9. If \( A=\left[\begin{array}{cc}x & -2 \\ 3 & 7\end{array}\right] \) and \( A^{-1}=\left[\begin{array}{cc}\frac{7}{34} & \frac{1}{17} \\ \frac{-3}{34} & \frac{2}{17}\end{array}\right] \), then the value of \( x \) is (a) 2 (b) 3 (c) \( -4 \) (d) 4
9. If \( A=\left[\begin{array}{cc}x & -2 \\ 3 & 7\end{array}\right] \) and \( A^{-1}=\left[\begin{array}{cc}\frac{7}{34} & \frac{1}{17} \\ \frac{-3}{34} & \frac{2}{17}\end{array}\right] \), then the value of \( x \) is (a) 2 (b) 3 (c) \( -4 \) (d) 4
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X=2 ans because
the second matrix is A^-1 therefore all terms in the matrix would be reversed
So now ans would go as
A^-1 = 7/34 so A=34 / 7
X=34/7
7x=34
X=34/7
X=2
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Answered
AA-1 = I
⇒ \(\begin{bmatrix}x&-2\\3&7\end{bmatrix}\) \(\begin{bmatrix}7/34&1/17\\-3/34&2/17\end{bmatrix}\) \(=\begin{bmatrix}1&0\\0&1\end{bmatrix}\)
⇒ \(\begin{bmatrix}7x+6/34&x-4/17\\0&1\end{bmatrix}\) \(=\begin{bmatrix}1&0\\0&1\end{bmatrix}\)
x-4/17 = 0
⇒ x - 4 = 0
⇒ x = 4 (By comparing a12 elements of both matrices)
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