Question

For the following matrix calculate the determinant:
{{5,3,7},{1,0,1},{9,6,4}}

Answer 1

Question #51654, Math, Matrix | Tensor Analysis | for completion

For the following matrix calculate the determinant:

$$\{\{5, 3, 7 \}, \{1, 0, 1 \}, \{9, 6, 4 \} \}$$

**Answer to Question#51654**

**Solution**

The determinant is a scalar value that can be calculated from a square matrix.

$$A = \left( \begin{array}{ccc} 5 & 3 & 7 \\ 1 & 0 & 1 \\ 9 & 6 & 4 \end{array} \right)$$

The standard way to compute the determinant of a square matrix is to apply the Laplace expansion along any one of its rows or columns. For instance, an expansion along the first row yields:

$$\begin{aligned} \det(A) &= |A| = \left| \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} \right| = a_{11} \cdot (-1)^{1+1} \cdot \left| \begin{array}{cc} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array} \right| + a_{12} \cdot (-1)^{1+2} \cdot \left| \begin{array}{cc} a_{21} & a_{23} \\ a_{31} & a_{33} \end{array} \right| + a_{13} \cdot (-1)^{1+3} \cdot \left| \begin{array}{cc} a_{21} & a_{22} \\ a_{31} & a_{32} \end{array} \right| = \\ &= a_{11} \left| \begin{array}{cc} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array} \right| - a_{12} \cdot \left| \begin{array}{cc} a_{21} & a_{23} \\ a_{31} & a_{33} \end{array} \right| + a_{13} \cdot \left| \begin{array}{cc} a_{21} & a_{22} \\ a_{31} & a_{32} \end{array} \right| = \\ &= a_{11} \cdot \left(a_{22} a_{33} - a_{23} a_{32}\right) - a_{12} \cdot \left(a_{21} a_{33} - a_{23} a_{31}\right) + a_{13} \cdot \left(a_{21} a_{32} - a_{22} a_{31}\right) \end{aligned}$$ $$\begin{aligned} \det(A) &= |A| = \left| \begin{array}{ccc} 5 & 3 & 7 \\ 1 & 0 & 1 \\ 9 & 6 & 4 \end{array} \right| = 5 \cdot (-1)^{1+1} \left| \begin{array}{cc} 0 & 1 \\ 6 & 4 \end{array} \right| + 3 \cdot (-1)^{1+2} \left| \begin{array}{cc} 1 & 1 \\ 9 & 4 \end{array} \right| + 1 \cdot (-1)^{1+3} \left| \begin{array}{cc} 1 & 0 \\ 9 & 6 \end{array} \right| = \\ &= 5 \cdot (0 \cdot 4 - 6 \cdot 1) - 3 \cdot (1 \cdot 4 - 1 \cdot 9) + 7 \cdot (1 \cdot 6 - 0 \cdot 9) = -30 + 15 + 42 = 27. \end{aligned}$$

However the simplest way to compute the determinant of this matrix $A$ is to use the Laplace expansion along the second row:

$$\begin{aligned} \det(A) &= |A| = \left| \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} \right| = a_{21} \cdot (-1)^{2+1} \cdot \left| \begin{array}{cc} a_{12} & a_{13} \\ a_{32} & a_{33} \end{array} \right| + a_{22} \cdot (-1)^{2+2} \cdot \left| \begin{array}{cc} a_{11} & a_{13} \\ a_{31} & a_{33} \end{array} \right| + a_{23} \cdot (-1)^{2+3} \cdot \left| \begin{array}{cc} a_{11} & a_{12} \\ a_{31} & a_{32} \end{array} \right| = \\ &= -a_{21} \cdot \left| \begin{array}{cc} a_{12} & a_{13} \\ a_{32} & a_{33} \end{array} \right| + a_{22} \cdot \left| \begin{array}{cc} a_{11} & a_{13} \\ a_{31} & a_{33} \end{array} \right| - a_{23} \cdot \left| \begin{array}{cc} a_{11} & a_{12} \\ a_{31} & a_{32} \end{array} \right| = \\ &= -a_{21} \cdot \left(a_{12} a_{33} - a_{13} a_{32}\right) + a_{22} \cdot \left(a_{11} a_{33} - a_{13} a_{31}\right) - a_{23} \cdot \left(a_{11} a_{32} - a_{12} a_{31}\right) \end{aligned}$$ $$\begin{aligned} \det(A) &= |A| = \left| \begin{array}{ccc} 5 & 3 & 7 \\ 1 & 0 & 1 \\ 9 & 6 & 4 \end{array} \right| = 1 \cdot (-1)^{2+1} \left| \begin{array}{cc} 3 & 7 \\ 6 & 4 \end{array} \right| + \underbrace{0 \cdot (-1)^{2+2} \left| \begin{array}{cc} 5 & 7 \\ 9 & 4 \end{array} \right|} _ {0} + 1 \cdot (-1)^{2+3} \left| \begin{array}{cc} 5 & 3 \\ 9 & 6 \end{array} \right| = -(3 \cdot 4 - 6 \cdot 7) - (5 \cdot 6 - 3 \cdot 9) = \\ &= -(12 - 42) - (30 - 27) = 30 - 3 = 27 \end{aligned}$$

**Answer:** $\operatorname{det}(A) = 27$

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