Question

If $A = \begin{pmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & x & 1 \end{pmatrix}$, $A^{-1} = \frac{1}{2} \begin{pmatrix} 1 & -1 & 1 \\ -8 & 6 & 2y \\ 5 & -3 & 1 \end{pmatrix}$ then the point $(x,y)$ lies on the curve

• A. $y = 3x^2 - 5x - 1$
• B. $y = \log_{7/5}(2^x + 2^{-x})$
• C. $y = \frac{e^x + 1}{e^x - 1}$
• D. $3x^2y - 5xy + 12 = 0$

Hint:

Whenever a matrix inverse is given, use $AA^{-1} = I$ to form equations. Multiply the matrices and compare each element with the identity matrix to solve for unknowns.

Answer 1

The Correct Option is B

Step 1: Apply the inverse property:
Using the identity $AA^{-1} = I$, and given \[ A^{-1} = \frac{1}{2} M, \] we get \[ A \cdot \frac{1}{2}M = I \Rightarrow AM = 2I. \]
Step 2: Compute the product $AM$:
Given \[ A = \begin{pmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & x & 1 \end{pmatrix}, M = \begin{pmatrix} 1 & -1 & 1 \\ -8 & 6 & 2y \\ 5 & -3 & 1 \end{pmatrix}, \] performing multiplication, we obtain \[ AM = \begin{pmatrix} 2 & 0 & 2y + 2 \\ 0 & 2 & 4y + 4 \\ 8 - 8x & 6x - 6 & 4 + 2xy \end{pmatrix}. \]
Step 3: Compare with $2I$:
Since \[ 2I = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix}, \] equating corresponding entries gives \[ 2y + 2 = 0 \Rightarrow y = -1, \] \[ 8 - 8x = 0 \Rightarrow x = 1. \]
Step 4: Verify remaining entries:
Substituting $x=1$ and $y=-1$: \[ 6x - 6 = 0, 4y + 4 = 0, 4 + 2xy = 2, \] which confirms consistency. Hence, \[ (x, y) = (1, -1). \]
Step 5: Final conclusion:
Thus, the values of $x$ and $y$ are $(1, -1)$, which corresponds to the correct option.