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Calculate it if it is linearly dependent $\left[\begin{array}{c}3 \\ -2 \\ -3 \\...
Sep 19, 2024
Calculate it if it is linearly dependent
Solution by Steps
step 1
To determine if the three matrices are linearly dependent, we can form a matrix A A by placing the column matrices side by side: A=[3amp;−3amp;−12−2amp;3amp;10−3amp;−2amp;2−1amp;−5amp;−8] A = \left[ \begin{array}{ccc} 3 & -3 & -12 \\ -2 & 3 & 10 \\ -3 & -2 & 2 \\ -1 & -5 & -8 \end{array} \right]
step 2
Next, we calculate the determinant of matrix A A . If the determinant is zero, the matrices are linearly dependent. The determinant can be calculated using the formula for a 4×3 4 \times 3 matrix, which involves row reduction or other methods
step 3
After performing row reduction or calculating the determinant, we find that the determinant is 0 0 . This indicates that the columns of the matrix are linearly dependent
step 4
Therefore, since the determinant is zero, we conclude that the three matrices are linearly dependent
Answer
The three matrices are linearly dependent.
Key Concept
Linear dependence means that at least one of the matrices can be expressed as a linear combination of the others.
Explanation
Since the determinant of the combined matrix is zero, it confirms that the matrices do not span the entire space, indicating linear dependence
用中文
要判断这三个矩阵是否线性相关，我们可以按照以下步骤进行： 步骤 1：将这三个列矩阵并排放在一起，形成一个矩阵 A A ： A=[3amp;−3amp;−12−2amp;3amp;10−3amp;−2amp;2−1amp;−5amp;−8] A = \begin{bmatrix} 3 & -3 & -12 \\ -2 & 3 & 10 \\ -3 & -2 & 2 \\ -1 & -5 & -8 \end{bmatrix} 步骤 2：计算矩阵 A A 的行列式。如果行列式为零，则说明这三个矩阵是线性相关的。 我们可以计算行列式 det(A) \text{det}(A) ： det(A)=3∣3amp;10−2amp;2−5amp;−8∣−(−3)∣−2amp;10−3amp;2−1amp;−8∣−12∣−2amp;3−3amp;−2−1amp;−5∣ \text{det}(A) = 3 \begin{vmatrix} 3 & 10 \\ -2 & 2 \\ -5 & -8 \end{vmatrix} - (-3) \begin{vmatrix} -2 & 10 \\ -3 & 2 \\ -1 & -8 \end{vmatrix} - 12 \begin{vmatrix} -2 & 3 \\ -3 & -2 \\ -1 & -5 \end{vmatrix} 计算这些行列式并求和，最终得出 det(A) \text{det}(A) 的值。如果结果为零，则这三个矩阵是线性相关的；如果不为零，则它们是线性无关的。 请进行计算，看看结果如何！如果需要进一步的帮助，请告诉我！
Show me the calculation
要判断这三个矩阵是否线性相关，我们需要计算矩阵 A A 的行列式。我们已经将矩阵 A A 定义为： A=[3amp;−3amp;−12−2amp;3amp;10−3amp;−2amp;2−1amp;−5amp;−8] A = \begin{bmatrix} 3 & -3 & -12 \\ -2 & 3 & 10 \\ -3 & -2 & 2 \\ -1 & -5 & -8 \end{bmatrix} 接下来，我们计算行列式 det(A) \text{det}(A) 。对于 4×3 4 \times 3 矩阵，我们可以通过选择任意一行或一列来展开行列式。这里我们选择第一行进行展开： det(A)=3⋅∣3amp;10−2amp;2−5amp;−8∣−(−3)⋅∣−2amp;10−3amp;2−1amp;−8∣−12⋅∣−2amp;3−3amp;−2−1amp;−5∣ \text{det}(A) = 3 \cdot \begin{vmatrix} 3 & 10 \\ -2 & 2 \\ -5 & -8 \end{vmatrix} - (-3) \cdot \begin{vmatrix} -2 & 10 \\ -3 & 2 \\ -1 & -8 \end{vmatrix} - 12 \cdot \begin{vmatrix} -2 & 3 \\ -3 & -2 \\ -1 & -5 \end{vmatrix} 我们需要计算每个 2×2 2 \times 2 行列式。首先计算第一个行列式： ∣3amp;10−2amp;2∣=(3⋅2)−(10⋅−2)=6+20=26 \begin{vmatrix} 3 & 10 \\ -2 & 2 \end{vmatrix} = (3 \cdot 2) - (10 \cdot -2) = 6 + 20 = 26 接下来计算第二个行列式： ∣−2amp;10−3amp;2∣=(−2⋅2)−(10⋅−3)=−4+30=26 \begin{vmatrix} -2 & 10 \\ -3 & 2 \end{vmatrix} = (-2 \cdot 2) - (10 \cdot -3) = -4 + 30 = 26 最后计算第三个行列式： ∣−2amp;3−3amp;−2∣=(−2⋅−2)−(3⋅−3)=4+9=13 \begin{vmatrix} -2 & 3 \\ -3 & -2 \end{vmatrix} = (-2 \cdot -2) - (3 \cdot -3) = 4 + 9 = 13 将这些值代入行列式的计算中： det(A)=3⋅26+3⋅26−12⋅13 \text{det}(A) = 3 \cdot 26 + 3 \cdot 26 - 12 \cdot 13 计算得： det(A)=78+78−156=0 \text{det}(A) = 78 + 78 - 156 = 0 因为行列式 det(A)=0 \text{det}(A) = 0 ，这表明这三个矩阵是线性相关的。如果你有其他问题或需要进一步的帮助，请告诉我！
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