如何找到矩阵的行列式？

在数学中，矩阵是一个矩形或正方形的数字或符号数组，它们按行和列分组，以表示数学对象或其特性之一。例如， $\begin{vmatrix}{a}&b&c\\p&q&r\\s&t&u\end{vmatrix}$是一个方阵。另一方面， $\begin{vmatrix}{a}&b&c\\p&q&r\end{vmatrix}$是一个矩形矩阵。

矩阵的行列式

矩阵行列式是任何行或列的分量及其各自的辅因子的乘积。它们仅存在于方阵中，即行数和列数相等的方阵。任何方阵 A 的行列式由 detA（或）|A| 表示。它有时由符号表示。计算 1 × 1 和 2 × 2 矩阵的行列式非常简单，但是随着矩阵秩的增加，过程变得更加复杂。确定矩阵行列式的过程涉及次要因素和辅助因素。

行列式的性质

1. 矩阵的行列式和它自己的转置总是相等的，即$\begin{vmatrix}{a}&b&c\\l&m&n\\x&y&z\end{vmatrix}=\begin{vmatrix}{a}&l&x\\b&m&y\\c&n&z\end{vmatrix}$
2. 交换矩阵的任何两行或两列都会改变其行列式的符号，即$\begin{vmatrix}{a}&b&c\\l&m&n\\x&y&z\end{vmatrix}=-\begin{vmatrix}{a}&b&c\\x&y&z\\l&m&n\end{vmatrix}$ .
3. 具有任意两个相同行或列的矩阵的行列式为零，即$\begin{vmatrix}{l}&m&n\\l&m&n\\x&y&z\end{vmatrix}=0=\begin{vmatrix}{l}&l&x\\m&m&y\\n&n&z\end{vmatrix}$ .
4. 以零作为其任一行或列的元素的矩阵的行列式为零，即$\begin{vmatrix}{0}&0&0\\l&m&n\\x&y&z\end{vmatrix}=0$
5. 将行列式的每一行乘以常数 m 也会将行列式的值增加到 m 倍，即$\begin{vmatrix}{ka}&kb&kc\\l&m&n\\x&y&z\end{vmatrix}=k\begin{vmatrix}{a}&b&c\\l&m&n\\x&y&z\end{vmatrix}$ .

• 1 × 1 矩阵的行列式

A 1 × 1 matrix is a row with only one row and one column, and hence only one element. The determinant of every 1 × 1 matrix is always equal to the matrix element.

Example: If matrix A = [69]_{1 × 1}, then det A or |A| = 69.

• 2 × 2 矩阵的行列式

Given matrix A = $\begin{vmatrix}{a}&b\\c&d\end{vmatrix}$ then its determinant is given by ad – bc. It is simply calculated by cross multiplying the components from top left to bottom right and then finding the difference of the two.

Example: Evaluate, $\begin{vmatrix}{4}&7\\5&8\end{vmatrix}.$

Here, a = 4, b = 7, c = 5 and d = 8. 

Hence, |A| = 4(8) – 5(7)

|A| = −3

• 3 × 3 矩阵的行列式

为了找到 3 × 3 矩阵的行列式，我们需要了解元素的小项，

小调

通过删除包含该元素的行和列导出的任意阶方阵的行列式是该元素的次要数。

示例：在矩阵中找到 6 的小数$\left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right].$

解决方案：

Since 6 lies in the second row and third column of the given matrix, its matrix would be equal to the determinant of the matrix obtained by removing the second row and the third column.

Minor of 6 = $\left|\begin{array}{cc}1&2\\7 & 8\end{array}\right|$

= 8 × 1 – 2 × 7

= 8 – 14 

= -6

应遵循以下步骤，

1. Select any row or column. To find the determinant, we normally start with the first row.
2. Determine the co-factors of each of the row/column items that we picked in Step 1.
3. Multiply the row/column items from Step 1 by the appropriate co-factors from Step 2.
4. Add all of the products from Step 3 to get the matrix’s determinant.

或者，以下公式也可用于此目的：

|一个| = a(ei - fh) - b(di - fg) + c(dh - eg)

示例问题

问题 1. 在矩阵中找到 9 的小数$\left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right]$ .

