Question

Solve the following system of equations using an inverse matrix. You must also indicate the inverse matrix, A^−1, that was used to solve the system. You may optionally write the inverse matrix with a scalar coefficient.

x−6y=−1
−2x+9y=5

Answer 1

The solution to the system of equations is x = −7 and y = −1.

Explanation:

To solve the given system of equations using an inverse matrix, we can represent the system in matrix form, Ax=B, then find the inverse of matrix A to solve for 'x'.

Given matrix:

`\left\{\begin{array}{ccc}x-6y=-1\\-2x+9y=5\end{array}\right`

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Step 1: Represent the System of Equations in Matrix Form
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First, we rewrite the system of equations in matrix for, Ax=B:

`A = \left[\begin{array}{cc}1&-6\\-2&9\\\end{array}\right], \ x = \left[\begin{array}{cc}x\\y\\\end{array}\right], \ B = \left[\begin{array}{cc}-1\\5\\\end{array}\right]`

So, Ax=B can be written as:

`\Longrightarrow \left[\begin{array}{cc}1&-6\\-2&9\\\end{array}\right] \left[\begin{array}{cc}x\\y\\\end{array}\right] = \left[\begin{array}{cc}-1\\5\\\end{array}\right]`

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Step 2: Find the Inverse of Matrix A
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`\boxed{\left\begin{array}{ccc}\text{The inverse of a 2x2 matrix,} \ \left(\begin{array}{ccc}a&b\\c&d\end{array}\right), \ \text{is given by:}\\\\\\A^(-1)= (1)/(ad-bc)\left(\begin{array}{ccc}d&-b\\-c&a\end{array}\right)\end{array}\right}`

Here,

`\Longrightarrow A = \left[\begin{array}{cc}1&-6\\-2&9\\\end{array}\right]`

The determinant ∣A∣ is:

`\Longrightarrow |A|=(1)(9)-(-6)(-2)=9-12\\\\\\\\\therefore |A|=-3`

Since the determinant is not zero, A has an inverse, which can be calculated as:

`\Longrightarrow A^(-1)= (1)/(-3)\left(\begin{array}{ccc}9&6\\2&1\end{array}\right)\\\\\\\\\therefore A^(-1)=\left(\begin{array}{ccc}-3&-2\\-2/3&-1/3\end{array}\right)`

`\hrulefill`

Step 3: Solve for 'x'
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We can solve for 'x' using x=A⁻¹B:

`\Longrightarrow x = \left(\begin{array}{ccc}-3&-2\\-2/3&-1/3\end{array}\right)\left(\begin{array}{cc}-1\\5\\\end{array}\right)`

Upon multiplication, we get:

`\Longrightarrow x = \left(\begin{array}{cc}(-3)(-1)+(-2)(5)\\(-2/3)(-1)+(-1/3)(5)\\\end{array}\right)\\\\\\\\\Longrightarrow x = \left(\begin{array}{cc}3-10\\2/3-5/3\\\end{array}\right)\\\\\\\\\Longrightarrow x = \left(\begin{array}{cc}-7\\-3/3\\\end{array}\right)\\\\\\\\\therefore \boxed{\boxed{x=\left(\begin{array}{cc}-7\\-1\\\end{array}\right)}}`

Thus, the solution to the system of equations is x = −7 and y = −1.