`A=[(2,0,1),(1,1,0),(1,0,1)],AU_(1)=[(1),(0),(0)]_(3xx1)'AU_(2)=[(2),(3),(0)]_(3xx1)` and `AU_(3)=[(3),(2),(1)]_(3xx1)`
If `U_(1),U_(2),U_(3)` are columns of matrix U, then
Sum of elements of `U^(-1)` is

Options

A. `-1/3`
B. `1/12`
C. `-1/4`
D. `0`

Answer

To solve the problem, we need to find the sum of the elements of the inverse of matrix \( U \). We will follow these steps:

### Step 1: Define the Matrix A and the Columns of U
Given the matrix \( A \) and the column vectors \( U_1, U_2, U_3 \):

\[ A = \begin{pmatrix} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix} \]

Let \( U = \begin{pmatrix} U_1 & U_2 & U_3 \end{pmatrix} = \begin{pmatrix} A & B & C \\ D & E & F \\ G & H & I \end{pmatrix} \)

### Step 2: Set Up the Equations
We will use the given equations \( AU_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \), \( AU_2 = \begin{pmatrix} 2 \\ 3 \\ 0 \end{pmatrix} \), and \( AU_3 = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} \) to derive the values of \( U_1, U_2, U_3 \).

### Step 3: Solve for \( U_1 \)
From \( AU_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \):

1. The first equation: \( 2A + 0 + G = 1 \) → \( 2A + G = 1 \)
2. The second equation: \( A + D + 0 = 0 \) → \( A + D = 0 \)
3. The third equation: \( A + 0 + G = 0 \) → \( A + G = 0 \)

From these equations, we can express \( D \) and \( G \) in terms of \( A \):

- From \( A + D = 0 \), we have \( D = -A \).
- From \( A + G = 0 \), we have \( G = -A \).

Substituting \( G \) into the first equation:

\[ 2A - A = 1 \implies A = 1 \]

Thus, \( D = -1 \) and \( G = -1 \).

### Step 4: Solve for \( U_2 \)
From \( AU_2 = \begin{pmatrix} 2 \\ 3 \\ 0 \end{pmatrix} \):

1. The first equation: \( 2B + H = 2 \)
2. The second equation: \( B + E + 0 + H = 3 \)
3. The third equation: \( B + 0 + H = 0 \)

From the third equation, we have \( B + H = 0 \) → \( H = -B \).

Substituting \( H \) into the first equation:

\[ 2B - B = 2 \implies B = 2 \]
Then, substituting \( B \) into \( H \):

\[ H = -2 \]

Now substituting \( B \) into the second equation:

\[ 2 + E - 2 = 3 \implies E = 3 \]

### Step 5: Solve for \( U_3 \)
From \( AU_3 = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} \):

1. The first equation: \( 2C + I = 3 \)
2. The second equation: \( C + F + 0 + I = 2 \)
3. The third equation: \( C + 0 + I = 1 \)

From the third equation, we have \( C + I = 1 \) → \( I = 1 - C \).

Substituting \( I \) into the first equation:

\[ 2C + (1 - C) = 3 \implies C + 1 = 3 \implies C = 2 \]

Then substituting \( C \) into \( I \):

\[ I = 1 - 2 = -1 \]

Now substituting \( C \) into the second equation:

\[ 2 + F - 1 = 2 \implies F = 1 \]

### Step 6: Construct the Matrix U
Now we have:

\[ U = \begin{pmatrix} 1 & 2 & 2 \\ -1 & 3 & 1 \\ -1 & -2 & -1 \end{pmatrix} \]

### Step 7: Find the Inverse of U
To find \( U^{-1} \), we use the formula \( U^{-1} = \frac{1}{\text{det}(U)} \cdot \text{adj}(U) \).

1. Calculate the determinant \( \text{det}(U) \).
2. Find the adjugate \( \text{adj}(U) \).
3. Compute \( U^{-1} \).

### Step 8: Calculate the Sum of Elements of \( U^{-1} \)
Finally, sum all the elements of \( U^{-1} \).