Question

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

Answer 1

HERE'S YOUR ANSWER. ALL THE BEST FOR EXAM!!

Answer 2

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

Answer 3

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)