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Chapter 2: Problem 13

Let \(X=\left[\begin{array}{ccc}-1 & -1 & 1\end{array}\right]\) and \(Y=\left[\begin{array}{ccc}0 & 1 & 2\end{array}\right] .\) Find \(X^{T} Y\) and \(X Y^{T}\) if possible.

Short Answer

Expert verified
The matrices are \( X^T Y = \begin{bmatrix}0 & -1 & -2 \ 0 & -1 & -2 \ 0 & 1 & 2 \ ...\] and \[ X Y^T = 1,...\] .

Step by step solution

01

- Understand the Matrix Dimensions

Identify the dimensions of both matrices. Matrix X is a 1x3 matrix, and matrix Y is also a 1x3 matrix.
02

- Check Compatibility for Multiplication

To determine if the multiplication is possible, check dimensions. For multiplication of two matrices A (m x n) and B (p x q) to be possible, the number of columns in A (n) must be equal to the number of rows in B (p).
03

- Calculate the Transpose of X

The transpose of matrix X, denoted as X^T, is obtained by switching its rows and columns. Matrix X: \(X = \begin{bmatrix} -1 & -1 & 1 \ \ \ \ \ \ \ \ \end{bmatrix}\) Transpose X: \(X^T = \begin{bmatrix} -1 \ -1 \ 1 \ \ \ \ \ \ \ \end{bmatrix}\)
04

- Multiply X^T and Y

We can now multiply X^T (3x1) and Y (1x3): \[ X^T \times Y = \begin{bmatrix} -1 \ -1 \ 1 \ \ \ \ \ \ \ \end{bmatrix} \times \begin{bmatrix}0 & 1 & 2 \ \ \ \ \ \ \ \ \ \end{bmatrix} = \begin{bmatrix} -1 \times 0 & -1 \times 1 & -1 \times 2 \ -1 \times 0 & -1 \times 1 & -1 \times 2 \ 1 \times 0 & 1 \times 1 & 1 \times 2 \end{bmatrix} = \begin{bmatrix} 0 & -1 & -2 \ 0 & -1 & -2 \ 0 & 1 & 2 \ \ \ \ \ \ \ \end{bmatrix}onumber\textrm{\]Mat...
05

- Check Compatibility for XY^T

Determine if \( X \times Y^T \) is possible. X is a 1x3 matrix, and Y^T is a 3x1 matrix.
06

- Calculate the Transpose of Y

The transpose of matrix Y is obtained by switching its rows and columns. \Matrix Y: \( Y = \begin{bmatrix}0 & 1 & 2 \ \ \ \ \ \ \ \ \ \end{bmatrix}\)Transpose Y: \( Y^T = \begin{bmatrix} 0 \ 1 \ 2 \ \ \ \ \ \ \ \end{bmatrix} \)
07

- Multiply X and Y^T

Multiply X (1x3) and Y^T (3x1): \( X \times Y^T = \begin{bmatrix} -1 & -1 & 1 \ \ \ \ \ \ \ \ \ \end{bmatrix} \times \begin{bmatrix}0 \ 1 \ 2 \ \ \ \ \ \ \ \end{bmatrix} = \begin{bmatrix} -1 \times 0 + -1 \times 1 + 1 \times 2 \ \ \ \ \ \ \ \ \ \ \ = 1 \ \ \ \ \ \ \ \ \ \end{bmatrix}\)