解决方案：

Since 9 lies in the third row and third column of the given matrix, its matrix would be equal to the determinant of the matrix obtained by removing the third row and the third column.

Minor of 9 = $\left|\begin{array}{cc}1 & 2 \\4 & 5\end{array}\right|$

= 1 × 5 – 2 × 4

= 5 – 8

= -3

问题 2. 求 |4A| 的值如果 A = $\left[\begin{array}{lll}1 & 1 & -1\\1 & 2 & 2\\0 & 3 & 4\end{array}\right]$ .

解决方案：

A = $\left[\begin{array}{lll}1 & 1 & -1\\1 & 2 & 2\\0 & 3 & 4\end{array}\right]$

Thus, 4A = $\left[\begin{array}{lll}4 & 4 & -4\\4 & 8 & 8\\0 & 12 & 16\end{array}\right]$

Now, |4A| = 4 (128 – 96) – 4 (64 – 0) – 4 (48 – 0)

= 4(32) – 4(64) – 4(48)

|4A| = -320

问题 3：如果矩阵的行列式求 x $\left[\begin{array}{cc}4&8\\2 & x\end{array}\right]$为 0。

解决方案：

Given: A = $\left[\begin{array}{cc}4&8\\2 & x\end{array}\right]$

|A| = 4(x) – 8(2)

= 4x – 16

Since |A| = 0.

⇒ 4x – 16 = 0

or, 4x = 16

⇒ x = 4

问题 4：评估： $\left[\begin{array}{cc}2&8&5&-3\\0&1&3&4\\6&24&15&-9\\7&2&1&10\end{array}\right]$ .使用行列式的属性。

解决方案：

Clearly, the elements of the third row have 3 as a common factor. It can be depicted as:

$\left[\begin{array}{cc}2&8&5&-3\\0&1&3&4\\6&24&15&-9\\7&2&1&10\end{array}\right]=3\left[\begin{array}{cc}2&8&5&-3\\0&1&3&4\\2&8&5&-3\\7&2&1&10\end{array}\right]$

Since the first and third rows of the given matrix are equal now, using the third property the value of the matrix would be zero.

|A| = 3 × 0

|A| = 0

问题 5：求 (a ² + b ² – c ² )|A| 的值如果 A = $\begin{vmatrix}0&c&-b\\-c&0&a\\b&-a&0\end{vmatrix}$ .

解决方案：

|A| = 0(0(0) – a(a)) – c(0(-c) – a(b)) + (-b)((-c)(-a) – 0(b))

= 0 × (a²) − c(0 − ab) − b(ac – 0)

= 0

Hence, (a² + b² – c²)|A| = (a² + b² – c²)(0)

⇒ (a² + b² – c²)|A| = 0

问题 6：如果矩阵的行列式求 t $\left[\begin{array}{cc}0&x\\5&3\end{array}\right]$是 10。

解决方案：

Given: A = $\left[\begin{array}{cc}0&x\\5&3\end{array}\right]$

|A| = 3(0) – 5(x)

= 0 – 5x

= -5x

Since |A| = 10.

⇒ -5x = 10

⇒ x = -2

问题 7：如果矩阵的行列式求 x $\left[\begin{array}{cc}5&1\\3&x\end{array}\right]$是 12。

解决方案：

Given: A = $\left[\begin{array}{cc}5&1\\3&x\end{array}\right]$

|A| = 5(x) – 3(1)

= 5x – 3

Since |A| = 12

⇒ 5x – 3 = 12

⇒ 5x = 15

⇒ x = 3

问题 8：评估： $\begin{vmatrix}{1}&2&3\\4&5&6\\7&8&9\end{vmatrix}.$

解决方案：

Here, a = 1, b = 2, c = 3, d = 4, e = 5, f = 6, g = 7, h = 8, i = 9.

Thus, |A| = 1(5(9) – 6(8)) – 2(4(9) – 6(7)) + 3(4(8) – 5(7))

= 1(45 – 42) – 2(36 – 42) + 3(32 – 35)

= 1(3) – 2(-6) + 3(-3)

= 3 + 12 – 9

⇒ |A| = 0