Answer :
Let's solve each part step-by-step.
### Part (a)
Given the matrices:
[tex]\[ A = \begin{pmatrix} 1 & 1 & 2 \\ 1 & 2 & 1 \\ 3 & 2 & 2 \end{pmatrix} \][/tex]
[tex]\[ B = \begin{pmatrix} 1 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & -1 & 2 \end{pmatrix} \][/tex]
#### (i) Calculate [tex]\( A + B \)[/tex]
We directly add the corresponding elements of matrices [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
[tex]\[ A + B = \begin{pmatrix} 1 + 1 & 1 + 2 & 2 + 1 \\ 1 + 2 & 2 + 1 & 1 + 1 \\ 3 + 1 & 2 - 1 & 2 + 2 \end{pmatrix} = \begin{pmatrix} 2 & 3 & 3 \\ 3 & 3 & 2 \\ 4 & 1 & 4 \end{pmatrix} \][/tex]
So, [tex]\[ A + B = \begin{pmatrix} 2 & 3 & 3 \\ 3 & 3 & 2 \\ 4 & 1 & 4 \end{pmatrix} \][/tex]
#### (ii) Calculate [tex]\( (AB)^x \)[/tex]
First, we need to find the product [tex]\(AB\)[/tex]:
[tex]\[ AB = \begin{pmatrix} 1 & 1 & 2 \\ 1 & 2 & 1 \\ 3 & 2 & 2 \end{pmatrix} \begin{pmatrix} 1 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & -1 & 2 \end{pmatrix} = \begin{pmatrix} (1 \cdot 1 + 1 \cdot 2 + 2 \cdot 1) & (1 \cdot 2 + 1 \cdot 1 + 2 \cdot -1) & (1 \cdot 1 + 1 \cdot 1 + 2 \cdot 2) \\ (1 \cdot 1 + 2 \cdot 2 + 1 \cdot 1) & (1 \cdot 2 + 2 \cdot 1 + 1 \cdot -1) & (1 \cdot 1 + 2 \cdot 1 + 1 \cdot 2) \\ (3 \cdot 1 + 2 \cdot 2 + 2 \cdot 1) & (3 \cdot 2 + 2 \cdot 1 + 2 \cdot -1) & (3 \cdot 1 + 2 \cdot 1 + 2 \cdot 2) \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 1 + 2 + 2 & 2 + 1 - 2 & 1 + 1 + 4 \\ 1 + 4 + 1 & 2 + 2 - 1 & 1 + 2 + 2 \\ 3 + 4 + 2 & 6 + 2 - 2 & 3 + 2 + 4 \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 5 & 1 & 6 \\ 6 & 3 & 5 \\ 9 & 6 & 9 \end{pmatrix} \][/tex]
The problem specifies the power [tex]\( x = 2 \)[/tex]. Thus, we calculate [tex]\( (AB)^2 \)[/tex] by squaring the matrix [tex]\(AB\)[/tex]:
[tex]\[ (AB)^2 = AB \cdot AB = \begin{pmatrix} 5 & 1 & 6 \\ 6 & 3 & 5 \\ 9 & 6 & 9 \end{pmatrix} \begin{pmatrix} 5 & 1 & 6 \\ 6 & 3 & 5 \\ 9 & 6 & 9 \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} (5 \cdot 5 + 1 \cdot 6 + 6 \cdot 9) & (5 \cdot 1 + 1 \cdot 3 + 6 \cdot 6) & (5 \cdot 6 + 1 \cdot 5 + 6 \cdot 9) \\ (6 \cdot 5 + 3 \cdot 6 + 5 \cdot 9) & (6 \cdot 1 + 3 \cdot 3 + 5 \cdot 6) & (6 \cdot 6 + 3 \cdot 5 + 5 \cdot 9) \\ (9 \cdot 5 + 6 \cdot 6 + 9 \cdot 9) & (9 \cdot 1 + 6 \cdot 3 + 9 \cdot 6) & (9 \cdot 6 + 6 \cdot 5 + 9 \cdot 9) \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} (25 + 6 + 54) & (5 + 3 + 36) & (30 + 5 + 54) \\ (30 + 18 + 45) & (6 + 9 + 30) & (36 + 15 + 45) \\ (45 + 36 + 81) & (9 + 18 + 54) & (54 + 30 + 81) \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 85 & 44 & 89 \\ 93 & 45 & 96 \\ 162 & 81 & 165 \end{pmatrix} \][/tex]
So, [tex]\( (AB)^2 = \begin{pmatrix} 85 & 44 & 89 \\ 93 & 45 & 96 \\ 162 & 81 & 165 \end{pmatrix} \)[/tex]
