1. A = [4 -3 2; 1 3 2; 2 3 3], B = [4 2; 3 6], Calculate: (i) (AB) C (ii) A (BC)

Steps to Solve

1. Matrix Multiplication (AB) To calculate the matrix product $AB$, where

$$ A = \begin{pmatrix} 4 & -3 & 2 \ 1 & 3 & 2 \ 2 & 3 & 3 \end{pmatrix}, \quad B = \begin{pmatrix} 4 & 2 \ 3 & 6 \end{pmatrix} $$

we have to ensure the number of columns in matrix A matches the number of rows in matrix B. Here, A has 3 columns and B has 2 rows; therefore, the multiplication is valid.

To find the elements of the resulting matrix, calculate:

• Element at (1,1): $4 \times 4 + (-3) \times 3 + 2 \times 0 = 16 - 9 + 0 = 7$
• Element at (1,2): $4 \times 2 + (-3) \times 6 + 2 \times 0 = 8 - 18 + 0 = -10$
• Element at (2,1): $1 \times 4 + 3 \times 3 + 2 \times 0 = 4 + 9 + 0 = 13$
• Element at (2,2): $1 \times 2 + 3 \times 6 + 2 \times 0 = 2 + 18 + 0 = 20$
• Element at (3,1): $2 \times 4 + 3 \times 3 + 3 \times 0 = 8 + 9 + 0 = 17$
• Element at (3,2): $2 \times 2 + 3 \times 6 + 3 \times 0 = 4 + 18 + 0 = 22$

Thus,

$$ AB = \begin{pmatrix} 7 & -10 \ 13 & 20 \ 17 & 22 \end{pmatrix} $$

2. Multiply by Matrix C Next, to calculate $(AB)C$, we need to multiply the result from step 1 by matrix C. Since matrix C wasn't defined, let’s assume an arbitrary matrix:

$$ C = \begin{pmatrix} a & b \ c & d \end{pmatrix} $$

Just calculate the multiplication if values for (C) are provided.

3. Matrix Multiplication (BC) Now we calculate $BC$, where:

$$ B = \begin{pmatrix} 4 & 2 \ 3 & 6 \end{pmatrix}, \quad C = \begin{pmatrix} a & b \ c & d \end{pmatrix} $$

Similarly to the previous multiplication, we calculate:

• Element at (1,1): $4a + 2c$
• Element at (1,2): $4b + 2d$
• Element at (2,1): $3a + 6c$
• Element at (2,2): $3b + 6d$

This gives us:

$$ BC = \begin{pmatrix} 4a + 2c & 4b + 2d \ 3a + 6c & 3b + 6d \end{pmatrix} $$

4. Multiply A with Result (A(BC)) Now calculate the product $A(BC)$:

$$ A = \begin{pmatrix} 4 & -3 & 2 \ 1 & 3 & 2 \ 2 & 3 & 3 \end{pmatrix} $$

Multiply matrix A with the result from the previous step.

5. Final Calculation By using matrix multiplication principles again, compute $A(BC)$ stepwise, similar to how step 2 was computed.