Question

1. write the system as a single matrix equation and solve:

$4x - 2y = -16$
$-5x + 3y = -2$

Explanation:

Step1: Write the matrix equation

A system of linear equations \(a_1x + b_1y = c_1\) and \(a_2x + b_2y = c_2\) can be written as \(

$$\begin{bmatrix}a_1&b_1\\a_2&b_2\end{bmatrix}$$

$$\begin{bmatrix}x\\y\end{bmatrix}$$

=

$$\begin{bmatrix}c_1\\c_2\end{bmatrix}$$

\). For the system \(4x - 2y=- 16\) and \(-5x + 3y=-2\), the matrix equation is \(

$$\begin{bmatrix}4&-2\\-5&3\end{bmatrix}$$

$$\begin{bmatrix}x\\y\end{bmatrix}$$

=

$$\begin{bmatrix}-16\\-2\end{bmatrix}$$

\).

Step2: Find the inverse of the coefficient matrix

The inverse of a \(2\times2\) matrix \(

$$\begin{bmatrix}a&b\\c&d\end{bmatrix}$$

\) is \(\frac{1}{ad - bc}

$$\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$$

\), provided \(ad - bc
eq0\). For the matrix \(A=

$$\begin{bmatrix}4&-2\\-5&3\end{bmatrix}$$

\), the determinant \(\det(A)=(4\times3)-(-2\times - 5)=12 - 10 = 2\). So the inverse \(A^{-1}=\frac{1}{2}

$$\begin{bmatrix}3&2\\5&4\end{bmatrix}$$

=

$$\begin{bmatrix}\frac{3}{2}&1\\\frac{5}{2}&2\end{bmatrix}$$

\).

Step3: Multiply both sides by the inverse

Multiply both sides of the matrix equation \(A

$$\begin{bmatrix}x\\y\end{bmatrix}$$

=

$$\begin{bmatrix}-16\\-2\end{bmatrix}$$

\) by \(A^{-1}\). We get \(A^{-1}A

$$\begin{bmatrix}x\\y\end{bmatrix}$$

=A^{-1}

$$\begin{bmatrix}-16\\-2\end{bmatrix}$$

\). Since \(A^{-1}A = I\) (the identity matrix), \(

$$\begin{bmatrix}x\\y\end{bmatrix}$$

=A^{-1}

$$\begin{bmatrix}-16\\-2\end{bmatrix}$$

\).

Now, calculate the product:
\[

$$\begin{bmatrix}\frac{3}{2}&1\\\frac{5}{2}&2\end{bmatrix}$$

$$\begin{bmatrix}-16\\-2\end{bmatrix}$$

=

$$\begin{bmatrix}\frac{3}{2}\times(-16)+1\times(-2)\\\frac{5}{2}\times(-16)+2\times(-2)\end{bmatrix}$$

=

$$\begin{bmatrix}-24 - 2\\-40 - 4\end{bmatrix}$$

=

$$\begin{bmatrix}-26\\-44\end{bmatrix}$$

?
\]
Wait, that's a mistake. Let's recalculate the product:

For the first element of the resulting vector:
\(\frac{3}{2}\times(-16)+1\times(-2)=\frac{-48}{2}-2=-24 - 2=-26\)? No, wait, the inverse calculation was wrong. Wait, the inverse of \(

$$\begin{bmatrix}a&b\\c&d\end{bmatrix}$$

\) is \(\frac{1}{ad - bc}

$$\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$$

\). So for \(A=

$$\begin{bmatrix}4&-2\\-5&3\end{bmatrix}$$

\), \(a = 4\), \(b=-2\), \(c=-5\), \(d = 3\). So \(ad - bc=4\times3-(-2)\times(-5)=12 - 10 = 2\). Then \(A^{-1}=\frac{1}{2}

$$\begin{bmatrix}3&2\\5&4\end{bmatrix}$$

\) (because \(-b = 2\), \(-c = 5\)). Now, multiply \(A^{-1}\) with \(

$$\begin{bmatrix}-16\\-2\end{bmatrix}$$

\):

First row: \(\frac{3}{2}\times(-16)+1\times(-2)=\frac{-48}{2}-2=-24 - 2=-26\)? Wait, that can't be right. Wait, let's check the original equations. Let's solve the system using elimination.

From the first equation \(4x-2y=-16\), we can simplify it to \(2x - y=-8\), so \(y = 2x + 8\). Substitute into the second equation: \(-5x+3(2x + 8)=-2\). \(-5x + 6x+24=-2\), \(x+24=-2\), \(x=-26\). Then \(y=2\times(-26)+8=-52 + 8=-44\). Wait, but let's check the matrix multiplication again.

Wait, the inverse calculation: \(A^{-1}=\frac{1}{2}

$$\begin{bmatrix}3&2\\5&4\end{bmatrix}$$

\). Then:

\(\frac{3}{2}\times(-16)=-24\), \(1\times(-2)=-2\), sum is \(-26\) (for \(x\))? Wait, but when we solved by substitution, we got \(x = -26\)? Wait, let's check the first equation: \(4\times(-26)-2y=-16\), \(-104-2y=-16\), \(-2y=88\), \(y=-44\). Then the second equation: \(-5\times(-26)+3\times(-44)=130 - 132=-2\), which is correct. So the calculation is correct.

Answer:

The matrix equation is \(

$$\begin{bmatrix}4&-2\\-5&3\end{bmatrix}$$

$$\begin{bmatrix}x\\y\end{bmatrix}$$

=

$$\begin{bmatrix}-16\\-2\end{bmatrix}$$

\) and the solution is \(x=-26\), \(y = -44\) (or in vector form \(

$$\begin{bmatrix}x\\y\end{bmatrix}$$

=

$$\begin{bmatrix}-26\\-44\end{bmatrix}$$

\))