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
\( A =
\begin{pmatrix}
-4 & 5 & -2 \\
7 & 0 & -3 \\
3 & 4 & 7
\end{pmatrix}
\)
este:
exercițiu nou
Determinantul matricei
\( A =
\begin{pmatrix}
-4 & 5 & -2 \\
7 & 0 & -3 \\
3 & 4 & 7
\end{pmatrix}
\)
este
\( \det A = -394\).
\( \det A = \)
\(
\begin{vmatrix}
-4 & 5 & -2 \\
7 & 0 & -3 \\
3 & 4 & 7
\end{vmatrix}
=
\)
\(
\begin{vmatrix}
\color{tomato} -4 & 5 & -2 \\
\color{orangered} 7 & \color{tomato} 0 & -3 \\
\color{red} 3 & \color{orangered} 4 & \color{tomato} 7 \\
-4 & \color{red} 5 & \color{orangered} -2 \\
7 & 0 & \color{red} -3\end{vmatrix}
=
\)
\(
\begin{vmatrix}
-4 & 5 & \color{turquoise} -2 \\
7 & \color{turquoise} 0 & \color{steelblue} -3 \\
\color{turquoise} 3 & \color{steelblue} 4 & \color{blue} 7 \\
\color{steelblue} -4 & \color{blue} 5 & -2 \\
\color{blue} 7 & 0 & -3
\end{vmatrix}
=
\)
\( = \color{tomato} (-4) \cdot 0 \cdot 7 \) \( + \) \( \color{orangered} 7 \cdot 4 \cdot (-2) \) \( + \) \( \color{red} 3 \cdot 5 \cdot (-3) \) \( - \) \( \color{turquoise} (-2) \cdot 0 \cdot 3 \) \( - \) \( \color{steelblue} (-3) \cdot 4 \cdot (-4) \) \( - \) \( \color{blue} 7 \cdot 5 \cdot 7 \) \( = \)
\( = 0 - 56 - 45 + 0 - 48 - 245 = \)
\( = -394 \).
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