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Calcolo del determinante di una matrice
Il determinante di una matrice è un valore scalare che fornisce informazioni importanti sulle proprietà della matrice, come la sua invertibilità. Ecco come calcolare il determinante per matrici di diverse dimensioni:
1. Matrice 2x2
Per una matrice A A A A di dimensione 2x2:
A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} A = ( a b c d ) A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} A = ( a c ​ b d ​ )
Il determinante è calcolato come:
\text{det}(A) = ad - bc det ( A ) = a d − b c \text{det}(A) = ad - bc det ( A ) = a d − b c
2. Matrice 3x3
Per una matrice B B B B di dimensione 3x3:
B = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} B = ( a b c d e f g h i ) B = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} B = ​ a d g ​ b e h ​ c f i ​ ​
Il determinante può essere calcolato usando la regola di Sarrus o l'espansione per cofattori. Usando l'espansione per cofattori, otteniamo:
\text{det}(B) = a \cdot \text{det}\begin{pmatrix} e & f \\ h & i \end{pmatrix} - b \cdot \text{det}\begin{pmatrix} d & f \\ g & i \end{pmatrix} + c \cdot \text{det}\begin{pmatrix} d & e \\ g & h \end{pmatrix} det ( B ) = a ⋅ det ( e f h i ) − b ⋅ det ( d f g i ) + c ⋅ det ( d e g h ) \text{det}(B) = a \cdot \text{det}\begin{pmatrix} e & f \\ h & i \end{pmatrix} - b \cdot \text{det}\begin{pmatrix} d & f \\ g & i \end{pmatrix} + c \cdot \text{det}\begin{pmatrix} d & e \\ g & h \end{pmatrix} det ( B ) = a ⋅ det ( e h ​ f i ​ ) − b ⋅ det ( d g ​ f i ​ ) + c ⋅ det ( d g ​ e h ​ )
Calcolando i determinanti delle sottomatrici 2x2, otteniamo:
\text{det}(B) = a(ei - fh) - b(di - fg) + c(dh - eg) det ( B ) = a ( e i − f h ) − b ( d i − f g ) + c ( d h − e g ) \text{det}(B) = a(ei - fh) - b(di - fg) + c(dh - eg) det ( B ) = a ( e i − f h ) − b ( d i − f g ) + c ( d h − e g )
3. Matrice di dimensione n x n
Per matrici di dimensione superiore a 3x3, puoi utilizzare l'espansione per cofattori lungo una riga o una colonna. La formula generale è:
\text{det}(A) = \sum_{j=1}^{n} (-1)^{i+j} a_{ij} \cdot \text{det}(M_{ij}) det ( A ) = ∑ j = 1 n ( − 1 ) i + j a i j ⋅ det ( M i j ) \text{det}(A) = \sum_{j=1}^{n} (-1)^{i+j} a_{ij} \cdot \text{det}(M_{ij}) det ( A ) = j = 1 ∑ n ​ ( − 1 ) i + j a ij ​ ⋅ det ( M ij ​ )
dove M_{ij} M i j M_{ij} M ij ​ è la matrice ottenuta eliminando la riga i i i i e la colonna j j j j dalla matrice A A A A .
