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We are given two matrices \( K \) and \( L \), and the problem states that \( K = L \), so we need to find the value of \( c \) that satisfies this condition.

The matrices are:
\[
K = \begin{pmatrix} 
a & 2 & 3 \\
5 & 4 & b \\
8 & 3c & 11 
\end{pmatrix}
\quad \text{and} \quad 
L = \begin{pmatrix} 
6 & 2 & 3 \\
5 & 4 & 21 \\
8 & 4b & 11 
\end{pmatrix}
\]

Since \( K = L \), corresponding elements of these matrices must be equal. Let's match them:

1. First row, first column: \( a = 6 \)
2. Second row, third column: \( b = 21 \)
3. Third row, second column: \( 3c = 4b \)

From the third equation, substitute \( b = 21 \) into \( 3c = 4b \):
\[
3c = 4 \times 21 = 84
\]
\[
c = \frac{84}{3} = 28
\]

Thus, the value of \( c \) is \( 28 \).

The correct answer is **E. 28**.

Would you like any further details or explanations?

Here are some related questions to expand your understanding:

1. What is the process for solving systems of equations from matrices?
2. How are matrices typically used to solve linear algebra problems?
3. What are the rules for matrix equality?
4. Can you explain how substitution works in algebraic equations?
5. How would the solution change if the matrices were not square?

**Tip:** In matrix equality, every corresponding element in both matrices must be exactly the same for the matrices to be considered equal.

Given two matrices K = \begin{pmatrix} a & 2 & 3 \\ 5 & 4 & b \\ 8 & 3c & 11 \end{pmatrix} and L = \begin{pmatrix} 6 & 2 & 3 \\ 5 & 4 & 21 \\ 8 & 4b & 11 \end{pmatrix}, find the value of c if K = L.

Learn how to solve for c in a matrix equality problem where two 3x3 matrices K and L are equal. The problem explores matrix equality and substitution techniques. Suitable for students in Grades 10-12.

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