Consider a n × n matrix such that n ≥ 3 : (1) You can select any row or column to expand its determinant. (2) Once the row/column is selected, multiply each element of that row/column by its cofactor, then add all these products. (3) The sum-product described above is equal to the determinant of the original matrix. \text{\color{#c34632}Consider a $n\times n$ matrix such that $n\geq3$:
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(1) You can select any row or column to expand its determinant.
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(2) Once the row/column is selected, multiply each element of that row/column by its cofactor, then add all these products.
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(3) The sum-product described above is equal to the determinant of the original matrix. }
Consider a n × n matrix such that n ≥ 3: (1) You can select any row or column to expand its determinant. (2) Once the row/column is selected, multiply each element of that row/column by its cofactor, then add all these products. (3) The sum-product described above is equal to the determinant of the original matrix.
Let us expand this determinant along column 2 because it has the maximum zeros.
∣ 2 0 1 2 1 0 2 4 0 − 5 ∣ = 0 ⋅ ( − 1 ) 1 + 2 ⋅ ∣ 1 2 4 − 5 ∣ + 0 ⋅ ( − 1 ) 2 + 2 ⋅ ∣ 2 1 2 4 − 5 ∣ + 0 ⋅ ( − 1 ) 3 + 2 ⋅ ∣ 2 1 2 1 2 ∣ \left|\begin{array}{ccc} 2 & 0 & \dfrac{1}{2}\\\\ 1 & 0 & 2 \\\\ 4 & 0 & -5\end{array}\right|=0\cdot \left(-1\right)^{1+2}\cdot\left|\begin{array}{cc} 1 & 2 \\ 4 & -5\end{array}\right|+0\cdot \left(-1\right)^{2+2}\cdot \left|\begin{array}{cc} 2 & \dfrac{1}{2} \\\\ 4 & -5\end{array}\right|+0\cdot \left(-1\right)^{3+2}\cdot \left|\begin{array}{cc} 2 & \dfrac{1}{2}\\\\ 1 & 2 \end{array}\right|
∣ ∣ 2 1 4 0 0 0 2 1 2 − 5 ∣ ∣ = 0 ⋅ ( − 1 ) 1 + 2 ⋅ ∣ ∣ 1 4 2 − 5 ∣ ∣ + 0 ⋅ ( − 1 ) 2 + 2 ⋅ ∣ ∣ 2 4 2 1 − 5 ∣ ∣ + 0 ⋅ ( − 1 ) 3 + 2 ⋅ ∣ ∣ 2 1 2 1 2 ∣ ∣
= 0 + 0 + 0 = 0 =0+0+0=0
= 0 + 0 + 0 = 0