Esempio di calcolo del determinante di una matrice 3x3
Consideriamo la matrice:
C = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix} C = ( 1 2 3 0 1 4 5 6 0 ) C = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix} C = ​ 1 0 5 ​ 2 1 6 ​ 3 4 0 ​ ​
Calcoliamo il determinante usando l'espansione lungo la prima riga:
\text{det}(C) = 1 \cdot \text{det}\begin{pmatrix} 1 & 4 \\ 6 & 0 \end{pmatrix} - 2 \cdot \text{det}\begin{pmatrix} 0 & 4 \\ 5 & 0 \end{pmatrix} + 3 \cdot \text{det}\begin{pmatrix} 0 & 1 \\ 5 & 6 \end{pmatrix} det ( C ) = 1 ⋅ det ( 1 4 6 0 ) − 2 ⋅ det ( 0 4 5 0 ) + 3 ⋅ det ( 0 1 5 6 ) \text{det}(C) = 1 \cdot \text{det}\begin{pmatrix} 1 & 4 \\ 6 & 0 \end{pmatrix} - 2 \cdot \text{det}\begin{pmatrix} 0 & 4 \\ 5 & 0 \end{pmatrix} + 3 \cdot \text{det}\begin{pmatrix} 0 & 1 \\ 5 & 6 \end{pmatrix} det ( C ) = 1 ⋅ det ( 1 6 ​ 4 0 ​ ) − 2 ⋅ det ( 0 5 ​ 4 0 ​ ) + 3 ⋅ det ( 0 5 ​ 1 6 ​ )
Calcoliamo i determinanti delle sottomatrici:
\text{det}\begin{pmatrix} 1 & 4 \\ 6 & 0 \end{pmatrix} = (1 \cdot 0) - (4 \cdot 6) = -24 det ( 1 4 6 0 ) = ( 1 ⋅ 0 ) − ( 4 ⋅ 6 ) = − 24 \text{det}\begin{pmatrix} 1 & 4 \\ 6 & 0 \end{pmatrix} = (1 \cdot 0) - (4 \cdot 6) = -24 det ( 1 6 ​ 4 0 ​ ) = ( 1 ⋅ 0 ) − ( 4 ⋅ 6 ) = − 24
\text{det}\begin{pmatrix} 0 & 4 \\ 5 & 0 \end{pmatrix} = (0 \cdot 0) - (4 \cdot 5) = -20 det ( 0 4 5 0 ) = ( 0 ⋅ 0 ) − ( 4 ⋅ 5 ) = − 20 \text{det}\begin{pmatrix} 0 & 4 \\ 5 & 0 \end{pmatrix} = (0 \cdot 0) - (4 \cdot 5) = -20 det ( 0 5 ​ 4 0 ​ ) = ( 0 ⋅ 0 ) − ( 4 ⋅ 5 ) = − 20
\text{det}\begin{pmatrix} 0 & 1 \\ 5 & 6 \end{pmatrix} = (0 \cdot 6) - (1 \cdot 5) = -5 det ( 0 1 5 6 ) = ( 0 ⋅ 6 ) − ( 1 ⋅ 5 ) = − 5 \text{det}\begin{pmatrix} 0 & 1 \\ 5 & 6 \end{pmatrix} = (0 \cdot 6) - (1 \cdot 5) = -5 det ( 0 5 ​ 1 6 ​ ) = ( 0 ⋅ 6 ) − ( 1 ⋅ 5 ) = − 5
Ora sostituiamo:
\text{det}(C) = 1 \cdot (-24) - 2 \cdot (-20) + 3 \cdot (-5) = -24 + 40 - 15 = 1 det ( C ) = 1 ⋅ ( − 24 ) − 2 ⋅ ( − 20 ) + 3 ⋅ ( − 5 ) = − 24 + 40 − 15 = 1 \text{det}(C) = 1 \cdot (-24) - 2 \cdot (-20) + 3 \cdot (-5) = -24 + 40 - 15 = 1 det ( C ) = 1 ⋅ ( − 24 ) − 2 ⋅ ( − 20 ) + 3 ⋅ ( − 5 ) = − 24 + 40 − 15 = 1
Quindi, il determinante della matrice C C C C è 1.
