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[[-22],[27]]=[[2,4],[-3,-7]]Y+[[-8],[7]]

[2227]=[2amp;43amp;7]Y+[87] \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]

Full solution

Q. [2227]=[2437]Y+[87] \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]
  1. 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 Y\mathbf{Y}. Let's denote the matrix \begin{bmatrix}2 & 4\-3 & -7\end{bmatrix} as A\mathbf{A} and the matrix Y\mathbf{Y} as Y\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}
  2. Subtracting Matrices: First, we need to subtract the matrix [[8],[7]][[-8],[7]] from the matrix [[22],[27]][[-22],[27]] to isolate AYA * Y on one side of the equation.\newline[[-22],[27]] - [[-8],[7]] = [[-22 - (-8)],[27 - 7]]\(\newline= [[-22 + 8],[20]]\newline= [[-14],[20]]\)
  3. Finding Inverse of Matrix: Now we have the equation AY=[14 20]A * Y = \left[\begin{array}{c} -14 \ 20 \end{array}\right]. We need to find the inverse of matrix AA, which is [2amp;4 3amp;7]\left[\begin{array}{cc} 2 & 4 \ -3 & -7 \end{array}\right], in order to solve for YY.
  4. Calculating Determinant: The inverse of a 2×22 \times 2 matrix [aamp;b camp;d]\left[\begin{array}{cc} a & b \ c & d \end{array}\right] is given by (1adbc)×[damp;b camp;a]\left(\frac{1}{ad-bc}\right) \times \left[\begin{array}{cc} d & -b \ -c & a \end{array}\right]. Let's calculate the determinant (adbc)(ad-bc) for matrix AA.\newlineDeterminant = (2×7)(4×3)(2 \times -7) - (4 \times -3)\newline= 14+12-14 + 12\newline= 2-2
  5. Calculating Inverse: Since the determinant is not zero, matrix AA is invertible. Now we can calculate the inverse of AA.A1=(1Determinant)[7amp;4 3amp;2]A^{-1} = \left(\frac{1}{\text{Determinant}}\right) * \begin{bmatrix}-7 & -4 \ 3 & 2\end{bmatrix}=(12)[7amp;4 3amp;2]= \left(-\frac{1}{2}\right) * \begin{bmatrix}-7 & -4 \ 3 & 2\end{bmatrix}=[72amp;42 32amp;22]= \begin{bmatrix}\frac{7}{2} & \frac{4}{2} \ -\frac{3}{2} & -\frac{2}{2}\end{bmatrix}=[3.5amp;2 1.5amp;1]= \begin{bmatrix}3.5 & 2 \ -1.5 & -1\end{bmatrix}
  6. Multiplying Matrices: Now that we have A1A^{-1}, we can multiply it by the matrix \begin{bmatrix}-14\20\end{bmatrix} to solve for YY.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}
  7. Matrix Multiplication: Let's perform the matrix multiplication to find YY.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}49-49 + 4040(-21\) - 2020\end{array}\right]= \left[\begin{array}{c}9-9(-41\)\end{array}\right]

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