Scalar multiples questions with answers

Questions

1. Question 1A whole-number multiple

   $A = \begin{pmatrix} 5 & -2 & 2 \\ 1 & 8 & 7 \end{pmatrix}$. Find $5A$.

   Show the worked solution

   1. Step 1: Row 1: multiply every element by the scalarColumn 1: $5 \times 5 = 25$Column 2: $5 \times (-2) = -10$Column 3: $5 \times 2 = 10$
   2. Step 2: Row 2: multiply every element by the scalarColumn 1: $5 \times 1 = 5$Column 2: $5 \times 8 = 40$Column 3: $5 \times 7 = 35$
   3. Step 3: Write the answer as a matrix$5A = \begin{pmatrix} 25 & -10 & 10 \\ 5 & 40 & 35 \end{pmatrix}$
2. Question 2A whole-number multiple

   $A = \begin{pmatrix} -1 & 8 & 2 \\ -1 & 2 & 1 \end{pmatrix}$. Find $-2A$.

   Show the worked solution

   1. Step 1: Row 1: multiply every element by the scalarColumn 1: $-2 \times (-1) = 2$Column 2: $-2 \times 8 = -16$Column 3: $-2 \times 2 = -4$
   2. Step 2: Row 2: multiply every element by the scalarColumn 1: $-2 \times (-1) = 2$Column 2: $-2 \times 2 = -4$Column 3: $-2 \times 1 = -2$
   3. Step 3: Write the answer as a matrix$-2A = \begin{pmatrix} 2 & -16 & -4 \\ 2 & -4 & -2 \end{pmatrix}$
3. Question 3A fraction of a matrix

   $A = \begin{pmatrix} -9 & -9 \\ -12 & 9 \end{pmatrix}$. Find $\dfrac{1}{3}A$.

   Show the worked solution

   1. Step 1: Row 1: multiply every element by the fractionColumn 1: $\dfrac{1}{3} \times (-9) = -3$Column 2: $\dfrac{1}{3} \times (-9) = -3$
   2. Step 2: Row 2: multiply every element by the fractionColumn 1: $\dfrac{1}{3} \times (-12) = -4$Column 2: $\dfrac{1}{3} \times 9 = 3$
   3. Step 3: Write the answer as a matrix$\dfrac{1}{3}A = \begin{pmatrix} -3 & -3 \\ -4 & 3 \end{pmatrix}$
4. Question 4A fraction of a matrix

   $A = \begin{pmatrix} -6 & -6 \\ 12 & -4 \end{pmatrix}$. Find $\dfrac{1}{2}A$.

   Show the worked solution

   1. Step 1: Row 1: multiply every element by the fractionColumn 1: $\dfrac{1}{2} \times (-6) = -3$Column 2: $\dfrac{1}{2} \times (-6) = -3$
   2. Step 2: Row 2: multiply every element by the fractionColumn 1: $\dfrac{1}{2} \times 12 = 6$Column 2: $\dfrac{1}{2} \times (-4) = -2$
   3. Step 3: Write the answer as a matrix$\dfrac{1}{2}A = \begin{pmatrix} -3 & -3 \\ 6 & -2 \end{pmatrix}$
5. Question 5Combining multiples

   $A = \begin{pmatrix} 5 & -4 \\ 5 & 6 \end{pmatrix}$ and $B = \begin{pmatrix} 5 & 7 \\ -1 & 7 \end{pmatrix}$. Find $A - 3B$.

   Show the worked solution

   1. Step 1: Row 1: multiply, then combineColumn 1: $5 - 3 \times 5 = -10$Column 2: $(-4) - 3 \times 7 = -25$
   2. Step 2: Row 2: multiply, then combineColumn 1: $5 - 3 \times (-1) = 8$Column 2: $6 - 3 \times 7 = -15$
   3. Step 3: Write the answer as a matrix$A - 3B = \begin{pmatrix} -10 & -25 \\ 8 & -15 \end{pmatrix}$
6. Question 6Combining multiples

   $A = \begin{pmatrix} -3 & -3 \\ -3 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} -3 & -2 \\ 5 & 4 \end{pmatrix}$. Find $A - 3B$.

   Show the worked solution

   1. Step 1: Row 1: multiply, then combineColumn 1: $(-3) - 3 \times (-3) = 6$Column 2: $(-3) - 3 \times (-2) = 3$
   2. Step 2: Row 2: multiply, then combineColumn 1: $(-3) - 3 \times 5 = -18$Column 2: $1 - 3 \times 4 = -11$
   3. Step 3: Write the answer as a matrix$A - 3B = \begin{pmatrix} 6 & 3 \\ -18 & -11 \end{pmatrix}$