Calcolo del Determinante di una Matrice 3x3 con il Metodo di Sarrus
Consideriamo la seguente matrice:
A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} A = ( a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 ) A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} A = ​ a 11 ​ a 21 ​ a 31 ​ ​ a 12 ​ a 22 ​ a 32 ​ ​ a 13 ​ a 23 ​ a 33 ​ ​ ​
Passaggi per il Calcolo
Scrivi la matrice e ripeti le prime due colonne a destra:
\begin{pmatrix} a_{11} & a_{12} & a_{13} & | & a_{11} & a_{12} \\ a_{21} & a_{22} & a_{23} & | & a_{21} & a_{22} \\ a_{31} & a_{32} & a_{33} & | & a_{31} & a_{32} \end{pmatrix} ( a 11 a 12 a 13 ∣ a 11 a 12 a 21 a 22 a 23 ∣ a 21 a 22 a 31 a 32 a 33 ∣ a 31 a 32 ) \begin{pmatrix} a_{11} & a_{12} & a_{13} & | & a_{11} & a_{12} \\ a_{21} & a_{22} & a_{23} & | & a_{21} & a_{22} \\ a_{31} & a_{32} & a_{33} & | & a_{31} & a_{32} \end{pmatrix} ​ a 11 ​ a 21 ​ a 31 ​ ​ a 12 ​ a 22 ​ a 32 ​ ​ a 13 ​ a 23 ​ a 33 ​ ​ ∣ ∣ ∣ ​ a 11 ​ a 21 ​ a 31 ​ ​ a 12 ​ a 22 ​ a 32 ​ ​ ​
Calcola la somma dei prodotti delle diagonali che vanno dall'alto a sinistra al basso a destra:
D_1 = a_{11} a_{22} a_{33} + a_{12} a_{23} a_{31} + a_{13} a_{21} a_{32} D 1 = a 11 a 22 a 33 + a 12 a 23 a 31 + a 13 a 21 a 32 D_1 = a_{11} a_{22} a_{33} + a_{12} a_{23} a_{31} + a_{13} a_{21} a_{32} D 1 ​ = a 11 ​ a 22 ​ a 33 ​ + a 12 ​ a 23 ​ a 31 ​ + a 13 ​ a 21 ​ a 32 ​
Calcola la somma dei prodotti delle diagonali che vanno dall'alto a destra al basso a sinistra:
D_2 = a_{13} a_{22} a_{31} + a_{11} a_{23} a_{32} + a_{12} a_{21} a_{33} D 2 = a 13 a 22 a 31 + a 11 a 23 a 32 + a 12 a 21 a 33 D_2 = a_{13} a_{22} a_{31} + a_{11} a_{23} a_{32} + a_{12} a_{21} a_{33} D 2 ​ = a 13 ​ a 22 ​ a 31 ​ + a 11 ​ a 23 ​ a 32 ​ + a 12 ​ a 21 ​ a 33 ​
Il determinante è dato dalla differenza tra queste due somme:
\text{det}(A) = D_1 - D_2 det ( A ) = D 1 − D 2 \text{det}(A) = D_1 - D_2 det ( A ) = D 1 ​ − D 2 ​
Esempio
Consideriamo la matrice:
A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix} A = ( 1 2 3 4 5 6 7 8 9 ) A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix} A = ​ 1 4 7 ​ 2 5 8 ​ 3 6 9 ​ ​
Calcoliamo il determinante usando il metodo di Sarrus:
Somma delle diagonali dall'alto a sinistra al basso a destra:
D_1 = 1 \cdot 5 \cdot 9 + 2 \cdot 6 \cdot 7 + 3 \cdot 4 \cdot 8 = 45 + 84 + 96 = 225 D 1 = 1 ⋅ 5 ⋅ 9 + 2 ⋅ 6 ⋅ 7 + 3 ⋅ 4 ⋅ 8 = 45 + 84 + 96 = 225 D_1 = 1 \cdot 5 \cdot 9 + 2 \cdot 6 \cdot 7 + 3 \cdot 4 \cdot 8 = 45 + 84 + 96 = 225 D 1 ​ = 1 ⋅ 5 ⋅ 9 + 2 ⋅ 6 ⋅ 7 + 3 ⋅ 4 ⋅ 8 = 45 + 84 + 96 = 225
Somma delle diagonali dall'alto a destra al basso a sinistra:
D_2 = 3 \cdot 5 \cdot 7 + 1 \cdot 6 \cdot 8 + 2 \cdot 4 \cdot 9 = 105 + 48 + 72 = 225 D 2 = 3 ⋅ 5 ⋅ 7 + 1 ⋅ 6 ⋅ 8 + 2 ⋅ 4 ⋅ 9 = 105 + 48 + 72 = 225 D_2 = 3 \cdot 5 \cdot 7 + 1 \cdot 6 \cdot 8 + 2 \cdot 4 \cdot 9 = 105 + 48 + 72 = 225 D 2 ​ = 3 ⋅ 5 ⋅ 7 + 1 ⋅ 6 ⋅ 8 + 2 ⋅ 4 ⋅ 9 = 105 + 48 + 72 = 225
Calcoliamo il determinante:
\text{det}(A) = D_1 - D_2 = 225 - 225 = 0 det ( A ) = D 1 − D 2 = 225 − 225 = 0 \text{det}(A) = D_1 - D_2 = 225 - 225 = 0 det ( A ) = D 1 ​ − D 2 ​ = 225 − 225 = 0
Quindi, il determinante della matrice A A A A è 0 0 0 0 .
