When performing matrix multiplication, especially with a column matrix and a row matrix, remember that the resulting matrix's dimensions are determined by the row count of the column matrix and the column count of the row matrix. The outer product of a \( m \times 1 \) and \( 1 \times n \) matrix results in a matrix of dimensions \( m \times n \). For transposing a matrix, simply swap the rows with columns, and this operation is essential for many matrix-related problems. Don't forget that matrix multiplication follows the distributive property, so you can multiply each element of the rows and columns accordingly.
To find \( (PQ)' \), we first compute the product \( PQ \) where:
The product \( PQ \) will be a \( 3 \times 3 \) matrix given by:
\[PQ = \begin{bmatrix}-1 \\2 \\1\end{bmatrix}\times\begin{bmatrix}2 & -4 & 1\end{bmatrix}=\begin{bmatrix}-2 & 4 & -1 \\4 & -8 & 2 \\2 & -4 & 1\end{bmatrix}\]
Next, we find the transpose \( (PQ)' \):
\[(PQ)' = \begin{bmatrix}-2 & 4 & -1 \\4 & -8 & 2 \\2 & -4 & 1\end{bmatrix}'=\begin{bmatrix}-2 & 4 & 2 \\4 & -8 & -4 \\-1 & 2 & 1\end{bmatrix}\]
Thus, the correct answer is:
\[\begin{bmatrix}-2 & 4 & 2 \\4 & -8 & -4 \\-1 & 2 & 1\end{bmatrix}\]
To solve for \((PQ)^T\), we need to perform several steps. First, compute the product of matrices \(P\) and \(Q\), then transpose the resulting matrix.
Given matrices:
\(P = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}\)
\(Q = \begin{bmatrix} 2 & -4 & 1 \end{bmatrix}\)
Step 1: Calculate \(PQ\).
Since \(P\) is a column matrix (3x1) and \(Q\) is a row matrix (1x3), \(PQ\) will be a 3x3 matrix:
\(PQ = \begin{bmatrix} -1 \times 2 & -1 \times -4 & -1 \times 1 \\ 2 \times 2 & 2 \times -4 & 2 \times 1 \\ 1 \times 2 & 1 \times -4 & 1 \times 1 \end{bmatrix}\)
Simplifying each element gives:
\(PQ = \begin{bmatrix} -2 & 4 & -1 \\ 4 & -8 & 2 \\ 2 & -4 & 1 \end{bmatrix}\)
Step 2: Transpose the matrix \(PQ\).
The transpose of a matrix is obtained by swapping its rows and columns, so \((PQ)^T\) is:
\((PQ)^T = \begin{bmatrix} -2 & 4 & 2 \\ 4 & -8 & -4 \\ -1 & 2 & 1 \end{bmatrix}\)
Therefore, the final result for \((PQ)^T\) is:
\(\begin{bmatrix} -2 & 4 & 2 \\ 4 & -8 & -4 \\ -1 & 2 & 1 \end{bmatrix}\)
Consider the following Linear Programming Problem $ P $: Minimize $ x_1 + 2x_2 $, subject to
$ 2x_1 + x_2 \leq 2 $,
$ x_1 + x_2 = 1 $,
$ x_1, x_2 \geq 0 $.
The optimal value of the problem $ P $ is equal to:
Let $D = \{(x, y) \in \mathbb{R}^2 : x > 0 \text{ and } y > 0\}$. If the following second-order linear partial differential equation
$y^2 \frac{\partial^2 u}{\partial x^2} - x^2 \frac{\partial^2 u}{\partial y^2} + y \frac{\partial u}{\partial y} = 0$ on $D$
is transformed to
$\left( \frac{\partial^2 u}{\partial \eta^2} - \frac{\partial^2 u}{\partial \xi^2} \right) + \left( \frac{\partial u}{\partial \eta} + \frac{\partial u}{\partial \xi} \right) \frac{1}{2\eta} + \left( \frac{\partial u}{\partial \eta} - \frac{\partial u}{\partial \xi} \right) \frac{1}{2\xi} = 0$ on $D$,
for some $a, b \in \mathbb{R}$, via the coordinate transform $\eta = \frac{x^2}{2}$ and $\xi = \frac{y^2}{2}$, then which one of the following is correct?
Let \( p_1<p_2 \) be the two fixed points of the function \( g(x) = e^x - 2 \), where \( x \in {R} \). For \( x_0 \in {R} \), let the sequence \( (x_n)_{n \geq 1} \) be generated by the fixed-point iteration \[ x_n = g(x_{n-1}), \quad n \geq 1. \] Which one of the following is/are correct?
The prisoners in the concentration camps in World War-II had lost faith in the future. Being in the camp, I felt disgusted with the state of affairs and I forced my thoughts to turn to another subject. ”Suddenly, I saw myself standing on the platform of a well-lit, warm and pleasant lecture room. In front of me, the attentive audience were seated in comfortable upholstered seats. I saw myself giving a lecture on hope, optimism, and resilience under difficult circumstances.” Suddenly, all that oppressed me stopped giving me pain and distress. This practice was so impactful that I could succeed in rising above the situation and the sufferings of the moment.”
Just as individuals compare themselves with others in terms of similarities and differences with respect to what they have and what others have, individuals also compare the group they belong to with groups of which they are not a member. It has been found that groups are more likely to take extreme decisions than individuals alone. Suppose there is an employee who has been caught taking a bribe or engaging in some other unethical act. His/her colleagues are asked to decide what punishment he/she should be given. They may let him/her go scot-free or decide to terminate his/her services instead of imposing a punishment which may commensurate with the unethical act he/she had engaged in. Whatever the initial position in the group, this position becomes much stronger as a result of discussions and interaction in the group.