Solve by using matrices. 2x – y +2 + w = -3 x + 2y – 3z + w = 12 3x – y – + 2w = 3 -2x + 3y + 2 – 3w = -3​

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Solve by using matrices. 2x – y +2 + w = -3 x + 2y – 3z + w = 12 3x – y – + 2w = 3 -2x + 3y + 2 – 3w = -3​

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RI SƠ 4 years 2021-08-09T10:33:46+00:00 1 Answers 15 views 0

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  1. diemthu
    0
    2021-08-09T10:35:16+00:00
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    Some symbols and numbers are missing. I assume the system is supposed to read

    2x – y + 2z + w = -3

    x + 2y – 3z + w = 12

    3x – y – z + 2w = 3

    -2x + 3y + 2z – 3w = -3

    In matrix form, this is

    \begin{bmatrix}2&-1&2&1\\1&2&-3&1\\3&-1&-1&2\\-2&3&2&-3\end{bmatrix}\begin{bmatrix}x\\y\\z\\w\end{bmatrix}=\begin{bmatrix}-3\\12\\3\-3\end{bmatrix}

    which we can strip down to the augmented matrix,

    \left[\begin{array}{cccc|c}2&-1&2&1&-3\\1&2&-3&1&12\\3&-1&-1&2&3\\-2&3&2&-3&-3\end{array}\right]

    Now for the row operations:

    • swap rows 1 and 2

    \left[\begin{array}{cccc|c}1&2&-3&1&12\\2&-1&2&1&-3\\3&-1&-1&2&3\\-2&3&2&-3&-3\end{array}\right]

    • add -2 (row 1) to row 2, -3 (row 1) to row 3, and 2 (row 1) to row 4

    \left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&-7&8&-1&-33\\0&7&-4&-1&21\end{array}\right]

    • add 7 (row 2) to -5 (row 3), and row 3 to row 4

    \left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&16&-2&-24\\0&0&4&-2&-12\end{array}\right]

    • multiply through rows 3 and 4 by 1/2

    \left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&8&-1&-12\\0&0&2&-1&-6\end{array}\right]

    • add -4 (row 4) to row 3

    \left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&0&3&12\\0&0&2&-1&-6\end{array}\right]

    • swap rows 3 and 4

    \left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&2&-1&-6\\0&0&0&3&12\end{array}\right]

    • multiply through row 4 by 1/3

    \left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&2&-1&-6\\0&0&0&1&4\end{array}\right]

    • add row 4 to row 3

    \left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&2&0&-2\\0&0&0&1&4\end{array}\right]

    • multiply through row 3 by 1/2

    \left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&1&0&-1\\0&0&0&1&4\end{array}\right]

    • add -8 (row 3) and row 4 to row 2

    \left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&0&0&-15\\0&0&1&0&-1\\0&0&0&1&4\end{array}\right]

    • multiply through row 2 by -1/5

    \left[\begin{array}{cccc|c}1&2&-3&1&12\\0&1&0&0&3\\0&0&1&0&-1\\0&0&0&1&4\end{array}\right]

    • add -2 (row 2) and 3 (row 3) and -1 (row 4) to row 1

    \left[\begin{array}{cccc|c}1&0&0&0&-1\\0&1&0&0&3\\0&0&1&0&-1\\0&0&0&1&4\end{array}\right]

    Then the solution to the system is (x, y, z, w) = (-1, 3, -1, 4).

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