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Matrices \(A\) and \(B\) are given below. Simplify the given expression. $$ A=\left[\begin{array}{l} 3 \\ 5 \end{array}\right] \quad B=\left[\begin{array}{c} -2 \\ 4 \end{array}\right] $$ $$ -B+3 B-2 B $$

Short Answer

Expert verified
The expression simplifies to the zero matrix \(\left[\begin{array}{c} 0 \\ 0 \end{array}\right]\).

Step by step solution

01

Understand the Matrix Expression

We are given the expression \(-B + 3B - 2B\) and need to simplify it. Notice that all terms involve matrix \(B\) just with different coefficients. Our task is to combine these terms.
02

Apply Scalar Multiplication

Each term in the expression involves scalar multiplication. We multiply the matrix \(B\) by each of the respective scalars: \(-1\), \(3\), and \(-2\). Perform each multiplication separately: \(-B = -1 \times \left[\begin{array}{c} -2 \ 4 \end{array}\right] = \left[\begin{array}{c} 2 \ -4 \end{array}\right]\)\(3B = 3 \times \left[\begin{array}{c} -2 \ 4 \end{array}\right] = \left[\begin{array}{c} -6 \ 12 \end{array}\right]\)\(-2B = -2 \times \left[\begin{array}{c} -2 \ 4 \end{array}\right] = \left[\begin{array}{c} 4 \ -8 \end{array}\right]\)
03

Combine Matrices

Now combine the resulting matrices obtained from each scalar multiplication: \((-B + 3B - 2B) = \left[\begin{array}{c} 2 \ -4 \end{array}\right] + \left[\begin{array}{c} -6 \ 12 \end{array}\right] + \left[\begin{array}{c} 4 \ -8 \end{array}\right]\). Add the corresponding elements:The first row: \(2 + (-6) + 4 = 0\).The second row: \(-4 + 12 - 8 = 0\).
04

Conclusion

After adding the elements of the resulting matrices, we find that the simplified expression is \(\left[\begin{array}{c} 0 \ 0 \end{array}\right]\), which is a zero matrix.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Scalar Multiplication
When working with matrices, scalar multiplication is a fundamental operation. It involves multiplying every element of the given matrix by a constant value, known as a scalar. This process influences the matrix's overall values but doesn't alter the matrix's structure.
  • For instance, in this exercise, matrix \( B \) has elements \(-2\) and \(4\). When we multiply \( B \) by \(-1\), \(3\), and \(-2\), we're applying scalar multiplication.
  • The results are computed by multiplying each element of \( B \) by these constants, resulting in new matrices.
It's like stretching or shrinking the values within the matrix. Understanding how each scalar affects the matrix individually is vital before proceeding to the next steps like combining matrices. By applying this concept, matrices transform into new forms which can be further combined or used in calculations.
Matrix Combination
Matrix combination involves adding or subtracting matrices that are compatible (i.e., they have the same dimensions). This usually occurs after applying scalar multiplication. The result of this operation is a single matrix, achieved by summing or subtracting the corresponding elements across the matrices involved.
In the exercise, after performing scalar multiplication on matrix \(B\), we obtained three matrices. The next goal is combining these matrices.
  • The expression \(-B + 3B - 2B\) becomes a single matrix through matrix addition or subtraction.
  • Each element from the first row is combined with the corresponding elements in the same position from the other matrices. The same applies to the second row.
By combining the results: - The first row: \(2 + (-6) + 4 = 0\)- The second row: \(-4 + 12 - 8 = 0\)The process shows that even with different coefficients, matrices can be simplified to find a solution, often revealing something insightful such as the Zero Matrix.
Zero Matrix
The Zero Matrix is an essential concept in matrix simplification and linear algebra. It is a matrix where all elements are zeros across all rows and columns. Understanding this outcome is vital when simplifying expressions involving matrices because it often signifies a neutral or balanced state in calculations.
In this exercise, after combining the matrices \(-B\), \(3B\), and \(-2B\), the resultant matrix is a zero matrix \(\left[\begin{array}{c} 0 \ 0 \end{array}\right]\).
  • This result indicates that the combined effect of the initial operations balances out completely, leaving no residual values.
  • The zero matrix holds particular importance, as it can be thought of as a neutral element in matrix addition, meaning any matrix added to a zero matrix remains unchanged.
Zero matrices often appear in equations as a result of cancellation, providing valuable insights while solving complex algebraic expressions. Recognizing them can help verify the correctness and understand the full picture of matrix operations in algebraic contexts.

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Most popular questions from this chapter

A matrix \(A\) and vectors \(\vec{b}, \vec{u}\) and \(\vec{v}\) are given. Verify that \(\vec{u}\) and \(\vec{v}\) are both solutions to the equation \(A \vec{x}=\vec{b} ;\) that is, show that \(A \vec{u}=A \vec{v}=\vec{b}\). $$ \begin{array}{l} A=\left[\begin{array}{cccc} 0 & -3 & -1 & -3 \\ -4 & 2 & -3 & 5 \end{array}\right], \\ \vec{b}=\left[\begin{array}{c} 0 \\ 0 \end{array}\right], \vec{u}=\left[\begin{array}{c} 11 \\ 4 \\ -12 \\ 0 \end{array}\right], \\ \vec{v}=\left[\begin{array}{c} 9 \\ -12 \\ 0 \\ 12 \end{array}\right] \end{array} $$

Row and column vectors \(\vec{u}\) and \(\vec{v}\) are defined. Find the product \(\vec{u} \vec{v},\) where possible. $$ \vec{u}=\left[\begin{array}{ll} 2 & -5 \end{array}\right] \quad \vec{v}=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right] $$

A matrix \(A\) and vector \(\vec{b}\) are given. (a) Solve the equation \(A \vec{x}=\vec{O}\) (b) Solve the equation \(A \vec{x}=\vec{b}\). In each of the above, be sure to write your answer in vector format. Also, when possible, give 2 particular solutions to each equation. $$ \begin{array}{l} A=\left[\begin{array}{ccccc} 3 & 0 & -2 & -4 & 5 \\ 2 & 3 & 2 & 0 & 2 \\ -5 & 0 & 4 & 0 & 5 \end{array}\right], \\ \vec{b}=\left[\begin{array}{c} -1 \\ -5 \\ 4 \end{array}\right] \end{array} $$

Let \(A=\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{array}\right] .\) Find \(A^{2}\) and \(A^{3}\).

Matrices \(A\) and \(B\) are given. Compute \((A B)^{-1}\) and \(B^{-1} A^{-1}\). $$ A=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right], \quad B=\left[\begin{array}{ll} 7 & 1 \\ 2 & 1 \end{array}\right] $$

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