Math Problem Statement

Tentukan nilai x, y, dan z dari kesamaan matriks berikut:

Solution

Let's work through the matrix equations provided in the image step by step to find the values of xx, yy, and zz.

a.

[x31yz]=[4x1z2]\begin{bmatrix} x & 3 \\ 1 - y & z \end{bmatrix} = \begin{bmatrix} 4 & x - 1 \\ z & -2 \end{bmatrix} Matching the elements of the two matrices:

  • First row, first column: x=4x = 4
  • First row, second column: 3=x13 = x - 1, so x=4x = 4 (consistent with the previous result)
  • Second row, first column: 1y=z1 - y = z
  • Second row, second column: z=2z = -2

From this, z=2z = -2, and substituting z=2z = -2 into 1y=z1 - y = z, we get: 1y=2y=31 - y = -2 \quad \Rightarrow \quad y = 3

Thus, the solution for part (a) is: x=4,y=3,z=2x = 4, \quad y = 3, \quad z = -2


b.

[92yz2]=[x2x1yz]\begin{bmatrix} 9 & 2 \\ -y & z^2 \end{bmatrix} = \begin{bmatrix} x^2 & x - 1 \\ y & z \end{bmatrix} Matching the elements:

  • First row, first column: 9=x29 = x^2, so x=3x = 3 (taking the positive square root)
  • First row, second column: 2=x12 = x - 1, so x=3x = 3 (consistent with the previous result)
  • Second row, first column: y=y-y = y, which implies y=0y = 0
  • Second row, second column: z2=zz^2 = z

The equation z2=zz^2 = z can be solved as: z(z1)=0z=0 or z=1z(z - 1) = 0 \quad \Rightarrow \quad z = 0 \text{ or } z = 1

Thus, the solution for part (b) is: x=3,y=0,z=0 or z=1x = 3, \quad y = 0, \quad z = 0 \text{ or } z = 1


c.

[x523yz]=[2x112y+9z]\begin{bmatrix} x - 5 & 2 \\ 3 - y & z \end{bmatrix} = \begin{bmatrix} 2x - 11 & 2 \\ y + 9 & z \end{bmatrix} Matching the elements:

  • First row, first column: x5=2x11x - 5 = 2x - 11, solving for xx: x5=2x115+11=xx=6x - 5 = 2x - 11 \quad \Rightarrow \quad -5 + 11 = x \quad \Rightarrow \quad x = 6
  • First row, second column: 2=22 = 2 (this is consistent)
  • Second row, first column: 3y=y+93 - y = y + 9, solving for yy: 3y=y+939=2yy=33 - y = y + 9 \quad \Rightarrow \quad 3 - 9 = 2y \quad \Rightarrow \quad y = -3
  • Second row, second column: z=zz = z (this is consistent)

Thus, the solution for part (c) is: x=6,y=3x = 6, \quad y = -3


Final Answers:

  • a: x=4x = 4, y=3y = 3, z=2z = -2
  • b: x=3x = 3, y=0y = 0, z=0 or 1z = 0 \text{ or } 1
  • c: x=6x = 6, y=3y = -3

Would you like further details on any of these steps or have additional questions?

Here are five related questions to deepen your understanding:

  1. What is the general method for solving a system of matrix equations?
  2. How would the solution change if the matrices were not square?
  3. Can there be multiple solutions to a matrix equation, and why?
  4. What does it mean for two matrices to be equal, in terms of their elements?
  5. How can matrices be applied to solve real-world problems, such as systems of linear equations?

Tip: When solving matrix equations, always ensure each corresponding element is compared directly, and watch for consistent variable values.

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Math Problem Analysis

Mathematical Concepts

Matrix Equality
Linear Algebra
Algebraic Equations

Formulas

Matrix equality: for two matrices A = B, all corresponding elements must be equal
Solving systems of equations derived from matrix equality

Theorems

Matrix Equality Theorem

Suitable Grade Level

Grades 10-12