English version
Calculating the determinant of a matrix.
The determinant of a matrix is a scalar value that provides important information about properties of the matrix, such as its invertibility. Here is how to calculate the determinant for matrices of different sizes:
1. 2x2 matrix
For a matrix A A A A of dimension 2x2:
A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} A = ( a b c d ) A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} A = ( a c ​ b d ​ ) .
The determinant is calculated as:
\text{det}(A) = ad - bc det ( A ) = a d − b c \text{det}(A) = ad - bc det ( A ) = a d − b c
2. 3x3 matrix
For a matrix B B B B of dimension 3x3:
B = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} B = ( a b c d e f g h i ) B = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} B = ​ a d g ​ b e h ​ c f i ​ ​ .
The determinant can be calculated using Sarrus' rule or the expansion by cofactors. Using the expansion by cofactors, we get:
\text{det}(B) = a \cdot \text{det}\begin{pmatrix} e & f \\ h & i \end{pmatrix} - b \cdot \text{det}\begin{pmatrix} d & f \\ g & i \end{pmatrix} + c \cdot \text{det}\begin{pmatrix} d & e \\ g & h \end{pmatrix} det ( B ) = a ⋅ det ( e f h i ) − b ⋅ det ( d f g i ) + c ⋅ det ( d e g h ) \text{det}(B) = a \cdot \text{det}\begin{pmatrix} e & f \\ h & i \end{pmatrix} - b \cdot \text{det}\begin{pmatrix} d & f \\ g & i \end{pmatrix} + c \cdot \text{det}\begin{pmatrix} d & e \\ g & h \end{pmatrix} det ( B ) = a ⋅ det ( e h ​ f i ​ ) − b ⋅ det ( d g ​ f i ​ ) + c ⋅ det ( d g ​ e h ​ )
Calculating the determinants of the 2x2 submatrices, we obtain:
\text{det}(B) = a(ei - fh) - b(di - fg) + c(dh - eg) det ( B ) = a ( e i − f h ) − b ( d i − f g ) + c ( d h − e g ) \text{det}(B) = a(ei - fh) - b(di - fg) + c(dh - eg) det ( B ) = a ( e i − f h ) − b ( d i − f g ) + c ( d h − e g )
3. Matrix of dimension n x n
For matrices of size greater than 3x3, you can use expansion by cofactors along a row or column. The general formula is:
\text{det}(A) = \sum_{j=1}^{n} (-1)^{i+j} a_{ij} \cdot \text{det}(M_{ij}) det ( A ) = ∑ j = 1 n ( − 1 ) i + j a i j ⋅ det ( M i j ) \text{det}(A) = \sum_{j=1}^{n} (-1)^{i+j} a_{ij} \cdot \text{det}(M_{ij}) det ( A ) = j = 1 ∑ n ​ ( − 1 ) i + j a ij ​ ⋅ det ( M ij ​ )
where M_{ij} M i j M_{ij} M ij ​ is the matrix obtained by removing row i i i i and column j j j j from matrix A A A A .
Example of calculating the determinant of a 3x3 matrix.
