Sure! Let's solve for the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] for each part of the problem step by step.
### Part (a)
We start with the equation:
[tex]\[
\begin{bmatrix} 2 & b \end{bmatrix} + \begin{bmatrix} a & -1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \end{bmatrix}
\][/tex]
This leads to two separate equations:
1. [tex]\(2 + a = 1\)[/tex]
2. [tex]\(b - 1 = 2\)[/tex]
Solving these:
1. [tex]\(2 + a = 1 \Rightarrow a = 1 - 2 \Rightarrow a = -1\)[/tex]
2. [tex]\(b - 1 = 2 \Rightarrow b = 2 + 1 \Rightarrow b = 3\)[/tex]
So the solution for part (a) is:
[tex]\[
a = -1, \quad b = 3
\][/tex]
### Part (b)
Next, we consider the equation:
[tex]\[
\begin{bmatrix} a - 1 & b + 1 \end{bmatrix} - \begin{bmatrix} 1 & 2 \end{bmatrix} = \begin{bmatrix} -2 & 1 \end{bmatrix}
\][/tex]
This leads to two equations:
1. [tex]\((a - 1) - 1 = -2\)[/tex]
2. [tex]\((b + 1) - 2 = 1\)[/tex]
Solving these:
1. [tex]\(a - 2 = -2 \Rightarrow a = -2 + 2 \Rightarrow a = 0\)[/tex]
2. [tex]\(b - 1 = 1 \Rightarrow b = 1 + 1 \Rightarrow b = 2\)[/tex]
So the solution for part (b) is:
[tex]\[
a = 0, \quad b = 2
\][/tex]
### Part (c)
Finally, we consider the equation:
[tex]\[
\begin{bmatrix} a + 2 & 3 \end{bmatrix} + \begin{bmatrix} -1 & 2 + b \end{bmatrix} = \begin{bmatrix} 3 & 0 \end{bmatrix}
\][/tex]
This leads to two equations:
1. [tex]\(a + 2 - 1 = 3\)[/tex]
2. [tex]\(3 + (2 + b) = 0\)[/tex]
Solving these:
1. [tex]\(a + 1 = 3 \Rightarrow a = 3 - 1 \Rightarrow a = 2\)[/tex]
2. [tex]\(3 + 2 + b = 0 \Rightarrow 5 + b = 0 \Rightarrow b = -5\)[/tex]
So the solution for part (c) is:
[tex]\[
a = 2, \quad b = -5
\][/tex]
### Summary
The values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] for each part are:
- Part (a): [tex]\((a, b) = (-1, 3)\)[/tex]
- Part (b): [tex]\((a, b) = (0, 2)\)[/tex]
- Part (c): [tex]\((a, b) = (2, -5)\)[/tex]
Hence, the solutions are:
[tex]\[
((-1, 3), (0, 2), (2, -5))
\][/tex]
