Finding x, y, z by Equating Matrices
Question:
Find \( x, y, z \) so that \( A = B \), where
\[ A = \begin{bmatrix} x – 2 & 3 & 2x \\ 18z & y + 2 & 6z \end{bmatrix}, \quad B = \begin{bmatrix} y & z & 6 \\ 6y & x & 2y \end{bmatrix} \]
Concept Used
Two matrices are equal if their corresponding elements are equal.
Step 1: Equate Corresponding Elements
\[ x – 2 = y \quad …(1) \]
\[ 3 = z \quad …(2) \]
\[ 2x = 6 \quad …(3) \]
\[ 18z = 6y \quad …(4) \]
\[ y + 2 = x \quad …(5) \]
\[ 6z = 2y \quad …(6) \]
Step 2: Solve the Equations
From (3):
\[ 2x = 6 \Rightarrow x = 3 \]
From (2):
\[ z = 3 \]
Substitute into (6):
\[ 6z = 2y \Rightarrow 18 = 2y \Rightarrow y = 9 \]
Check consistency using (1):
\[ x – 2 = 3 – 2 = 1 \neq 9 \]
❌ This shows inconsistency, so we must instead use consistent equations:
From (5):
\[ y + 2 = x \]
Using (3): \( x = 3 \)
\[ y + 2 = 3 \Rightarrow y = 1 \]
Now check with (4):
\[ 18z = 6y \Rightarrow 18z = 6 \Rightarrow z = \frac{1}{3} \]
But from (2): \( z = 3 \), contradiction again.
Step 3: Correct Approach
Use consistent pair:
From (3): \( x = 3 \)
From (5): \( y + 2 = 3 \Rightarrow y = 1 \)
From (2): \( z = 3 \)
Check remaining equations:
\[ 18z = 54 \neq 6y = 6 \]
❌ Not satisfied ⇒ System is inconsistent.
Final Answer
No solution exists (matrices cannot be equal).