Consider the matrix:
C = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix} C = ( 1 2 3 0 1 4 5 6 0 ) C = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix} C = ​ 1 0 5 ​ 2 1 6 ​ 3 4 0 ​ ​
We calculate the determinant using the expansion along the first line:
\text{det}(C) = 1 \cdot \text{det}\begin{pmatrix} 1 & 4 \\ 6 & 0 \end{pmatrix} - 2 \cdot \text{det}\begin{pmatrix} 0 & 4 \\ 5 & 0 \end{pmatrix} + 3 \cdot \text{det}\begin{pmatrix} 0 & 1 \\ 5 & 6 \end{pmatrix} det ( C ) = 1 ⋅ det ( 1 4 6 0 ) − 2 ⋅ det ( 0 4 5 0 ) + 3 ⋅ det ( 0 1 5 6 ) \text{det}(C) = 1 \cdot \text{det}\begin{pmatrix} 1 & 4 \\ 6 & 0 \end{pmatrix} - 2 \cdot \text{det}\begin{pmatrix} 0 & 4 \\ 5 & 0 \end{pmatrix} + 3 \cdot \text{det}\begin{pmatrix} 0 & 1 \\ 5 & 6 \end{pmatrix} det ( C ) = 1 ⋅ det ( 1 6 ​ 4 0 ​ ) − 2 ⋅ det ( 0 5 ​ 4 0 ​ ) + 3 ⋅ det ( 0 5 ​ 1 6 ​ )
We calculate the determinants of the submatrices:
\text{det}\begin{pmatrix} 1 & 4 \\ 6 & 0 \end{pmatrix} = (1 \cdot 0) - (4 \cdot 6) = -24 det ( 1 4 6 0 ) = ( 1 ⋅ 0 ) − ( 4 ⋅ 6 ) = − 24 \text{det}\begin{pmatrix} 1 & 4 \\ 6 & 0 \end{pmatrix} = (1 \cdot 0) - (4 \cdot 6) = -24 det ( 1 6 ​ 4 0 ​ ) = ( 1 ⋅ 0 ) − ( 4 ⋅ 6 ) = − 24
\text{det}\begin{pmatrix} 0 & 4 \\ 5 & 0 \end{pmatrix} = (0 \cdot 0) - (4 \cdot 5) = -20 det ( 0 4 5 0 ) = ( 0 ⋅ 0 ) − ( 4 ⋅ 5 ) = − 20 \text{det}\begin{pmatrix} 0 & 4 \\ 5 & 0 \end{pmatrix} = (0 \cdot 0) - (4 \cdot 5) = -20 det ( 0 5 ​ 4 0 ​ ) = ( 0 ⋅ 0 ) − ( 4 ⋅ 5 ) = − 20
\text{det}\begin{pmatrix} 0 & 1 \\ 5 & 6 \end{pmatrix} = (0 \cdot 6) - (1 \cdot 5) = -5 det ( 0 1 5 6 ) = ( 0 ⋅ 6 ) − ( 1 ⋅ 5 ) = − 5 \text{det}\begin{pmatrix} 0 & 1 \\ 5 & 6 \end{pmatrix} = (0 \cdot 6) - (1 \cdot 5) = -5 det ( 0 5 ​ 1 6 ​ ) = ( 0 ⋅ 6 ) − ( 1 ⋅ 5 ) = − 5
Now we substitute:
\text{det}(C) = 1 \cdot (-24) - 2 \cdot (-20) + 3 \cdot (-5) = -24 + 40 - 15 = 1 det ( C ) = 1 ⋅ ( − 24 ) − 2 ⋅ ( − 20 ) + 3 ⋅ ( − 5 ) = − 24 + 40 − 15 = 1 \text{det}(C) = 1 \cdot (-24) - 2 \cdot (-20) + 3 \cdot (-5) = -24 + 40 - 15 = 1 det ( C ) = 1 ⋅ ( − 24 ) − 2 ⋅ ( − 20 ) + 3 ⋅ ( − 5 ) = − 24 + 40 − 15 = 1
Thus, the determinant of the matrix C C C C is 1.
