$\left[\begin{array}{c}-22 \\ 27\end{array}\right]=\left[\begin{array}{cc}2 & 4 \\ -3 & -7\end{array}\right] Y+\left[\begin{array}{c}-8 \\ 7\end{array}\right]$

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\left[\begin{array}{c}-22 \\ 27\end{array}\right]=\left[\begin{array}{cc}2 & 4 \\ -3 & -7\end{array}\right] Y+\left[\begin{array}{c}-8 \\ 7\end{array}\right]

Given Matrix Equation: We are given the matrix equation \begin{bmatrix}-22\27\end{bmatrix} = \begin{bmatrix}2 & 4\-3 & -7\end{bmatrix}\mathbf{Y} + \begin{bmatrix}-8\7\end{bmatrix}, and we need to solve for the matrix $\mathbf{Y}$ . Let's denote the matrix \begin{bmatrix}2 & 4\-3 & -7\end{bmatrix} as $\mathbf{A}$ and the matrix $\mathbf{Y}$ as $\mathbf{Y}$ . The equation can be rewritten as: \mathbf{A} \cdot \mathbf{Y} = \begin{bmatrix}-22\27\end{bmatrix} - \begin{bmatrix}-8\7\end{bmatrix}
Subtracting Matrices: First, we need to subtract the matrix $[[-8],[7]]$ from the matrix $[[-22],[27]]$ to isolate $A * Y$ on one side of the equation. $\newline$ [[-22],[27]] - [[-8],[7]] = [[-22 - (-8)],[27 - 7]]$\newline= [[-22 + 8],[20]] $\newline$ = [[-14],[20]]$
Finding Inverse of Matrix: Now we have the equation $A * Y = \left[\begin{array}{c} -14 \ 20 \end{array}\right]$ . We need to find the inverse of matrix $A$ , which is $\left[\begin{array}{cc} 2 & 4 \ -3 & -7 \end{array}\right]$ , in order to solve for $Y$ .
Calculating Determinant: The inverse of a $2 \times 2$ matrix $\left[\begin{array}{cc} a & b \ c & d \end{array}\right]$ is given by $\left(\frac{1}{ad-bc}\right) \times \left[\begin{array}{cc} d & -b \ -c & a \end{array}\right]$ . Let's calculate the determinant $(ad-bc)$ for matrix $A$ . $\newline$ Determinant = $(2 \times -7) - (4 \times -3)$ $\newline$ = $-14 + 12$ $\newline$ = $-2$
Calculating Inverse: Since the determinant is not zero, matrix $A$ is invertible. Now we can calculate the inverse of $A$ . $A^{-1} = \left(\frac{1}{\text{Determinant}}\right) * \begin{bmatrix}-7 & -4 \ 3 & 2\end{bmatrix}$ $= \left(-\frac{1}{2}\right) * \begin{bmatrix}-7 & -4 \ 3 & 2\end{bmatrix}$ $= \begin{bmatrix}\frac{7}{2} & \frac{4}{2} \ -\frac{3}{2} & -\frac{2}{2}\end{bmatrix}$ $= \begin{bmatrix}3.5 & 2 \ -1.5 & -1\end{bmatrix}$
Multiplying Matrices: Now that we have $A^{-1}$ , we can multiply it by the matrix \begin{bmatrix}-14\20\end{bmatrix} to solve for $Y$ .Y = A^{-1} \times \begin{bmatrix}-14\20\end{bmatrix} = \begin{bmatrix}3.5 & 2\-1.5 & -1\end{bmatrix} \times \begin{bmatrix}-14\20\end{bmatrix}
Matrix Multiplication: Let's perform the matrix multiplication to find $Y$ .Y = \left[\begin{array}{c}3.5 \times -14 + 2 \times 20\-1.5 \times -14 - 1 \times 20\end{array}\right]= \left[\begin{array}{c} $-49$ + $40$ (-21\) - $20$ \end{array}\right]= \left[\begin{array}{c} $-9$ (-41\)\end{array}\right]