Given three matrices $X$, $Y$, and $Z$ as follows: $X = \begin{bmatrix} 5 & 4 & -7 \\ -3 & p & 5 \end{bmatrix}$, $Y = \begin{bmatrix} 7 & 2 & 1 \\ 1 & 0 & 3 \\ 2 & 1 & 4 \end{bmatrix}$, $Z = \begin{bmatrix} 3 & q \\ 1 & 4 \\ 8 & -9 \end{bmatrix}$. We are asked to: I. Find the values of $p$ and $q$ if $X = Z^T$. II. Evaluate the following, if possible: i. $XY$ ii. $X^TZ$ iii. $3Z + X^T$

1. Problem Description

Given three matrices $X$, $Y$, and $Z$ as follows:

$X = \begin{bmatrix} 5 & 4 & -7 \\ -3 & p & 5 \end{bmatrix}$, $Y = \begin{bmatrix} 7 & 2 & 1 \\ 1 & 0 & 3 \\ 2 & 1 & 4 \end{bmatrix}$, $Z = \begin{bmatrix} 3 & q \\ 1 & 4 \\ 8 & -9 \end{bmatrix}$.

We are asked to:

I. Find the values of $p$ and $q$ if $X = Z^T$.

II. Evaluate the following, if possible:

i. $XY$

ii. $X^TZ$

iii. $3Z + X^T$

2. Solution Steps

I. Find $p$ and $q$ such that $X = Z^T$:

First, find the transpose of $Z$:

$Z^T = \begin{bmatrix} 3 & 1 & 8 \\ q & 4 & -9 \end{bmatrix}$

Since $X = Z^T$, we have:

$\begin{bmatrix} 5 & 4 & -7 \\ -3 & p & 5 \end{bmatrix} = \begin{bmatrix} 3 & 1 & 8 \\ q & 4 & -9 \end{bmatrix}$

By equating the corresponding elements, we find that this equality is impossible, since $5 \ne 3$, $4 \ne 1$, and $-7 \ne 8$. So there is no solution. However, there is an error in the problem, it should be $X^T = Z$.

So let's find $X^T$:

$X^T = \begin{bmatrix} 5 & -3 \\ 4 & p \\ -7 & 5 \end{bmatrix}$

Since $X^T = Z$, we have:

$\begin{bmatrix} 5 & -3 \\ 4 & p \\ -7 & 5 \end{bmatrix} = \begin{bmatrix} 3 & q \\ 1 & 4 \\ 8 & -9 \end{bmatrix}$

This is still impossible. However, if $Z = X^T$ we have $Z = \begin{bmatrix} 5 & -3 \\ 4 & p \\ -7 & 5 \end{bmatrix} = \begin{bmatrix} 3 & q \\ 1 & 4 \\ 8 & -9 \end{bmatrix}$. This also leads to no solution.

Assuming the question meant $X = Z^T$: This implies that $X$ should have the same dimension with $Z^T$, i.e., $2 \times 3$. Then we have

$\begin{bmatrix} 5 & 4 & -7 \\ -3 & p & 5 \end{bmatrix} = \begin{bmatrix} 3 & 1 & 8 \\ q & 4 & -9 \end{bmatrix}$. This can never be true.

Instead, assume $Z^T = \begin{bmatrix} 5 & 4 \\ -3 & p \\ -7 & 5\end{bmatrix}$. Therefore $Z = \begin{bmatrix} 5 & -3 \\ 4 & p \\ -7 & 5\end{bmatrix}$

Thus $\begin{bmatrix} 3 & q \\ 1 & 4 \\ 8 & -9 \end{bmatrix} = \begin{bmatrix} 5 & -3 \\ 4 & p \\ -7 & 5\end{bmatrix}$ This will lead to similar error.

It must be $Z^T = X$, which also means $Z=X^T$, so we would have

$X = \begin{bmatrix} 3 & 1 & 8 \\ q & 4 & -9 \end{bmatrix}$. Since we are given different values, there is no solution.

If we assume that the matrices should have the same sizes, then let's assume that

$X=\begin{bmatrix}3 & q \\ 1 & 4 \\ 8 & -9 \end{bmatrix} = Z^T$

Then $Z = \begin{bmatrix} 3 & 1 & 8 \\ q & 4 & -9 \end{bmatrix}^T$

$Z = \begin{bmatrix} 3 & q \\ 1 & 4 \\ 8 & -9 \end{bmatrix}$

We are given that $X=\begin{bmatrix} 5 & 4 & -7 \\ -3 & p & 5\end{bmatrix}$. Since we are supposed to be given dimensions, then these are wrong. However, the only possibility is to assume $X = Z^T$. Then $Z = X^T = \begin{bmatrix} 5 & -3 \\ 4 & p \\ -7 & 5 \end{bmatrix}$.

If $Z = \begin{bmatrix} 3 & q \\ 1 & 4 \\ 8 & -9 \end{bmatrix}$, then let's suppose that

$q=-3$, $p=4$.

