If A=[ 21 32 43
 41 70 10 ], B =[
   15
   76
 −28
 ], C = [11 91 −1   27] then prove that  
(AB)C=A(BC).

To prove that \((AB)C = A(BC)\) for the given matrices \( A \), \( B \), and \( C \), we need to perform the matrix multiplications on both sides of the equation and show that they result in the same matrix.

### Given Matrices:

\[
A = \begin{bmatrix} 
21 & 32 & 43 \\ 
41 & 70 & 10 
\end{bmatrix}
\]
\[
B = \begin{bmatrix} 
15 \\ 
76 \\ 
-28 
\end{bmatrix}
\]
\[
C = \begin{bmatrix} 
11 & 91 & -1 & 27 
\end{bmatrix}
\]

### Step 1: Calculate \( AB \) (Matrix multiplication of \( A \) and \( B \)).

Matrix \( A \) is \( 2 \times 3 \), and matrix \( B \) is \( 3 \times 1 \). So the result of \( AB \) will be a \( 2 \times 1 \) matrix.

\[
AB = \begin{bmatrix} 
21 & 32 & 43 \\ 
41 & 70 & 10 
\end{bmatrix}
\begin{bmatrix} 
15 \\ 
76 \\ 
-28 
\end{bmatrix}
\]

To compute the elements of the resulting matrix \( AB \):

\[
(AB)_{1,1} = 21(15) + 32(76) + 43(-28) = 315 + 2432 - 120

If A=[ 21 32 43; 41 70 10 ], B =[ 15; 76; −28 ], C = [11 91 −1 27] then prove that (AB)C=A(BC).

Learn how to prove the associative property of matrix multiplication (AB)C = A(BC) with detailed step-by-step matrix multiplication. This solution involves the use of 2x3 and 3x1 matrices and demonstrates how to compute both sides to verify equality. Suitable for students in Grades 11-12.

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Given Matrices:

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