Question

If the inverse of $$ \begin{bmatrix} -x & 14x & 7x \\ 0 & 1 & 0 \\ x & -4x & -2x \end{bmatrix} $$ is $$ \begin{bmatrix} 2 & 0 & 7 \\ 0 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix} $$ then the value of $$ \begin{vmatrix} x & x+1 & x+2 \\ x+1 & x+2 & x+3 \\ x+2 & x+3 & x+4 \end{vmatrix} $$

• A. \( \frac{x}{5} \)
• B. \( x - 5 \)
• C. \( 5x - 1 \)
• D. \( x + 5 \)

Hint:

Use properties of inverse matrices: \( A A^{-1} = I \), and for pattern matrices like Toeplitz form, try row/column operations to simplify determinants quickly.

Answer 1

The Correct Option is C

Let the given matrix be \( A \), and its inverse be \( A^{-1} \). We know: \[ AA^{-1} = I \Rightarrow \text{Use this to solve for } x \] Given that matrix multiplication of \( A \) with its inverse equals identity matrix, one can use suitable matrix multiplication or comparison methods to find the value of \( x \). Once \( x \) is found, substitute it into the 3x3 matrix: \[ \begin{vmatrix} x & x+1 & x+2 \\ x+1 & x+2 & x+3 \\ x+2 & x+3 & x+4 \end{vmatrix} \] This is a known determinant form: \[ \text{Determinant} = 0 \text{ for linear dependent rows, otherwise use expansion} \] On expansion or substitution, it simplifies to \( 5x - 1 \)