Then $Z = \begin{bmatrix} 3 & -3 \\ 1 & 4 \\ 8 & -9\end{bmatrix}$

This is no longer right.

Then let us assume $X = \begin{bmatrix} 5 & 4 \\ -3 & p \end{bmatrix}$, and $Z = \begin{bmatrix} 3 & q \\ 1 & 4 \end{bmatrix}$. We have $Z^T = \begin{bmatrix} 3 & 1 \\ q & 4 \end{bmatrix}$, $X=Z^T$. Then $5 = 3$, $4 = 1$, $-3 = q$, $p = 4$. So this is again incorrect.

If $X = Z^T$, then $Z=X^T$.

$X=\begin{bmatrix} 3 & 1 \\ 8 & -9 \end{bmatrix}^T$, and $Z = \begin{bmatrix} 3 & 8 \\ 1 & -9 \end{bmatrix}$. Again wrong.

II. Evaluate the following, if possible:

Since we could not find the correct values for $p$ and $q$, we would have to consider them separately. Let us suppose $X = \begin{bmatrix} 5 & 4 & -7 \\ -3 & p & 5 \end{bmatrix}$, $Y = \begin{bmatrix} 7 & 2 & 1 \\ 1 & 0 & 3 \\ 2 & 1 & 4 \end{bmatrix}$, $Z = \begin{bmatrix} 3 & q \\ 1 & 4 \\ 8 & -9 \end{bmatrix}$.

i. $XY$

$XY = \begin{bmatrix} 5 & 4 & -7 \\ -3 & p & 5 \end{bmatrix} \begin{bmatrix} 7 & 2 & 1 \\ 1 & 0 & 3 \\ 2 & 1 & 4 \end{bmatrix} = \begin{bmatrix} 35+4-14 & 10+0-7 & 5+12-28 \\ -21+p+10 & -6+0+5 & -3+3p+20 \end{bmatrix} = \begin{bmatrix} 25 & 3 & -11 \\ p-11 & -1 & 3p+17 \end{bmatrix}$

ii. $X^TZ$

$X^T = \begin{bmatrix} 5 & -3 \\ 4 & p \\ -7 & 5 \end{bmatrix}$.

$X^TZ = \begin{bmatrix} 5 & -3 \\ 4 & p \\ -7 & 5 \end{bmatrix} \begin{bmatrix} 3 & q \\ 1 & 4 \\ 8 & -9 \end{bmatrix}$ Matrix product is not defined.

Since $X$ has dimension $2 \times 3$ and $Z$ has dimension $3 \times 2$, then $X^T$ has dimension $3 \times 2$. Thus, the size of $Z$ needs to be $2 \times n$, it can not have 3 rows. Therefore $X^T Z$ can't be computed because $X^T$ size is $3 \times 2$. and $Z$ size is $3 \times 2$.

iii. $3Z + X^T$

$3Z = 3\begin{bmatrix} 3 & q \\ 1 & 4 \\ 8 & -9 \end{bmatrix} = \begin{bmatrix} 9 & 3q \\ 3 & 12 \\ 24 & -27 \end{bmatrix}$.

$X^T = \begin{bmatrix} 5 & -3 \\ 4 & p \\ -7 & 5 \end{bmatrix}$.

Therefore $3Z+X^T = \begin{bmatrix} 9 & 3q \\ 3 & 12 \\ 24 & -27 \end{bmatrix} + \begin{bmatrix} 5 & -3 \\ 4 & p \\ -7 & 5 \end{bmatrix} = \begin{bmatrix} 14 & 3q-3 \\ 7 & p+12 \\ 17 & -22 \end{bmatrix}$.

3. Final Answer

I. There are no values for $p$ and $q$ that satisfy $X = Z^T$. If we consider that $X = \begin{bmatrix} 5 & 4 \\ -3 & p \end{bmatrix}$ and $Z = \begin{bmatrix} 3 & q \\ 1 & 4 \end{bmatrix}$, there is still no solution for p, q. If we assume $q=-3$ and $p =4$, and $X = \begin{bmatrix} 5 & 4 & -7 \\ -3 & 4 & 5 \end{bmatrix}$, $Y = \begin{bmatrix} 7 & 2 & 1 \\ 1 & 0 & 3 \\ 2 & 1 & 4 \end{bmatrix}$, $Z = \begin{bmatrix} 3 & -3 \\ 1 & 4 \\ 8 & -9 \end{bmatrix}$

II.

i. $XY = \begin{bmatrix} 25 & 3 & -11 \\ p-11 & -1 & 3p+17 \end{bmatrix} = \begin{bmatrix} 25 & 3 & -11 \\ -7 & -1 & 29 \end{bmatrix}$.

ii. $X^TZ$ is not possible.

iii. $3Z + X^T = \begin{bmatrix} 14 & 3q-3 \\ 7 & p+12 \\ 17 & -22 \end{bmatrix} = \begin{bmatrix} 14 & -12 \\ 7 & 16 \\ 17 & -22 \end{bmatrix}$.