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Ymorist [56]
3 years ago
14

Find the inverse of this matrix. 1 -1 -1 -1 2 3 1 1 4

Mathematics
1 answer:
zheka24 [161]3 years ago
3 0

Let's use Gaussian elimination. Consider the augmented matrix,

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

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

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

• Add -2 (row 2) to row 3:

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

• Add -2 (row 3) to row 2:

\left[\begin{array}{ccc|ccc}1 & -1 & -1 & 1 & 0 & 0\\0 & 1 & 0 & 7 & 5 & -2\\0 & 0 & 1 & -3 & -2 & 1\end{array}\right]

• Add row 2 and row 3 to row 1:

\left[\begin{array}{ccc|ccc}1 & 0 & 0 & 5 & 3 & -1\\0 & 1 & 0 & 7 & 5 & -2\\0 & 0 & 1 & -3 & -2 & 1\end{array}\right]

So the inverse is

\begin{bmatrix}1&-1&-1\\-1&2&3\\1&1&4\end{bmatrix}^{-1} = \boxed{\begin{bmatrix}5&3&-1\\7&5&-2\\-3&-2&1\end{bmatrix}}

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What does i^739 equal? explain how you found your answer.
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<h3>Answer:   -i</h3>

========================================================

Explanation:

i = sqrt(-1)

Lets list out the first few powers of i

  • i^0 = 1
  • i^1 = i
  • i^2 = -1
  • i^3 = i*i^2 = i*(-1) = -i
  • i^4 = (i^2)^2 = (-1)^2 = 1

By the time we reach the fourth power, we repeat the cycle over again (since i^0 is also equal to 1). The cycle is of length 4, which means we'll divide the exponent over 4 to find the remainder. Ignore the quotient. That remainder will determine if we go for i^0, i^1, i^2 or i^3.

For example, i^5 = i^1 because 5/4 leads to a remainder 1.

Another example: i^6 = i^2 since 6/4 = 1 remainder 2

Again, we only care about the remainder to find out which bin we land on.

-------------

Turning to the question your teacher gave you, we have,

739/4 = 184 remainder 3

So i^739 = i^3 = -i

<h3>-i is the final answer</h3>

--------------

Side notes:

  • if i^a = i^b, then a-b is a multiple of 4
  • Recall that the divisibility by 4 trick involves looking at the last two digits of the number. So i^739 is identical to i^39.
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