If $A = \begin{bmatrix} 2 & 7 \\ 1 & 5 \end{bmatrix}$, then what is $A + 3A^{-1}$ equal to, where $A$ is a matrix of order 2?
7I
Given the matrix:
$$A = \begin{bmatrix} 2 & 7 \\ 1 & 5 \end{bmatrix}$$
Step 1: Find the determinant of \(A\)
$$|A| = (2 \times 5) - (7 \times 1) = 10 - 7 = 3$$
Step 2: Find the inverse of \(A\) (\(A^{-1}\))
For a \(2 \times 2\) matrix, \(A^{-1} = \frac{1}{|A|} \text{adj}(A)\):
$$A^{-1} = \frac{1}{3} \begin{bmatrix} 5 & -7 \\ -1 & 2 \end{bmatrix}$$
Step 3: Calculate \(3A^{-1}\)
$$3A^{-1} = 3 \times \frac{1}{3} \begin{bmatrix} 5 & -7 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & -7 \\ -1 & 2 \end{bmatrix}$$
Step 4: Calculate \(A + 3A^{-1}\)
$$A + 3A^{-1} = \begin{bmatrix} 2 & 7 \\ 1 & 5 \end{bmatrix} + \begin{bmatrix} 5 & -7 \\ -1 & 2 \end{bmatrix}$$
$$A + 3A^{-1} = \begin{bmatrix} 2+5 & 7-7 \\ 1-1 & 5+2 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$$
Final Answer:
$$A + 3A^{-1} = 7I$$
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