Find the values of x,y and z from the following matrix equations :
`[(x+y,2),(5+z,xy)]=[(6,2),(5,8)]`

Answer

To solve the matrix equation

\[ \begin{pmatrix} x+y & 2 \\ 5+z & xy \end{pmatrix} = \begin{pmatrix} 6 & 2 \\ 5 & 8 \end{pmatrix} \]

we will compare the corresponding elements of the matrices.

### Step 1: Set up the equations from the matrix equality

From the equality of the matrices, we can derive the following equations:

1. \( x + y = 6 \) (from the first row, first column)
2. \( 2 = 2 \) (from the first row, second column, which is always true)
3. \( 5 + z = 5 \) (from the second row, first column)
4. \( xy = 8 \) (from the second row, second column)

### Step 2: Solve for \( z \)

From equation (3):

\[ 5 + z = 5 \]

Subtracting 5 from both sides gives:

\[ z = 0 \]

### Step 3: Solve for \( x \) and \( y \)

Now we have two equations to work with:

1. \( x + y = 6 \) (equation 1)
2. \( xy = 8 \) (equation 4)

From equation (1), we can express \( y \) in terms of \( x \):

\[ y = 6 - x \]

### Step 4: Substitute \( y \) in equation (4)

Substituting \( y \) in equation (4):

\[ x(6 - x) = 8 \]

Expanding this gives:

\[ 6x - x^2 = 8 \]

Rearranging the equation:

\[ -x^2 + 6x - 8 = 0 \]

Multiplying through by -1 to make the leading coefficient positive:

\[ x^2 - 6x + 8 = 0 \]

### Step 5: Factor the quadratic equation

Now we can factor the quadratic:

\[ (x - 2)(x - 4) = 0 \]

Setting each factor to zero gives us:

\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]
\[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \]

### Step 6: Find corresponding values of \( y \)

Now we can find the corresponding values of \( y \) for each \( x \):

1. If \( x = 2 \):
\[ y = 6 - 2 = 4 \]

2. If \( x = 4 \):
\[ y = 6 - 4 = 2 \]

### Step 7: Summarize the solutions

We have two pairs of solutions for \( (x, y) \):

1. \( (x, y) = (2, 4) \)
2. \( (x, y) = (4, 2) \)

And we already found \( z = 0 \).

### Final Answer

The values are:
- \( x = 2, y = 4, z = 0 \)
- or
- \( x = 4, y = 2, z = 0 \)