Question

Please help! Thanks in advance!

Answer 1

Answer:

  1)  $\left[\begin{array}{c}x&y\end{array}\right] =\left[\begin{array}{c}\dfrac{1}{2}&&\dfrac{3}{4}\end{array}\right]$

  2)  $\dfrac{1}{26}\left[\begin{array}{cc}2&6\\-3&4\end{array}\right]$

Step-by-step explanation:

1) The system of equations can be written as the augmented matrix ...

  $\left[\begin{array}{cc|c}2&-4&-2\\3&2&3\end{array}\right]$

Dividing the first row by 2 makes it ...

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

Then subtracting 3 times the first row from the second gives ...

  $\left[\begin{array}{cc|c}1&-2&-1\\0&8&6\end{array}\right]$

And dividing the second row by 8 puts 1s on the diagonal.

  $\left[\begin{array}{cc|c}1&-2&-1\\0&1&\dfrac{3}{4}\end{array}\right]$

This row echelon form tells us ...

  y = 3/4

  x -2y = -1

So, substituting for y in the second of these equations, we find ...

  x = -1 +2(3/4) = 1/2

The solution is (x, y) = (1/2, 3/4).

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2) For a 2×2 matrix, finding the inverse is not so difficult by hand. The inverse is the transpose of the cofactor matrix divided by the determinant.

The cofactor matrix is ...

  $\left[\begin{array}{cc}2&-3\\6&4\end{array}\right]$

and its transpose is ...

  $\left[\begin{array}{cc}2&6\\-3&4\end{array}\right]$

The determinant is the product of diagonal elements less the product of off-diagonal elements, so is (4)(2) -(3)(-6) = 8+18 = 26.

Dividing the transpose of the cofactor matrix by this gives the inverse ...

  $\dfrac{1}{26}\left[\begin{array}{cc}2&6\\-3&4\end{array}\right]$