Question

Can you help solve this home work problem in the picture I included

Answer 1

We are given the following system of equations:

`\begin{gathered} 5x+4y=-9 \\ -3x-2y=-4 \end{gathered}`

We can rewrite the system of equations using matrix notations. Matrix A is the matrix of coefficients on the left side, therefore:

`A=\begin{bmatrix}{5} & {4} \\ {-3} & -{2}\end{bmatrix}`

We get a 2x2 matrix of the form:

`A=\begin{bmatrix}{a} & {b} \\ {c} & {d}\end{bmatrix}`

The inverse of this type of matrix is:

`A^(-1)=(1)/(ad-bc)\begin{bmatrix}{d} & {-b} \\ {-c} & {a}\end{bmatrix}`

Applying the formula we get:

`A^(-1)=(1)/((5)(-2)-(4)(-3))\begin{bmatrix}{-2} & {-4} \\ {3} & {5}\end{bmatrix}`

Solving the operations:

`A^(-1)=(1)/(2)\begin{bmatrix}{-2} & {-4} \\ {3} & {5}\end{bmatrix}`

And thus we get the inverse of the matrix A.

The system of equations in matrix form is:

`Ax=b`

Where:

`\begin{gathered} x=\begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix} \\ \\ b=\begin{bmatrix}{-9} & {} \\ {-4} & {}\end{bmatrix} \end{gathered}`

The solution of the system is determined by left-multiplying both sides by the inverse of A

`x=A^(-1)b`

Substituting the matrices we get:

`\begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix}=(1)/(2)\begin{bmatrix}{-2} & {-4} \\ {3} & {5}\end{bmatrix}\begin{bmatrix}{-9} & {} \\ {-4} & {}\end{bmatrix}`

Now, we multiply the matrices. We multiply each row element of the inverse of "A" by its corresponding column element of the matrix "b".

`\begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix}=(1)/(2)\begin{bmatrix}{(-2)(-9)+(-4)(-4)} & {} \\ {(3)(-9)+(5)(-4)} & {}\end{bmatrix}`

Solving the operations we get:

`\begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix}=(1)/(2)\begin{bmatrix}{34} & {} \\ {-47} & {}\end{bmatrix}`

Therefore, the value of "x" is:

`x=(34)/(2)=17`

The value of "y" is:

`y=-(47)/(2)`