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Ceneralize the result (4.11) to the case of a product of three matrices by proving that, for any conformable matrices \(A, B,\) and \(C,\) the equation \((A B C)^{\prime}=C^{\prime} B^{\prime} A^{\prime}\) holds.

Short Answer

Expert verified
The transpose of the product \((ABC)'\) is \(C'B'A'\).

Step by step solution

01

Understand the Matrix Transpose Property

Recall the transpose property for matrices: For any two matrices \( A \) and \( B \), the transpose of their product is the product of their transposes in reverse order, i.e., \((AB)' = B'A'\). This property will be fundamental in extending the result for three matrices.
02

Apply the Transpose Property to Two Matrices

Consider the matrices \( B \) and \( C \). Using the transpose property, we have \((BC)' = C'B'\). This forms the basis for further extension to three matrices.
03

Extend the Transpose Property to Three Matrices

Now consider the product \( ABC \). Using the property from Step 2, apply it again: the transpose \((AB)C)' = C'(AB)' = C'(B'A')\). Applying the same property to \((AB)'\), this becomes \(C' B' A'\).
04

Conclude the Generalization

By Step 3, we proved that \((ABC)' = C' B' A'\), confirming that the transpose of the product of matrices is the product of their individual transposes ordered in reverse.

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Key Concepts

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

Matrix Transpose
The concept of a matrix transpose is fundamental in matrix theory. Transposing a matrix involves flipping it over its diagonal, effectively switching its rows with its columns. For instance, if you have a matrix \( A \) with elements \( a_{ij} \), its transpose, denoted as \( A' \) or \( A^T \), would have elements \( a_{ji} \). This operation is particularly useful when dealing with certain properties of matrices, such as symmetry. For symmetric matrices, the matrix is equal to its transpose. Moreover, the transpose operation follows several intuitive properties:
  • (A')' = A: Taking the transpose twice yields the original matrix.
  • (A+B)' = A'+B': The transpose of a sum is the sum of the transposes.
  • (cA)' = cA': The transpose of a scalar multiplication is the scalar times the transpose.
These rules are handy to remember and make operations with matrices more predictable. Understanding the transpose property is crucial for grasping more complex topics, such as matrix multiplication and conformity, which involve manipulating multiple matrices.
Matrix Multiplication
Matrix multiplication is an essential operation in linear algebra, allowing us to combine two matrices to form a new one. Unlike regular multiplication, matrix multiplication involves a more complex set of rules. For two matrices \( A \) and \( B \) to be multiplied, the number of columns in \( A \) must equal the number of rows in \( B \). If \( A \) is an \( m \times n \) matrix and \( B \) is an \( n \times p \) matrix, their product \( AB \) is an \( m \times p \) matrix.The element in the \( i \)-th row and \( j \)-th column of the resulting matrix is calculated by taking the dot product of the \( i \)-th row of \( A \) and the \( j \)-th column of \( B \). Write this as:\[(AB)_{ij} = \sum_{k=1}^{n} a_{ik}b_{kj}.\]One important aspect of matrix multiplication is that it is generally not commutative; that is, \( AB eq BA \). This is vital when performing calculations and simplifications.Additionally, matrix multiplication can be quite intricate, especially when dealing with more than two matrices. However, it is associative, meaning you can group matrices when multiplying without altering the result, i.e., \( (AB)C = A(BC) \). This property allows for flexibility in computation and is essential when analyzing the transpose of a matrix product.
Conformable Matrices
When discussing matrix operations, the term 'conformable matrices' often comes up. Conformable matrices are simply matrices that can be combined under certain operations due to their compatible dimensions. This compatibility is primarily crucial in operations such as addition and multiplication.- **Addition:** Two matrices are conformable for addition if they have the same dimensions, meaning their rows and columns must match exactly.- **Multiplication:** For two matrices \( A \) and \( B \), \( A \) must have as many columns as \( B \) has rows. This ensures that the elements align appropriately for multiplication.Understanding whether matrices are conformable for various operations allows us to determine whether a calculation can be performed and what the resulting dimensions will be. In the generalized transpose property described in the exercise, the matrices \( A \), \( B \), and \( C \) are considered conformable, which means they can be multiplied in succession, allowing us to apply the concept of taking transposes in sequence. Remembering these basic rules about conformity is key to ensuring matrix operations are correctly performed and interpreted.

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Most popular questions from this chapter

Civen two nonzero vectors \(w_{1}\) and \(w_{2},\) the angle \(\theta\left(0 \leq \theta \leq 180^{\circ}\right)\) they form is related to the scalar product \(w_{1}^{\prime} w_{2}\left(=w_{2}^{\prime} w_{1}\right)\) as follows: $$\theta \text { is } a(n)\left\\{\begin{array}{l} \text { acute } \\ \text { right } \\ \text { obtuse } \end{array}\right\\} \text { angle if and only if } w_{1}^{\prime} w_{2}\left\\{\begin{array}{l} > \\ = \\ < \end{array}\right\\} 0$$ Verify this by computing the scalar product for each of the following pair of vectors (see Figs. \(4.2 \text { and } 4.3)\) (a) \(w_{1}=\left[\begin{array}{l}3 \\ 2\end{array}\right], w_{2}=\left[\begin{array}{l}1 \\ 4\end{array}\right]\) (b) \(w_{1}=\left[\begin{array}{l}1 \\ 4\end{array}\right], w_{2}=\left[\begin{array}{l}-3 \\ -2\end{array}\right]\) (c) \(w_{1}=\left[\begin{array}{l}3 \\ 2\end{array}\right], w_{2}=\left[\begin{array}{c}-3 \\ -2\end{array}\right]\) \((d) w_{1}=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right], w_{2}=\left[\begin{array}{l}0 \\ 2 \\ 0\end{array}\right]\) \((e) w_{1}=\left[\begin{array}{l}1 \\ 2 \\ 2\end{array}\right], w_{2}=\left[\begin{array}{l}1 \\ 2 \\ 0\end{array}\right]\)

Prove that for any two scalars \(g\) and \(k\) (a) \(k(A+B)=k A+k B\) \((b)(g+k) A=g A+k A\) (Note: To prove a result, you cannot use specific examples.)

The triangular inequality is written with the weak inequality sign \(\leq\), rather than the strict inequality sign \(<.\) Under what circumstances would the \(^{\prime \prime}=^{\prime \prime}\) part of the inequality apply?

The subtraction of a matrix \(B\) may be considered as the addition of the matrix (-1)\(B\). Does the commutative law of addition permit us to state that \(A-B=B-A ?\) If not, how would you correct the statement?

Having sold \(n\) items of merchandise at quantities \(Q_{1}, \ldots, Q_{n}\) and prices \(P_{1}, \ldots, P_{n}\) how would you express the total revenue in \((a) \sum\) notation and (b) vector notation?

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