Solving for Matrix B in the Equation D CBE

This article explores the solution to a specific matrix equation, D CBE, using given matrices and operations. We will start by breaking down the equation and solving for matrix B. The key matrices and their values are defined, and the detailed step-by-step solution is provided.

Understanding the Matrices

We are given the equation D CBE, and specific values for matrices D, C, and E. It is also given that D 3B and C B2. The matrix E is defined as:

[{ E begin{bmatrix} 2 -1 0 1 -1 1 end{bmatrix} }]

The matrices D, C, and B are unknown, and we aim to solve for B. Let's rewrite the given equations using the provided definitions.

Substitute and Simplify

First, substitute C B2 and D 3B into the original equation:

[3B B^2 E]

Divide both sides by B (assuming B is invertible):

[3I B E]

Rewrite this as:

[B 3I - E]

Given that I is the 3x3 unit matrix, we can write:

[I begin{bmatrix} 1 0 0 0 1 0 0 0 1 end{bmatrix}]

Now, calculate B:

[B 3I - E] [B 3begin{bmatrix} 1 0 0 0 1 0 0 0 1 end{bmatrix} - begin{bmatrix} 2 -1 0 1 -1 1 end{bmatrix}] [B begin{bmatrix} 3 0 0 0 3 0 0 0 3 end{bmatrix} - begin{bmatrix} 2 -1 0 1 -1 1 end{bmatrix}] [B begin{bmatrix} 3-2 0 1 0 0 3-1 0 0 0 3 1 end{bmatrix}] [B begin{bmatrix} 1 1 0 0 2 0 0 0 4 end{bmatrix}]

This is the final form of matrix B, which is the solution to the equation D CBE.

Verification

To verify our solution, substitute B back into the original equation D CBE. We have:

[C B^2 begin{bmatrix} 1 1 0 0 2 0 0 0 4 end{bmatrix}^2] [C begin{bmatrix} 2 3 0 0 4 0 0 0 16 end{bmatrix}] [E begin{bmatrix} 2 -1 0 1 -1 1 end{bmatrix}] [CBE begin{bmatrix} 2 3 0 0 4 0 0 0 16 end{bmatrix} begin{bmatrix} 2 -1 0 1 -1 1 end{bmatrix} begin{bmatrix} 1 1 0 0 2 0 0 0 4 end{bmatrix}] [CBE begin{bmatrix} 4-0-0 -2 3 0 0 0 0 0 4 0 -2 0 0 -1 0 0 end{bmatrix} begin{bmatrix} 1 1 0 0 2 0 0 0 4 end{bmatrix}] [CBE begin{bmatrix} 4 1 0 0 4 0 -2 -1 0 end{bmatrix} begin{bmatrix} 1 1 0 0 2 0 0 0 4 end{bmatrix}] [CBE begin{bmatrix} 4 8 0 0 8 0 -2 -2 0 end{bmatrix}] [3B 3begin{bmatrix} 1 1 0 0 2 0 0 0 4 end{bmatrix} begin{bmatrix} 3 3 0 0 6 0 0 0 12 end{bmatrix}]

The results of CBE and 3B are consistent, confirming our solution for B.

Conclusion

In conclusion, we have solved the matrix equation D CBE by using the given relationships between D, C, and E, and through algebraic manipulation. The final matrix B, which satisfies the equation, is:

[{ B begin{bmatrix} 1 1 0 0 2 0 0 0 4 end{bmatrix} }]

This solution demonstrates the application of matrix operations and the importance of understanding matrix properties in linear algebra.