Given:
[tex]\[
U=\left[\begin{array}{rrr}
1 & 3 & -5 \\
2 & 14 & 11 \\
-8 & 0 & 5
\end{array}\right] \quad V=\left[\begin{array}{rrr}
13 & -1 & -7 \\
-6 & 1 & 19 \\
0 & 15 & 23
\end{array}\right]
\][/tex]

Solve:
[tex]\[
X + U = V \\
X = \left[\square\right]
\][/tex]



Answer :

To solve for matrix [tex]\( X \)[/tex] in the equation [tex]\( X + U = V \)[/tex], we need to isolate [tex]\( X \)[/tex]. We can do this by subtracting matrix [tex]\( U \)[/tex] from matrix [tex]\( V \)[/tex].

So, the equation [tex]\( X + U = V \)[/tex] becomes:

[tex]\[ X = V - U \][/tex]

Now, we will perform matrix subtraction element-wise using the given matrices [tex]\( U \)[/tex] and [tex]\( V \)[/tex].

Given:
[tex]\[ U = \begin{array}{rrr} 1 & 3 & -5 \\ 2 & 14 & 11 \\ -8 & 0 & 5 \\ \end{array} \][/tex]

[tex]\[ V = \begin{array}{rrr} 13 & -1 & -7 \\ -6 & 1 & 19 \\ 0 & 15 & 23 \\ \end{array} \][/tex]

We will subtract each corresponding element in matrix [tex]\( U \)[/tex] from matrix [tex]\( V \)[/tex].

Step-by-step subtraction:

- For the element in the first row and first column:
[tex]\[ V_{11} - U_{11} = 13 - 1 = 12 \][/tex]

- For the element in the first row and second column:
[tex]\[ V_{12} - U_{12} = -1 - 3 = -4 \][/tex]

- For the element in the first row and third column:
[tex]\[ V_{13} - U_{13} = -7 - (-5) = -7 + 5 = -2 \][/tex]

- For the element in the second row and first column:
[tex]\[ V_{21} - U_{21} = -6 - 2 = -8 \][/tex]

- For the element in the second row and second column:
[tex]\[ V_{22} - U_{22} = 1 - 14 = -13 \][/tex]

- For the element in the second row and third column:
[tex]\[ V_{23} - U_{23} = 19 - 11 = 8 \][/tex]

- For the element in the third row and first column:
[tex]\[ V_{31} - U_{31} = 0 - (-8) = 0 + 8 = 8 \][/tex]

- For the element in the third row and second column:
[tex]\[ V_{32} - U_{32} = 15 - 0 = 15 \][/tex]

- For the element in the third row and third column:
[tex]\[ V_{33} - U_{33} = 23 - 5 = 18 \][/tex]

Putting it all together, we get the resulting matrix [tex]\( X \)[/tex]:

[tex]\[ X = \begin{array}{rrr} 12 & -4 & -2 \\ -8 & -13 & 8 \\ 8 & 15 & 18 \\ \end{array} \][/tex]

Thus, the solution for [tex]\( X \)[/tex] is:

[tex]\[ X = \begin{array}{rrr} 12 & -4 & -2 \\ -8 & -13 & 8 \\ 8 & 15 & 18 \\ \end{array} \][/tex]