### Part (b)
To solve the simultaneous equations using the inverse matrix method, we first write the system of equations as a matrix equation: [tex]\( C \mathbf{x} = \mathbf{d} \)[/tex], where:
[tex]\[ C = \begin{pmatrix} 3 & 2 & 2 \\ 1 & 2 & 3 \\ 4 & 1 & 3 \end{pmatrix} \][/tex]
[tex]\[ \mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \][/tex]
[tex]\[ \mathbf{d} = \begin{pmatrix} 13 \\ 13 \\ 13 \end{pmatrix} \][/tex]
We need to find [tex]\( C^{-1} \)[/tex], the inverse of matrix [tex]\( C \)[/tex]:
[tex]\[ C^{-1} = \begin{pmatrix} 0.23076923 & -0.30769231 & 0.15384615 \\ 0.69230769 & 0.07692308 & -0.53846154 \\ -0.53846154 & 0.38461538 & 0.30769231 \end{pmatrix} \][/tex]
The solution to the system of equations is given by:
[tex]\[ \mathbf{x} = C^{-1} \mathbf{d} \][/tex]
[tex]\[ \mathbf{x} = \begin{pmatrix} 0.23076923 & -0.30769231 & 0.15384615 \\ 0.69230769 & 0.07692308 & -0.53846154 \\ -0.53846154 & 0.38461538 & 0.30769231 \end{pmatrix} \begin{pmatrix} 13 \\ 13 \\ 13 \end{pmatrix} \][/tex]
Multiplying these matrices, we obtain:
[tex]\[ \mathbf{x} = \begin{pmatrix} (0.23076923 \cdot 13 + -0.30769231 \cdot 13 + 0.15384615 \cdot 13) \\ (0.69230769 \cdot 13 + 0.07692308 \cdot 13 + -0.53846154 \cdot 13) \\ (-0.53846154 \cdot 13 + 0.38461538 \cdot 13 + 0.30769231 \cdot 13) \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} \][/tex]
Therefore, the solutions to the system of equations are:
[tex]\[ x = 1, \; y = 3, \; z = 2 \][/tex]
### Part (a)
Given the matrices:
[tex]\[ A = \begin{pmatrix} 1 & 1 & 2 \\ 1 & 2 & 1 \\ 3 & 2 & 2 \end{pmatrix} \][/tex]
[tex]\[ B = \begin{pmatrix} 1 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & -1 & 2 \end{pmatrix} \][/tex]
#### (i) Calculate [tex]\( A + B \)[/tex]
We directly add the corresponding elements of matrices [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
[tex]\[ A + B = \begin{pmatrix} 1 + 1 & 1 + 2 & 2 + 1 \\ 1 + 2 & 2 + 1 & 1 + 1 \\ 3 + 1 & 2 - 1 & 2 + 2 \end{pmatrix} = \begin{pmatrix} 2 & 3 & 3 \\ 3 & 3 & 2 \\ 4 & 1 & 4 \end{pmatrix} \][/tex]
So, [tex]\[ A + B = \begin{pmatrix} 2 & 3 & 3 \\ 3 & 3 & 2 \\ 4 & 1 & 4 \end{pmatrix} \][/tex]
#### (ii) Calculate [tex]\( (AB)^x \)[/tex]
First, we need to find the product [tex]\(AB\)[/tex]:
[tex]\[ AB = \begin{pmatrix} 1 & 1 & 2 \\ 1 & 2 & 1 \\ 3 & 2 & 2 \end{pmatrix} \begin{pmatrix} 1 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & -1 & 2 \end{pmatrix} = \begin{pmatrix} (1 \cdot 1 + 1 \cdot 2 + 2 \cdot 1) & (1 \cdot 2 + 1 \cdot 1 + 2 \cdot -1) & (1 \cdot 1 + 1 \cdot 1 + 2 \cdot 2) \\ (1 \cdot 1 + 2 \cdot 2 + 1 \cdot 1) & (1 \cdot 2 + 2 \cdot 1 + 1 \cdot -1) & (1 \cdot 1 + 2 \cdot 1 + 1 \cdot 2) \\ (3 \cdot 1 + 2 \cdot 2 + 2 \cdot 1) & (3 \cdot 2 + 2 \cdot 1 + 2 \cdot -1) & (3 \cdot 1 + 2 \cdot 1 + 2 \cdot 2) \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 1 + 2 + 2 & 2 + 1 - 2 & 1 + 1 + 4 \\ 1 + 4 + 1 & 2 + 2 - 1 & 1 + 2 + 2 \\ 3 + 4 + 2 & 6 + 2 - 2 & 3 + 2 + 4 \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 5 & 1 & 6 \\ 6 & 3 & 5 \\ 9 & 6 & 9 \end{pmatrix} \][/tex]
