2x + y = 7
x - 2y = 1
A. Write the equation in matrix form.
B. Determine the inverse of the matrix
C. Hence solve the equations.
D. x and y are matrices
X=[1537]Y=[3421]X=\begin{bmatrix} 1 & 5 \\ 3 & 7 \end{bmatrix} Y=\begin{bmatrix} 3 & 4 \\ 2 & 1 \end{bmatrix}X=[1357]Y=[3241]
Evaluate X2 + Y
A:[211−2][xy]=[71]B:[211−2]=12⋅(−2)−1⋅1[−2−1−12]=[0.40.20.2−0.4]C:[xy]=[0.40.20.2−0.4][71]=[31]D:X2+Y=[1537][1537]+[3421]=[16402464]+[3421]=[19442665]A:\\\left[ \begin{matrix} 2& 1\\ 1& -2\\\end{matrix} \right] \left[ \begin{array}{c} x\\ y\\\end{array} \right] =\left[ \begin{array}{c} 7\\ 1\\\end{array} \right] \\B:\\\left[ \begin{matrix} 2& 1\\ 1& -2\\\end{matrix} \right] =\frac{1}{2\cdot \left( -2 \right) -1\cdot 1}\left[ \begin{matrix} -2& -1\\ -1& 2\\\end{matrix} \right] =\left[ \begin{matrix} 0.4& 0.2\\ 0.2& -0.4\\\end{matrix} \right] \\C:\\\left[ \begin{array}{c} x\\ y\\\end{array} \right] =\left[ \begin{matrix} 0.4& 0.2\\ 0.2& -0.4\\\end{matrix} \right] \left[ \begin{array}{c} 7\\ 1\\\end{array} \right] =\left[ \begin{array}{c} 3\\ 1\\\end{array} \right] \\D:\\X^2+Y=\left[ \begin{matrix} 1& 5\\ 3& 7\\\end{matrix} \right] \left[ \begin{matrix} 1& 5\\ 3& 7\\\end{matrix} \right] +\left[ \begin{matrix} 3& 4\\ 2& 1\\\end{matrix} \right] =\left[ \begin{matrix} 16& 40\\ 24& 64\\\end{matrix} \right] +\left[ \begin{matrix} 3& 4\\ 2& 1\\\end{matrix} \right] =\left[ \begin{matrix} 19& 44\\ 26& 65\\\end{matrix} \right]A:[211−2][xy]=[71]B:[211−2]=2⋅(−2)−1⋅11[−2−1−12]=[0.40.20.2−0.4]C:[xy]=[0.40.20.2−0.4][71]=[31]D:X2+Y=[1357][1357]+[3241]=[16244064]+[3241]=[19264465]
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