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Eva8 [605]
2 years ago
11

Please solve this!! Matrices.

Mathematics
1 answer:
mel-nik [20]2 years ago
6 0

It sounds like "elementary transformation method" refers to Gauss-Jordan elimination.

Start with the augmented matrix,

\left[\begin{array}{ccc|ccc}2&3&1&1&0&0\\2&8&1&0&1&0\\3&7&2&0&0&1\end{array}\right]

• Add (-1)(row 1) to row 2, and (-3)(row 1) to 2(row 3). In other words,

-(2, 3, 1, 1, 0, 0) + (2, 8, 1, 0, 1, 0) = (0, 5, 0, -1, 1, 0)

-3 (2, 3, 1, 1, 0, 0) + 2 (3, 7, 2, 0, 0, 1) = (0, 5, 1, -3, 0, 2)

I'll omit these details in the next several steps. The augmented matrix then transforms to

\left[\begin{array}{ccc|ccc}2&3&1&1&0&0\\0&5&0&-1&1&0\\0&5&1&-3&0&2\end{array}\right]

• Add (-1)(row 2) to row 3:

\left[\begin{array}{ccc|ccc}2&3&1&1&0&0\\0&5&0&-1&1&0\\0&0&1&-2&-1&2\end{array}\right]

• Add (-1)(row 3) to row 1:

\left[\begin{array}{ccc|ccc}2&3&0&3&1&-2\\0&5&0&-1&1&0\\0&0&1&-2&-1&2\end{array}\right]

• Add (-3)(row 2) to 5(row 1):

\left[\begin{array}{ccc|ccc}10&0&0&18&2&-10\\0&5&0&-1&1&0\\0&0&1&-2&-1&2\end{array}\right]

• Multiply row 1 by 1/10 and row 2 by 1/5:

\left[\begin{array}{ccc|ccc}1&0&0&1.8&0.2&-1\\0&1&0&-0.2&0.2&0\\0&0&1&-2&-1&2\end{array}\right]

We're done, and we find that

\begin{bmatrix}2&3&1\\2&8&1\\3&7&2\end{bmatrix}^{-1} = \begin{bmatrix}1.8&0.2&-1\\-0.2&0.2&0\\-2&-1&2\end{bmatrix} = \boxed{\dfrac15\begin{bmatrix}9&1&-5\\-1&1&0\\-10&-5&10\end{bmatrix}}

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