The problem specifies the power [tex]\( x = 2 \)[/tex]. Thus, we calculate [tex]\( (AB)^2 \)[/tex] by squaring the matrix [tex]\(AB\)[/tex]:
[tex]\[ (AB)^2 = AB \cdot AB = \begin{pmatrix} 5 & 1 & 6 \\ 6 & 3 & 5 \\ 9 & 6 & 9 \end{pmatrix} \begin{pmatrix} 5 & 1 & 6 \\ 6 & 3 & 5 \\ 9 & 6 & 9 \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} (5 \cdot 5 + 1 \cdot 6 + 6 \cdot 9) & (5 \cdot 1 + 1 \cdot 3 + 6 \cdot 6) & (5 \cdot 6 + 1 \cdot 5 + 6 \cdot 9) \\ (6 \cdot 5 + 3 \cdot 6 + 5 \cdot 9) & (6 \cdot 1 + 3 \cdot 3 + 5 \cdot 6) & (6 \cdot 6 + 3 \cdot 5 + 5 \cdot 9) \\ (9 \cdot 5 + 6 \cdot 6 + 9 \cdot 9) & (9 \cdot 1 + 6 \cdot 3 + 9 \cdot 6) & (9 \cdot 6 + 6 \cdot 5 + 9 \cdot 9) \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} (25 + 6 + 54) & (5 + 3 + 36) & (30 + 5 + 54) \\ (30 + 18 + 45) & (6 + 9 + 30) & (36 + 15 + 45) \\ (45 + 36 + 81) & (9 + 18 + 54) & (54 + 30 + 81) \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 85 & 44 & 89 \\ 93 & 45 & 96 \\ 162 & 81 & 165 \end{pmatrix} \][/tex]
So, [tex]\( (AB)^2 = \begin{pmatrix} 85 & 44 & 89 \\ 93 & 45 & 96 \\ 162 & 81 & 165 \end{pmatrix} \)[/tex]
### Part (b)
To solve the simultaneous equations using the inverse matrix method, we first write the system of equations as a matrix equation: [tex]\( C \mathbf{x} = \mathbf{d} \)[/tex], where:
[tex]\[ C = \begin{pmatrix} 3 & 2 & 2 \\ 1 & 2 & 3 \\ 4 & 1 & 3 \end{pmatrix} \][/tex]
[tex]\[ \mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \][/tex]
[tex]\[ \mathbf{d} = \begin{pmatrix} 13 \\ 13 \\ 13 \end{pmatrix} \][/tex]
We need to find [tex]\( C^{-1} \)[/tex], the inverse of matrix [tex]\( C \)[/tex]:
[tex]\[ C^{-1} = \begin{pmatrix} 0.23076923 & -0.30769231 & 0.15384615 \\ 0.69230769 & 0.07692308 & -0.53846154 \\ -0.53846154 & 0.38461538 & 0.30769231 \end{pmatrix} \][/tex]
The solution to the system of equations is given by:
[tex]\[ \mathbf{x} = C^{-1} \mathbf{d} \][/tex]
[tex]\[ \mathbf{x} = \begin{pmatrix} 0.23076923 & -0.30769231 & 0.15384615 \\ 0.69230769 & 0.07692308 & -0.53846154 \\ -0.53846154 & 0.38461538 & 0.30769231 \end{pmatrix} \begin{pmatrix} 13 \\ 13 \\ 13 \end{pmatrix} \][/tex]
Multiplying these matrices, we obtain:
[tex]\[ \mathbf{x} = \begin{pmatrix} (0.23076923 \cdot 13 + -0.30769231 \cdot 13 + 0.15384615 \cdot 13) \\ (0.69230769 \cdot 13 + 0.07692308 \cdot 13 + -0.53846154 \cdot 13) \\ (-0.53846154 \cdot 13 + 0.38461538 \cdot 13 + 0.30769231 \cdot 13) \end{pmatrix} \][/tex]
[tex]\[ = \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} \][/tex]
Therefore, the solutions to the system of equations are:
[tex]\[ x = 1, \; y = 3, \; z = 2 \][/tex]
