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What is the volume of the parallelepiped defined by the columns of matrix \( \left[\begin{array}{ccc}0 & 3 & 6 \\ -6 & 9 & 5 \\ 0 & 0 & -9\end{array}\right] \) Volume \( = \) Ex. 1

Ask by Gardner Evans. in the United States
Mar 20,2025

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The volume of the parallelepiped is 162.

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To find the volume of the parallelepiped defined by the columns of the given matrix, we can use the formula for the volume of a parallelepiped which is given by the absolute value of the determinant of the matrix formed by the columns of the given matrix. Given matrix: \[ \left[\begin{array}{ccc}0 & 3 & 6 \\ -6 & 9 & 5 \\ 0 & 0 & -9\end{array}\right] \] The volume of the parallelepiped is given by the absolute value of the determinant of the matrix: \[ \text{Volume} = \left| \det \left[\begin{array}{ccc}0 & 3 & 6 \\ -6 & 9 & 5 \\ 0 & 0 & -9\end{array}\right] \right| \] Let's calculate the determinant of the matrix to find the volume of the parallelepiped. Matrices by following steps: - step0: Evaluate the determinant: \(\left[\begin{array}{lll}{0}&{3}&{6}\\{-6}&{9}&{5}\\{0}&{0}&{-9}\end{array}\right]\) - step1: Use the appropriate notation: \(\left|\begin{array}{cc}{0}&{3}&{6}\\{-6}&{9}&{5}\\{0}&{0}&{-9}\end{array}\right|\) - step2: Use cofactor expansion along a row: \(0\times \left|\begin{array}{cc}{3}&{6}\\{9}&{5}\end{array}\right|-0\times \left|\begin{array}{cc}{0}&{6}\\{-6}&{5}\end{array}\right|-9\times \left|\begin{array}{cc}{0}&{3}\\{-6}&{9}\end{array}\right|\) - step3: Multiply: \(0-0-9\times 18\) - step4: Evaluate: \(0-0-162\) - step5: Remove 0: \(-162\) The determinant of the matrix is -162. Therefore, the volume of the parallelepiped defined by the columns of the given matrix is \(| -162 | = 162\).

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To find the volume of the parallelepiped defined by the columns of the matrix \( \left[\begin{array}{ccc}0 & 3 & 6 \\ -6 & 9 & 5 \\ 0 & 0 & -9\end{array}\right] \), you can use the absolute value of the determinant of the matrix. First, let's calculate the determinant using the formula for a 3x3 matrix: \[ \text{det}(A) = a(ei-fh) - b(di-fg) + c(dh-eg) \] For our matrix \( A = \left[\begin{array}{ccc}0 & 3 & 6 \\ -6 & 9 & 5 \\ 0 & 0 & -9\end{array}\right] \): 1. \( a=0, b=3, c=6 \) 2. \( d=-6, e=9, f=5 \) 3. \( g=0, h=0, i=-9 \) Plugging these values into the formula gives: \[ \text{det}(A) = 0(9 \cdot -9 - 5 \cdot 0) - 3(-6 \cdot -9 - 5 \cdot 0) + 6(-6 \cdot 0 - 9 \cdot 0) \] \[ = 0 - 3(54) + 0 = -162 \] Thus, the volume of the parallelepiped is given by the absolute value of the determinant: \[ \text{Volume} = | -162 | = 162 \] So, the final volume \( = 162 \). For a quick way to visualize this, think of the parallelepiped as a 3D shape extending from one corner of a rectangular box, defined by the vectors represented by the columns of this matrix. When you determine the volume, you find out how much space this shape takes up in that box! Another fun fact is that the concept of determinants is used in several fields! For instance, in physics, determinants can help solve systems of equations, determine rotational effects in mechanics, and are essential in computer graphics for transformations like scaling and rotating shapes. It's like a behind-the-scenes wizard that makes 3D world-building work seamlessly!

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