Let $$A = \begin{pmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{pmatrix}$$ and $$10 B = \begin{pmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{pmatrix}$$. If $$B$$ is the inverse of matrix $$A$$, then $$\alpha$$ is

We are given that matrix $$ B $$ is the inverse of matrix $$ A $$.

By definition, multiplying a matrix by its inverse results in the identity matrix:

$$ A \cdot B = I $$

We are given the expression for $$ 10B $$. To simplify the calculation and avoid fractions, we can multiply both sides of the identity equation by 10:

$$ A \cdot (10B) = 10I $$

Step 1: Setting up the Matrix Equation

We substitute the given matrices into the equation:

$$ \begin{pmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{pmatrix} = \begin{pmatrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{pmatrix} $$

Step 2: Isolate the Equation containing $$ \alpha $$

To find the value of $$ \alpha $$, we only need to compute an entry in the product matrix where $$ \alpha $$ is involved.

Let us look at the element in Row 2, Column 3 of the resulting product matrix.

According to the matrix multiplication rules, Row 2 of matrix $$ A $$ multiplies Column 3 of matrix $$ 10B $$:

$$ \text{Row 2 of } A = \begin{pmatrix} 2 & 1 & -3 \end{pmatrix} $$

$$ \text{Column 3 of } 10B = \begin{pmatrix} 2 \\ \alpha \\ 3 \end{pmatrix} $$

The entry in Row 2, Column 3 of the identity matrix $$ 10I $$ is $$ 0 $$.

Step 3: Solving for $$ \alpha $$

We compute the dot product and set it equal to 0:

$$ (2)(2) + (1)(\alpha) + (-3)(3) = 0 $$

$$ 4 + \alpha - 9 = 0 $$

$$ \alpha - 5 = 0 $$

$$ \alpha = 5 $$

Therefore, the value of $$ \alpha $$ is 5.