https://file.aiquickdraw.com/mathbot/file/1733297666653l0z3wpmz.jpg

Let's solve the given problem step by step.

### Problem Overview:
We have matrices:
\[
A = \begin{pmatrix} -1 & 0 \\ 3 & 5 \end{pmatrix}, 
B = \begin{pmatrix} 2 & 0 \\ 4 & -3 \\ 0 & -1 \end{pmatrix}, 
C = \begin{pmatrix} 6 \\ -2 \end{pmatrix}.
\]

We need to solve the following:
1. \(\frac{1}{2}C\)
2. \(CA\)
3. \(A^2\)
4. \(2BA\)

---

### Step 1: \(\frac{1}{2}C\)
\[
\frac{1}{2}C = \frac{1}{2} \begin{pmatrix} 6 \\ -2 \end{pmatrix}
= \begin{pmatrix} 3 \\ -1 \end{pmatrix}.
\]

---

### Step 2: \(CA\)
Since \(C\) is \(2 \times 1\) and \(A\) is \(2 \times 2\), the multiplication \(CA\) is **not defined** because the number of columns in \(C\) does not match the number of rows in \(A\).

---

### Step 3: \(A^2\)
\[
A^2 = A \cdot A = \begin{pmatrix} -1 & 0 \\ 3 & 5 \end{pmatrix} \cdot \begin{pmatrix} -1 & 0 \\ 3 & 5 \end{pmatrix}.
\]
Performing the matrix multiplication:
\[
A^2 = \begin{pmatrix}
(-1)(-1) + (0)(3) & (-1)(0) + (0)(5) \\
(3)(-1) + (5)(3) & (3)(0) + (5)(5)
\end{pmatrix}
= \begin{pmatrix}
1 & 0 \\
12 & 25
\end{pmatrix}.
\]

---

### Step 4: \(2BA\)
\[
2BA = 2 \cdot \left(B \cdot A\right).
\]
First, calculate \(B \cdot A\):
\[
B \cdot A = \begin{pmatrix} 2 & 0 \\ 4 & -3 \\ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} -1 & 0 \\ 3 & 5 \end{pmatrix}.
\]
Performing the multiplication:
\[
B \cdot A = \begin{pmatrix}
(2)(-1) + (0)(3) & (2)(0) + (0)(5) \\
(4)(-1) + (-3)(3) & (4)(0) + (-3)(5) \\
(0)(-1) + (-1)(3) & (0)(0) + (-1)(5)
\end{pmatrix}
= \begin{pmatrix}
-2 & 0 \\
-13 & -15 \\
-3 & -5
\end{pmatrix}.
\]
Now multiply by 2:
\[
2BA = 2 \cdot \begin{pmatrix} -2 & 0 \\ -13 & -15 \\ -3 & -5 \end{pmatrix}
= \begin{pmatrix} -4 & 0 \\ -26 & -30 \\ -6 & -10 \end{pmatrix}.
\]

---

### Final Answers:
1. \(\frac{1}{2}C = \begin{pmatrix} 3 \\ -1 \end{pmatrix}\)
2. \(CA\): **Not defined**
3. \(A^2 = \begin{pmatrix} 1 & 0 \\ 12 & 25 \end{pmatrix}\)
4. \(2BA = \begin{pmatrix} -4 & 0 \\ -26 & -30 \\ -6 & -10 \end{pmatrix}\)

Let me know if you want detailed steps for each calculation!

---

#### Further Questions:
1. How can matrix multiplication rules help identify undefined products quickly?
2. What are the implications of squaring a matrix, and why does \(A^2\) result in this specific pattern?
3. Could \(CA\) be redefined if \(C\) were reshaped? How?
4. How is scalar multiplication applied uniformly across matrices like in \(2BA\)?
5. What happens when you try to compute higher powers of matrix \(A\), like \(A^3\)?

#### Tip:
Always check matrix dimensions before performing multiplication to avoid errors in computation!

Diberi matriks A = [[-1, 0], [3, 5]], B = [[2, 0], [4, -3], [0, -1]], dan C = [6, -2]. Kirakan: (i) (1/2)C, (ii) CA, (iii) A^2, (iv) 2BA.

Solve matrix operations including scalar multiplication, matrix multiplication, and squaring. This problem involves matrices A, B, and C with calculations such as (1/2)C, A^2, and 2BA. Detailed solution provided, suitable for Grades 10-12.

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