Question

Solve the system by using a matrix equation

Answer 1

a. (17,11)

Explanation:

The given system is ;

`6x-9y=3`

`3x-4y=7`

The augmented matrices is

`\left[\begin{array}{ccc}6&-9&|3\\3&-4&|7\end{array}\right]`

Divide Row 1 by 6

`\left[\begin{array}{ccc}1&-(3)/(2)&|(1)/(2)\\3&-4&|7\end{array}\right]`

Subtract 3 times Row 1 from Row 2

`\left[\begin{array}{ccc}1&-(3)/(2)&|(1)/(2)\\0&(1)/(2)&|(11)/(2)\end{array}\right]`

Divide Row 2 by
`(1)/(2)`

`\left[\begin{array}{ccc}1&-(3)/(2)&|(1)/(2)\\0&1}&|11\end{array}\right]`

Add
`(3)/(2)` times Row 2 to Row 1

`\left[\begin{array}{ccc}1&0&|17\\0&1}&|11\end{array}\right]`

Hence the solution is (17,11)

Answer 2

Option A is correct (17,11).

Explanation:

6x - 9y = 3

3x - 4y =7

it can be represented in matrix form as
`\left[\begin{array}{cc}6&-9\\3&4\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{c}3\\7\end{array}\right]`

A=
`\left[\begin{array}{cc}6&-9\\3&4\end{array}\right]`

X=
`\left[\begin{array}{c}x\\y\end{array}\right]`

B=
`\left[\begin{array}{c}3\\7\end{array}\right]`

i.e, AX=B

or X= A⁻¹ B

A⁻¹ = 1/|A| * Adj A

determinant of A = |A|= (6*-4) - (-9*3)

= (-24)-(-27)

= (-24) + 27 = 3

so, |A| = 3

Adj A=
`\left[\begin{array}{cc}-4&9\\-3&6\end{array}\right]`

A⁻¹ =
`\left[\begin{array}{cc}-4&9\\-3&6\end{array}\right]`/3

A⁻¹ =
`\left[\begin{array}{cc}-4/3&3\\-1&2\end{array}\right]`

X= A⁻¹ B

X=
`\left[\begin{array}{cc}-4/3&3\\-1&2\end{array}\right] *\left[\begin{array}{c}3\\7\end{array}\right]`

X=
`\left[\begin{array}{c}(-4/3*3) + (3*7)\\(-1*3) + (2*7)\end{array}\right]`

X=
`\left[\begin{array}{c}-4+21\\-3+14\end{array}\right]`

X=
`\left[\begin{array}{c}17\\11\end{array}\right]`

x= 17, y= 11

solution set= (17,11).