-2&3&2\end{array}\right]$

   det (C . B)

   $\left[\begin{array}{ccc}-4&0&0\\12&3&-4\\-2&3&2\end{array}\right]\left[\begin{array}{ccc}-4&0\\12&3\\-2&3\end{array}\right]$

   diagonal principal

   -4 × 3 × 2 + 0 × (-4) × (-2) + 0 × 12 × 3

   -24 + 0 + 0 = -24

   diagonal secundária (comece com o sinal negativo)

   -0 × 3 × (-2) - (-4) × (-4) × 3 - 0 × 12 × 2

   0 - 48 - 0 = -48

   det (C . B) = -24 - 48  →  det (C . B) = -72