Questions: Find A if (I2 + 1/18 A)^-1 = [[3 0] [-1 1]]

Transcript text: Find $A$ if \[ \left(I_{2}+\frac{1}{18} A\right)^{-1}=\left[\begin{array}{cc} 3 & 0 \\ -1 & 1 \end{array}\right] \]

Solution

Solution Steps

Step 1: Identify the Given Equation

We are given the equation: \[ \left(I_{2}+\frac{1}{18} A\right)^{-1}=\begin{bmatrix} 3 & 0 \\ -1 & 1 \end{bmatrix} \] where $ I_2 $ is the $2 \times 2$ identity matrix.

Step 2: Find the Inverse of the Given Matrix

The inverse of the matrix on the right-hand side is: \[ B = \begin{bmatrix} 3 & 0 \\ -1 & 1 \end{bmatrix}^{-1} = \begin{bmatrix} \frac{1}{3} & 0 \\ \frac{1}{3} & 1 \end{bmatrix} \]

Step 3: Set Up the Equation for $ A $

Substitute $ B $ back into the equation: \[ I_2 + \frac{1}{18} A = B \] \[ A = 18(B - I_2) \]

Step 4: Calculate $ A $

Substitute the values of $ B $ and $ I_2 $: \[ A = 18 \left( \begin{bmatrix} \frac{1}{3} & 0 \\ \frac{1}{3} & 1 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right) \] \[ A = 18 \begin{bmatrix} \frac{1}{3} - 1 & 0 \\ \frac{1}{3} - 0 & 1 - 1 \end{bmatrix} \] \[ A = 18 \begin{bmatrix} -\frac{2}{3} & 0 \\ \frac{1}{3} & 0 \end{bmatrix} \] \[ A = \begin{bmatrix} -12 & 0 \\ 6 & 0 \end{bmatrix} \]