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+1 vote
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in Applications of Matrices and Determinants by (48.8k points)

Given A = \(\begin{pmatrix} 1 &-2\\[0.3em] -5&7 \\[0.3em] \end{pmatrix}\) then A (adj A) = ...

(a) \(\begin{pmatrix} 1 &0 \\[0.3em] 0&1 \\[0.3em] \end{pmatrix}\)

(b) (1/17)\(\begin{pmatrix} 1 &0 \\[0.3em] 0&1 \\[0.3em] \end{pmatrix}\)

(c) (1/3)\(\begin{pmatrix} 1 &0 \\[0.3em] 0&1 \\[0.3em] \end{pmatrix}\)

(d) (1/-3)\(\begin{pmatrix} 1 &0 \\[0.3em] 0&1 \\[0.3em] \end{pmatrix}\)

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by (49.5k points)

(d) (1/-3)\(\begin{pmatrix} 1 &0 \\[0.3em] 0&1 \\[0.3em] \end{pmatrix}\)

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