Question

If \( \begin{bmatrix} x+y & 2 \\ 5+z & xy \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix} \), then find the values of \(x, y, z\).

Hint:

When solving systems of equations derived from matrices, be systematic. Solve for the simplest variables first (like 'z' in this case). For a system like \(x+y=a\) and \(xy=b\), you are essentially looking for two numbers whose sum is 'a' and product is 'b'. This corresponds to finding the roots of the quadratic equation \(t^2 - at + b = 0\).

Answer 1

Step 1: Understanding the Concept:
This problem is based on the principle of equality of matrices. Two matrices are equal if and only if they have the same dimensions and their corresponding elements are equal.
Step 2: Key Formula or Approach:
Given the matrix equation \( A = B \), where \( A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \) and \( B = \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix} \), the equality implies:
- \( a_{11} = b_{11} \)
- \( a_{12} = b_{12} \)
- \( a_{21} = b_{21} \)
- \( a_{22} = b_{22} \)
We will set up a system of equations based on this principle and solve for the variables \(x\), \(y\), and \(z\).
Step 3: Detailed Explanation or Calculation:
The given matrix equation is:
\[ \begin{bmatrix} x+y & 2 \\ 5+z & xy \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix} \] By equating the corresponding elements, we get the following system of equations:
1. \( x + y = 6 \)
2. \( 2 = 2 \) (This is consistent and provides no new information)
3. \( 5 + z = 5 \)
4. \( xy = 8 \)
Solve for z:
From equation (3):
\[ 5 + z = 5 \]
\[ z = 5 - 5 \]
\[ z = 0 \]
Solve for x and y:
We have a system of two equations with two variables:
(1) \( x + y = 6 \)
(4) \( xy = 8 \)
From equation (1), express \( y \) in terms of \( x \):
\[ y = 6 - x \]
Substitute this expression for \( y \) into equation (4):
\[ x(6 - x) = 8 \]
\[ 6x - x^2 = 8 \]
Rearrange into a standard quadratic equation form \( ax^2 + bx + c = 0 \):
\[ x^2 - 6x + 8 = 0 \]
Factor the quadratic equation:
\[ (x - 4)(x - 2) = 0 \]
This gives two possible values for \( x \):
\( x = 4 \) or \( x = 2 \).
Now find the corresponding values for \( y \) using \( y = 6 - x \):
Case 1: If \( x = 4 \)
\[ y = 6 - 4 = 2 \]
Case 2: If \( x = 2 \)
\[ y = 6 - 2 = 4 \]
So, we have two possible sets of solutions for \( (x, y) \).
Step 4: Final Answer:
The values of the variables are:
- \( z = 0 \)
- Either \( x = 4, y = 2 \) or \( x = 2, y = 4 \).
The complete solutions are \( (x=4, y=2, z=0) \) and \( (x=2, y=4, z=0) \).