Multiplying matrices questions with answers

Questions

1. Question 1Can AB be found? Its order

   $A$ is a $3 \times 1$ matrix and $B$ is a $3 \times 2$ matrix. State the order of $AB$, or write "not possible" if $AB$ cannot be found.

   Show the worked solution

   1. Step 1: Write the orders side by side$(3 \mathbin{\times} 1)(3 \mathbin{\times} 2)$
   2. Step 2: Compare the middle numbers$1 \ne 3$Can $AB$ be found? $\text{No}$
   3. Step 3: So the answer isSo $AB$ is not possible.
2. Question 2A row times a column

   $A = \begin{pmatrix} -3 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} -2 \\ 4 \end{pmatrix}$. Find $AB$.

   Show the worked solution

   1. Step 1: The order of AB$(1 \times 2)(2 \times 1) \rightarrow 1 \times 1$
   2. Step 2: Row 1 of A times each column of B$(-3) \times (-2) + 2 \times 4 = 14$
   3. Step 3: Write the answer as a matrix$AB = \begin{pmatrix} 14 \end{pmatrix}$
3. Question 32 × 2 times 2 × 1

   $A = \begin{pmatrix} 1 & 3 \\ 4 & -2 \end{pmatrix}$ and $B = \begin{pmatrix} 3 \\ -1 \end{pmatrix}$. Find $AB$.

   Show the worked solution

   1. Step 1: The order of AB$(2 \times 2)(2 \times 1) \rightarrow 2 \times 1$
   2. Step 2: Row 1 of A times each column of B$1 \times 3 + 3 \times (-1) = 0$
   3. Step 3: Row 2 of A times each column of B$4 \times 3 + (-2) \times (-1) = 14$
   4. Step 4: Write the answer as a matrix$AB = \begin{pmatrix} 0 \\ 14 \end{pmatrix}$
4. Question 42 × 2 times 2 × 2

   $A = \begin{pmatrix} 5 & 4 \\ -2 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} -3 & 2 \\ 3 & -2 \end{pmatrix}$. Find $AB$.

   Show the worked solution

   1. Step 1: The order of AB$(2 \times 2)(2 \times 2) \rightarrow 2 \times 2$
   2. Step 2: Row 1 of A times each column of BColumn 1: $5 \times (-3) + 4 \times 3 = -3$Column 2: $5 \times 2 + 4 \times (-2) = 2$
   3. Step 3: Row 2 of A times each column of BColumn 1: $(-2) \times (-3) + 4 \times 3 = 18$Column 2: $(-2) \times 2 + 4 \times (-2) = -12$
   4. Step 4: Write the answer as a matrix$AB = \begin{pmatrix} -3 & 2 \\ 18 & -12 \end{pmatrix}$
5. Question 5Other orders

   $A = \begin{pmatrix} 6 \\ 1 \end{pmatrix}$ and $B = \begin{pmatrix} -1 & -1 \end{pmatrix}$. Find $AB$.

   Show the worked solution

   1. Step 1: The order of AB$(2 \times 1)(1 \times 2) \rightarrow 2 \times 2$
   2. Step 2: Row 1 of A times each column of BColumn 1: $6 \times (-1) = -6$Column 2: $6 \times (-1) = -6$
   3. Step 3: Row 2 of A times each column of BColumn 1: $1 \times (-1) = -1$Column 2: $1 \times (-1) = -1$
   4. Step 4: Write the answer as a matrix$AB = \begin{pmatrix} -6 & -6 \\ -1 & -1 \end{pmatrix}$