Question

Please help! Thanks in advance!

Answer 1

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

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

Explanation:

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

`\left[\begin{array}cc2&-4&-2\\3&2&3\end{array}\right]`

Dividing the first row by 2 makes it ...

`\left[\begin{array}c1&-2&-1\\3&2&3\end{array}\right]`

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

`\left[\begin{array}c1&-2&-1\\0&8&6\end{array}\right]`

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

`\left[\begin{array}c1&-2&-1\\0&1&(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 ...

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