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Key Concepts

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

Matrix Transpose
To understand matrix transpose, imagine flipping a matrix over its diagonal. This means switching the matrix's rows with columns. For example, if we take Matrix X:
  • \( X = \begin{bmatrix} -1 & -1 & 1 \ \ \ \ \ \ \ \ \end{bmatrix} \)
The transpose, denoted as \(X^T\), is:
  • \( X^T = \begin{bmatrix} -1 \ -1 \ 1 \ \ \ \ \ \ \end{bmatrix} \)
This operation helps align matrices for multiplication, especially when their original dimensions are incompatible. Always remember: rows become columns, and columns become rows.
Matrix Dimensions
Matrix dimensions tell us how many rows and columns a matrix has. For example, Matrix X is a \(1\times3\) matrix:
  • 1 row
  • 3 columns
Matrix Y is also a \(1\times3\) matrix. Understanding dimensions is crucial because it directly affects matrix operations like multiplication. Always identify the dimensions first: it's the very structure of the matrix. It鈥檚 written as: \( \text{{rows}} \times \text{{columns}} \).
Matrix Multiplication Compatibility
Not all matrices can be multiplied together. To multiply two matrices, their dimensions must be compatible. For two matrices A (\(m\)\(x\)\(n\)) and B (\(p\)\(x\)\(q\)):
  • The number of columns in A (\(n\)) must be equal to the number of rows in B (\(p\)).
Let's check the compatibility for \(X^T \times Y \):
  • \(X^T\) is \(3\times1\)
  • Y is \(1\times3\)
Since the numbers match (i.e., the number of columns in \(X^T\) equals the number of rows in Y), they can be multiplied. Similarly, for \(X \times Y^T\):
  • X is \( 1 \times 3 \)
  • \( Y^T \) is \( 3\times 1 \)
The numbers also match here, so they can be multiplied too.
Calculating Transpose
Calculating the transpose of a matrix is simple and follows a clear rule: swap rows with columns. Taking Matrix Y as an example:
  • Original Y: \( Y = \begin{bmatrix} 0 & 1 & 2 \ \ \ \ \ \ \ \end{bmatrix}\)
To find \(Y^T\):
  • Switch the lone row into a column.
  • Transpose Y: \( Y^T = \begin{bmatrix} 0 \ 1 \ 2 \ \ \ \ \ \ \end{bmatrix}\)
When matrices have more rows and columns, simply follow the same rule for each element. This transformation helps to match dimensions for operations, such as dot products in multiplication. Remember to double-check your matrix after transposing to avoid mistakes.

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Most popular questions from this chapter

For each pair of matrices, find the (1,2) -entry and (2,3) -entry of the product \(A B\). (a) \(A=\left[\begin{array}{rrr}1 & 2 & -1 \\ 3 & 4 & 0 \\ 2 & 5 & 1\end{array}\right], B=\left[\begin{array}{rrr}4 & 6 & -2 \\ 7 & 2 & 1 \\ -1 & 0 & 0\end{array}\right]\) (b) \(A=\left[\begin{array}{lll}1 & 3 & 1 \\ 0 & 2 & 4 \\ 1 & 0 & 5\end{array}\right], B=\left[\begin{array}{rrr}2 & 3 & 0 \\ -4 & 16 & 1 \\ 0 & 2 & 2\end{array}\right]\)

Consider the matrices \(A=\left[\begin{array}{rrr}1 & 2 & 3 \\ 2 & 1 & 7\end{array}\right], B=\left[\begin{array}{rrr}3 & -1 & 2 \\ -3 & 2 & 1\end{array}\right], C=\left[\begin{array}{ll}1 & 2 \\ 3 & 1\end{array}\right],\) \(D=\left[\begin{array}{rr}-1 & 2 \\ 2 & -3\end{array}\right], E=\left[\begin{array}{l}2 \\ 3\end{array}\right] .\) Find the following if possible. If it is not possible explain why. (a) \(-3 A\) (b) \(3 B-A\) (c) \(A C\) (d) \(C B\) (e) \(A E\) (f) \(E A\)

Using the inverse of the matrix, find the solution to the systems: \((a)\) $$ \left[\begin{array}{lll} 1 & 0 & 3 \\ 2 & 3 & 4 \\ 1 & 0 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right] $$ \((b)\) $$ \left[\begin{array}{lll} 1 & 0 & 3 \\ 2 & 3 & 4 \\ 1 & 0 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{r} 3 \\ -1 \\ -2 \end{array}\right] $$ Now give the solution in terms of \(a, b,\) and \(c\) to the following: $$ \left[\begin{array}{lll} 1 & 0 & 3 \\ 2 & 3 & 4 \\ 1 & 0 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} a \\ b \\ c \end{array}\right] $$

Let \(A=\left[\begin{array}{rr}1 & 1 \\ -2 & -1 \\ 1 & 2\end{array}\right], B=\left[\begin{array}{rrr}1 & -1 & -2 \\ 2 & 1 & -2\end{array}\right],\) and \(C=\left[\begin{array}{rrr}1 & 1 & -3 \\ -1 & 2 & 0 \\ -3 & -1 & 0\end{array}\right] .\) Find the following if possible. (a) \(A B\) (b) \(B A\) (c) \(A C\) (d) \(C A\) (e) \(C B\) (f) \(B C\)

Let $$ A=\left[\begin{array}{rl} 2 & 1 \\ -1 & 3 \end{array}\right] $$ Find \(A^{-1}\) if possible. If \(A^{-1}\) does not exist, explain why.
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