Matematică >> matrice și determinanți >> 4
\( \color{red}A = \color{dimgray} \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} \)
se poate calcula utilizând regula lui Sarrus:
\( \det A = \) \( \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} = \) \( \begin{vmatrix} \color{tomato} a_{11} & a_{12} & a_{13} \\ \color{orangered} a_{21} & \color{tomato} a_{22} & a_{23} \\ \color{red} a_{31} & \color{orangered} a_{32} & \color{tomato} a_{33} \\ a_{11} & \color{red} a_{12} & \color{orangered} a_{13} \\ a_{21} & a_{22} & \color{red} a_{23} \end{vmatrix} = \) \( \begin{vmatrix} a_{11} & a_{12} & \color{turquoise} a_{13} \\ a_{21} & \color{turquoise} a_{22} & \color{steelblue} a_{23} \\ \color{turquoise} a_{31} & \color{steelblue} a_{32} & \color{blue} a_{33} \\ \color{steelblue} a_{11} & \color{blue} a_{12} & a_{13} \\ \color{blue} a_{21} & a_{22} & a_{23} \end{vmatrix} = \)
\( = \color{tomato} a_{11} \cdot a_{22} \cdot a_{33} \) \( + \) \( \color{orangered} a_{21} \cdot a_{32} \cdot a_{13} \) \( + \) \( \color{red} a_{31} \cdot a_{12} \cdot a_{23} \)
\( - \) \( \color{turquoise} a_{13} \cdot a_{22} \cdot a_{31} \) \( - \) \( \color{steelblue} a_{23} \cdot a_{32} \cdot a_{11} \) \( - \) \( \color{blue} a_{33} \cdot a_{12} \cdot a_{21} \).
determinantul matricei
\( A = \begin{pmatrix} 3 & 6 & -4 \\ 0 & 5 & 1 \\ -3 & -1 & 2 \end{pmatrix} \)
este:
\( \det A = \) \( \begin{vmatrix} 3 & 6 & -4 \\ 0 & 5 & 1 \\ -3 & -1 & 2 \end{vmatrix} = \) \( \begin{vmatrix} \color{tomato} 3 & 6 & -4 \\ \color{orangered} 0 & \color{tomato} 5 & 1 \\ \color{red} -3 & \color{orangered} -1 & \color{tomato} 2 \\ 3 & \color{red} 6 & \color{orangered} -4 \\ 0 & 5 & \color{red} 1 \end{vmatrix} = \) \( \begin{vmatrix} 3 & 6 & \color{turquoise} -4 \\ 0 & \color{turquoise} 5 & \color{steelblue} 1 \\ \color{turquoise} -3 & \color{steelblue} -1 & \color{blue} 2 \\ \color{steelblue} 3 & \color{blue} 6 & -4 \\ \color{blue} 0 & 5 & 1 \end{vmatrix} = \)
\( = \color{tomato} 3 \cdot 5 \cdot 2 \) \( + \) \( \color{orangered} 0 \cdot (-1) \cdot (-4) \) \( + \) \( \color{red} (-3) \cdot 6 \cdot 1 \) \( - \) \( \color{turquoise} (-4) \cdot 5 \cdot (-3) \) \( - \) \( \color{steelblue} 1 \cdot (-1) \cdot 3 \) \( - \) \( \color{blue} 2 \cdot 6 \cdot 0 \) \( = \)
\( = 30 + 0 - 18 - 60 + 3 - 0 = \)
\( = - 45 \).
Determinantul matricei
\( A =
\begin{pmatrix}
-1 & 7 & 5 \\
-10 & -2 & -8 \\
-3 & 1 & -6
\end{pmatrix}
\)
este:
exercițiu nou
Determinantul matricei
\( A =
\begin{pmatrix}
-1 & 7 & 5 \\
-10 & -2 & -8 \\
-3 & 1 & -6
\end{pmatrix}
\)
este
\( \det A = -352\).
\( \det A = \)
\(
\begin{vmatrix}
-1 & 7 & 5 \\
-10 & -2 & -8 \\
-3 & 1 & -6
\end{vmatrix}
=
\)
\(
\begin{vmatrix}
\color{tomato} -1 & 7 & 5 \\
\color{orangered} -10 & \color{tomato} -2 & -8 \\
\color{red} -3 & \color{orangered} 1 & \color{tomato} -6 \\
-1 & \color{red} 7 & \color{orangered} 5 \\
-10 & -2 & \color{red} -8\end{vmatrix}
=
\)
\(
\begin{vmatrix}
-1 & 7 & \color{turquoise} 5 \\
-10 & \color{turquoise} -2 & \color{steelblue} -8 \\
\color{turquoise} -3 & \color{steelblue} 1 & \color{blue} -6 \\
\color{steelblue} -1 & \color{blue} 7 & 5 \\
\color{blue} -10 & -2 & -8
\end{vmatrix}
=
\)
\( = \color{tomato} (-1) \cdot (-2) \cdot (-6) \) \( + \) \( \color{orangered} (-10) \cdot 1 \cdot 5 \) \( + \) \( \color{red} (-3) \cdot 7 \cdot (-8) \) \( - \) \( \color{turquoise} 5 \cdot (-2) \cdot (-3) \) \( - \) \( \color{steelblue} (-8) \cdot 1 \cdot (-1) \) \( - \) \( \color{blue} (-6) \cdot 7 \cdot (-10) \) \( = \)
\( = -12 - 50 + 168 - 30 - 8 - 420 = \)
\( = -352 \).
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