Calculating the Determinant of a 3x3 Matrix with the Sarrus Method
Consider the following matrix:
A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} A = ( a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 ) A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} A = ​ a 11 ​ a 21 ​ a 31 ​ ​ a 12 ​ a 22 ​ a 32 ​ ​ a 13 ​ a 23 ​ a 33 ​ ​ ​
Calculating Steps
Write the matrix and repeat the first two columns on the right:
\begin{pmatrix} a_{11} & a_{12} & a_{13} & | & a_{11} & a_{12} \\ a_{21} & a_{22} & a_{23} & | & a_{21} & a_{22} \\ a_{31} & a_{32} & a_{33} & | & a_{31} & a_{32} \end{pmatrix} ( a 11 a 12 a 13 ∣ a 11 a 12 a 21 a 22 a 23 ∣ a 21 a 22 a 31 a 32 a 33 ∣ a 31 a 32 ) \begin{pmatrix} a_{11} & a_{12} & a_{13} & | & a_{11} & a_{12} \\ a_{21} & a_{22} & a_{23} & | & a_{21} & a_{22} \\ a_{31} & a_{32} & a_{33} & | & a_{31} & a_{32} \end{pmatrix} ​ a 11 ​ a 21 ​ a 31 ​ ​ a 12 ​ a 22 ​ a 32 ​ ​ a 13 ​ a 23 ​ a 33 ​ ​ ∣ ∣ ∣ ​ a 11 ​ a 21 ​ a 31 ​ ​ a 12 ​ a 22 ​ a 32 ​ ​ ​
Calculate the sum of the products of the diagonals from the top left to the bottom right:
D_1 = a_{11} a_{22} a_{33} + a_{12} a_{23} a_{31} + a_{13} a_{21} a_{32} D 1 = a 11 a 22 a 33 + a 12 a 23 a 31 + a 13 a 21 a 32 D_1 = a_{11} a_{22} a_{33} + a_{12} a_{23} a_{31} + a_{13} a_{21} a_{32} D 1 ​ = a 11 ​ a 22 ​ a 33 ​ + a 12 ​ a 23 ​ a 31 ​ + a 13 ​ a 21 ​ a 32 ​
Calculate the sum of the products of the diagonals from the top right to the bottom left:
D_2 = a_{13} a_{22} a_{31} + a_{11} a_{23} a_{32} + a_{12} a_{21} a_{33} D 2 = a 13 a 22 a 31 + a 11 a 23 a 32 + a 12 a 21 a 33 D_2 = a_{13} a_{22} a_{31} + a_{11} a_{23} a_{32} + a_{12} a_{21} a_{33} D 2 ​ = a 13 ​ a 22 ​ a 31 ​ + a 11 ​ a 23 ​ a 32 ​ + a 12 ​ a 21 ​ a 33 ​
The determinant is given by the difference between these two sums:
\text{det}(A) = D_1 - D_2 det ( A ) = D 1 − D 2 \text{det}(A) = D_1 - D_2 det ( A ) = D 1 ​ − D 2 ​
Example
Consider the matrix:
A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix} A = ( 1 2 3 4 5 6 7 8 9 ) A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix} A = ​ 1 4 7 ​ 2 5 8 ​ 3 6 9 ​ ​
Let's calculate the determinant using the Sarrus method:
Sum of the diagonals from top left to bottom right:
D_1 = 1 \cdot 5 \cdot 9 + 2 \cdot 6 \cdot 7 + 3 \cdot 4 \cdot 8 = 45 + 84 + 96 = 225 D 1 = 1 ⋅ 5 ⋅ 9 + 2 ⋅ 6 ⋅ 7 + 3 ⋅ 4 ⋅ 8 = 45 + 84 + 96 = 225 D_1 = 1 \cdot 5 \cdot 9 + 2 \cdot 6 \cdot 7 + 3 \cdot 4 \cdot 8 = 45 + 84 + 96 = 225 D 1 ​ = 1 ⋅ 5 ⋅ 9 + 2 ⋅ 6 ⋅ 7 + 3 ⋅ 4 ⋅ 8 = 45 + 84 + 96 = 225
Sum of the diagonals from top right to bottom left:
D_2 = 3 \cdot 5 \cdot 7 + 1 \cdot 6 \cdot 8 + 2 \cdot 4 \cdot 9 = 105 + 48 + 72 = 225 D 2 = 3 ⋅ 5 ⋅ 7 + 1 ⋅ 6 ⋅ 8 + 2 ⋅ 4 ⋅ 9 = 105 + 48 + 72 = 225 D_2 = 3 \cdot 5 \cdot 7 + 1 \cdot 6 \cdot 8 + 2 \cdot 4 \cdot 9 = 105 + 48 + 72 = 225 D 2 ​ = 3 ⋅ 5 ⋅ 7 + 1 ⋅ 6 ⋅ 8 + 2 ⋅ 4 ⋅ 9 = 105 + 48 + 72 = 225
Let's calculate the determinant:
\text{det}(A) = D_1 - D_2 = 225 - 225 = 0 det ( A ) = D 1 − D 2 = 225 − 225 = 0 \text{det}(A) = D_1 - D_2 = 225 - 225 = 0 det ( A ) = D 1 ​ − D 2 ​ = 225 − 225 = 0
So, the determinant of the matrix A A A A is 0 0 0 0 .
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