Question
Find the value of \(x\) for which \[ \begin{bmatrix} 2 & 0 & 7 \\ 0 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix} \begin{bmatrix} -x & 14 & 7x \\ 0 & 1 & 0 \\ x & -4x & -2x \end{bmatrix} = I \]
Solution
Step 1: Multiply Matrices
\[ = \begin{bmatrix} 2(-x)+0+7x & 2(14)+0+7(-4x) & 2(7x)+0+7(-2x) \\ 0+1+0 & 0+1+0 & 0+0+0 \\ 1(-x)+(-2)0+1(x) & 1(14)+(-2)(1)+1(-4x) & 1(7x)+0+1(-2x) \end{bmatrix} \]Step 2: Simplify
\[ = \begin{bmatrix} 5x & 28 – 28x & 0 \\ 0 & 1 & 0 \\ 0 & 12 – 4x & 5x \end{bmatrix} \]Step 3: Compare with Identity Matrix
\[ \begin{bmatrix} 5x & 28 – 28x & 0 \\ 0 & 1 & 0 \\ 0 & 12 – 4x & 5x \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]Step 4: Solve
\[ 5x = 1 \Rightarrow x = \frac{1}{5} \] \[ 28 – 28x = 0 \Rightarrow x = 1 \] \[ 12 – 4x = 0 \Rightarrow x = 3 \]Final Answer
No common value of \(x\) satisfies all conditions.
Hence, no such value of \(x\) exists.