Assuming that the sums and products given below are defined which of the following is not true for matrices.

Options

A. AB=AC does not imply B=C
B. `A+B=B+A`
C. `(AB)'=B'A'`
D. `AB=OimpliesA=O` or B=O

Answer

To determine which of the given statements about matrices is not true, we will analyze each option step by step.

### Step 1: Define the Matrices
Let's define the matrices as follows:
- Matrix A = \(\begin{pmatrix} 1 & 2 \\ 3 & 6 \end{pmatrix}\)
- Matrix B = \(\begin{pmatrix} 1 & 7 \\ 6 & 4 \end{pmatrix}\)
- Matrix C = \(\begin{pmatrix} 5 & 11 \\ 4 & 2 \end{pmatrix}\)

### Step 2: Check Option 1
We need to check if \(AB = AC\) implies \(B = C\).

1. **Calculate \(AB\)**:
\[ AB = \begin{pmatrix} 1 & 2 \\ 3 & 6 \end{pmatrix} \begin{pmatrix} 1 & 7 \\ 6 & 4 \end{pmatrix} \]
- First row, first column: \(1 \cdot 1 + 2 \cdot 6 = 1 + 12 = 13\)
- First row, second column: \(1 \cdot 7 + 2 \cdot 4 = 7 + 8 = 15\)
- Second row, first column: \(3 \cdot 1 + 6 \cdot 6 = 3 + 36 = 39\)
- Second row, second column: \(3 \cdot 7 + 6 \cdot 4 = 21 + 24 = 45\)

Thus,
\[ AB = \begin{pmatrix} 13 & 15 \\ 39 & 45 \end{pmatrix} \]

2. **Calculate \(AC\)**:
\[ AC = \begin{pmatrix} 1 & 2 \\ 3 & 6 \end{pmatrix} \begin{pmatrix} 5 & 11 \\ 4 & 2 \end{pmatrix} \]
- First row, first column: \(1 \cdot 5 + 2 \cdot 4 = 5 + 8 = 13\)
- First row, second column: \(1 \cdot 11 + 2 \cdot 2 = 11 + 4 = 15\)
- Second row, first column: \(3 \cdot 5 + 6 \cdot 4 = 15 + 24 = 39\)
- Second row, second column: \(3 \cdot 11 + 6 \cdot 2 = 33 + 12 = 45\)

Thus,
\[ AC = \begin{pmatrix} 13 & 15 \\ 39 & 45 \end{pmatrix} \]

Since \(AB = AC\) but \(B \neq C\), this option is true.

### Step 3: Check Option 2
We need to check if \(A + B = B + A\).

1. **Calculate \(A + B\)**:
\[ A + B = \begin{pmatrix} 1 & 2 \\ 3 & 6 \end{pmatrix} + \begin{pmatrix} 1 & 7 \\ 6 & 4 \end{pmatrix} = \begin{pmatrix} 1+1 & 2+7 \\ 3+6 & 6+4 \end{pmatrix} = \begin{pmatrix} 2 & 9 \\ 9 & 10 \end{pmatrix} \]

2. **Calculate \(B + A\)**:
\[ B + A = \begin{pmatrix} 1 & 7 \\ 6 & 4 \end{pmatrix} + \begin{pmatrix} 1 & 2 \\ 3 & 6 \end{pmatrix} = \begin{pmatrix} 1+1 & 7+2 \\ 6+3 & 4+6 \end{pmatrix} = \begin{pmatrix} 2 & 9 \\ 9 & 10 \end{pmatrix} \]

Since \(A + B = B + A\), this option is true.

### Step 4: Check Option 3
We need to verify if \((AB)^T = B^T A^T\).

1. **Calculate \((AB)^T\)**:
We already found \(AB = \begin{pmatrix} 13 & 15 \\ 39 & 45 \end{pmatrix}\).
\[ (AB)^T = \begin{pmatrix} 13 & 39 \\ 15 & 45 \end{pmatrix} \]

2. **Calculate \(B^T\) and \(A^T\)**:
\[ B^T = \begin{pmatrix} 1 & 6 \\ 7 & 4 \end{pmatrix}, \quad A^T = \begin{pmatrix} 1 & 3 \\ 2 & 6 \end{pmatrix} \]

3. **Calculate \(B^T A^T\)**:
\[ B^T A^T = \begin{pmatrix} 1 & 6 \\ 7 & 4 \end{pmatrix} \begin{pmatrix} 1 & 3 \\ 2 & 6 \end{pmatrix} \]
- First row, first column: \(1 \cdot 1 + 6 \cdot 2 = 1 + 12 = 13\)
- First row, second column: \(1 \cdot 3 + 6 \cdot 6 = 3 + 36 = 39\)
- Second row, first column: \(7 \cdot 1 + 4 \cdot 2 = 7 + 8 = 15\)
- Second row, second column: \(7 \cdot 3 + 4 \cdot 6 = 21 + 24 = 45\)

Thus,
\[ B^T A^T = \begin{pmatrix} 13 & 39 \\ 15 & 45 \end{pmatrix} \]

Since \((AB)^T \neq B^T A^T\), this option is false.

### Step 5: Check Option 4
We need to check if \(AB = 0\) implies \(A = 0\) and \(B = 0\).

1. **Assume \(A\) and \(B\)**:
Let \(A = \begin{pmatrix} 0 & 0 \\ 4 & 0 \end{pmatrix}\) and \(B = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}\).

2. **Calculate \(AB\)**:
\[ AB = \begin{pmatrix} 0 & 0 \\ 4 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]

Here, \(AB = 0\) does not imply that both \(A\) and \(B\) are zero. Therefore, this option is false.

### Conclusion
The option that is not true is **Option 4**: \(AB = 0\) implies \(A = 0\) and \(B = 0\).