Question

Find $x, y, z$ and $w$ given that $3 \begin{bmatrix} x & y \\z & w \end{bmatrix} = \begin{bmatrix} x & 5 \\-1 & 2w \end{bmatrix} +\begin{bmatrix} 6 & x+y \\z+w & 5 \end{bmatrix}$

• A. $x = 3,\ y = 4,\ z = 3,\ w = 5$
• B. $x = 3,\ y = 4,\ z = 5,\ w = 2$
• C. $x = 2,\ y = 4,\ z = 2,\ w = 5$
• D. $x = 3,\ y = 4,\ z = 2,\ w = 5$

Hint:

To solve matrix equations, equate corresponding elements to form a system of equations, solve systematically, and verify by substitution.

Answer 1

The Correct Option is D

Step 1: Set up the matrix equation. 
Given: \[ 3 \begin{bmatrix} x & y \\z & w \end{bmatrix} = \begin{bmatrix} x & 5 \\-1 & 2w \end{bmatrix} + \begin{bmatrix} 6 & x+y \\z+w & 5 \end{bmatrix}, \] first compute the right-hand side by adding the two matrices: \[ \begin{bmatrix} x & 5 \\-1 & 2w \end{bmatrix} + \begin{bmatrix} 6 & x+y \\z+w & 5 \end{bmatrix} = \begin{bmatrix} x + 6 & 5 + (x + y) \\-1 + (z + w) & 2w + 5 \end{bmatrix} = \begin{bmatrix} x + 6 & x + y + 5 \\z + w - 1 & 2w + 5 \end{bmatrix}. \] The equation becomes: \[ 3 \begin{bmatrix} x & y \\z & w \end{bmatrix} = \begin{bmatrix} x + 6 & x + y + 5 \\z + w - 1 & 2w + 5 \end{bmatrix}. \] 
Step 2: Apply the scalar multiplication and equate matrices. 
Multiply the left-hand side by 3: \[ 3 \begin{bmatrix} x & y \\z & w \end{bmatrix} = \begin{bmatrix} 3x & 3y \\3z & 3w \end{bmatrix}. \] Equate the two matrices: \[ \begin{bmatrix} 3x & 3y \\3z & 3w \end{bmatrix} = \begin{bmatrix} x + 6 & x + y + 5 \\z + w - 1 & 2w + 5 \end{bmatrix}. \] This gives four equations by equating corresponding elements:
1. \( 3x = x + 6 \),
2. \( 3y = x + y + 5 \),
3. \( 3z = z + w - 1 \),
4. \( 3w = 2w + 5 \).
Step 3: Solve the equations. 
- Equation 1: \( 3x = x + 6 \): \[ 3x - x = 6 \implies 2x = 6 \implies x = 3. \] - Equation 4: \( 3w = 2w + 5 \): \[ 3w - 2w = 5 \implies w = 5. \] - Equation 2: \( 3y = x + y + 5 \), substitute \( x = 3 \): \[ 3y = 3 + y + 5 \implies 3y = y + 8 \implies 3y - y = 8 \implies 2y = 8 \implies y = 4. \] - Equation 3: \( 3z = z + w - 1 \), substitute \( w = 5 \): \[ 3z = z + 5 - 1 \implies 3z = z + 4 \implies 3z - z = 4 \implies 2z = 4 \implies z = 2. \] 
Step 4: Verify the solution. 
The values are \( x = 3 \), \( y = 4 \), \( z = 2 \), \( w = 5 \). Substitute back into the original equation: Left-hand side: \( 3 \begin{bmatrix} 3 & 4 \\2 & 5 \end{bmatrix} = \begin{bmatrix} 9 & 12 \\6 & 15 \end{bmatrix} \),
Right-hand side: \( \begin{bmatrix} 3 & 5 \\-1 & 2 \cdot 5 \end{bmatrix} + \begin{bmatrix} 6 & 3 + 4 \\2 + 5 & 5 \end{bmatrix} = \begin{bmatrix} 3 & 5 \\-1 & 10 \end{bmatrix} + \begin{bmatrix} 6 & 7 \\7 & 5 \end{bmatrix} = \begin{bmatrix} 9 & 12 \\6 & 15 \end{bmatrix} \), 
Both sides match, confirming the solution. 
Step 5: Select the correct answer. 
The values \( x = 3 \), \( y = 4 \), \( z = 2 \), \( w = 5 \) match option (4).