Math Snap

STEP 1

1. Matrices $A$ and $B$ are both $3 \times 2$ matrices.
2. Matrix addition and subtraction require matrices to have the same dimensions.
3. Scalar multiplication of a matrix involves multiplying every element of the matrix by the scalar.

STEP 2

1. Check if $A + B$ is possible and calculate it.
2. Check if $A - B$ is possible and calculate it.
3. Calculate $3A$.
4. Check if $3A - 4B$ is possible and calculate it.

STEP 3

Check if $A + B$ is possible by verifying that both matrices have the same dimensions. Since both $A$ and $B$ are $3 \times 2$ matrices, addition is possible.
Calculate $A + B$ by adding corresponding elements:
\[ A + B = \left[\begin{array}{rr} 6 & -2 \\ 2 & 5 \\ -3 & 2 \end{array}\right] + \left[\begin{array}{rr}
4 & 8 \\
-2 & -4 \\
1 & 10
\end{array}\right] = \left[\begin{array}{rr}
6+4 & -2+8 \\
2+(-2) & 5+(-4) \\
-3+1 & 2+10
\end{array}\right] = \left[\begin{array}{rr}
10 & 6 \\
0 & 1 \\
-2 & 12
\end{array}\right] \]

STEP 4

Check if $A - B$ is possible by verifying that both matrices have the same dimensions. Since both $A$ and $B$ are $3 \times 2$ matrices, subtraction is possible.
Calculate $A - B$ by subtracting corresponding elements:
\[ A - B = \left[\begin{array}{rr} 6 & -2 \\ 2 & 5 \\ -3 & 2 \end{array}\right] - \left[\begin{array}{rr}
4 & 8 \\
-2 & -4 \\
1 & 10
\end{array}\right] = \left[\begin{array}{rr}
6-4 & -2-8 \\
2-(-2) & 5-(-4) \\
-3-1 & 2-10
\end{array}\right] = \left[\begin{array}{rr}
2 & -10 \\
4 & 9 \\
-4 & -8
\end{array}\right] \]

STEP 5

Calculate $3A$ by multiplying each element of matrix $A$ by 3:
\[ 3A = 3 \times \left[\begin{array}{rr} 6 & -2 \\ 2 & 5 \\ -3 & 2 \end{array}\right] = \left[\begin{array}{rr}
3 \times 6 & 3 \times (-2) \\
3 \times 2 & 3 \times 5 \\
3 \times (-3) & 3 \times 2
\end{array}\right] = \left[\begin{array}{rr}
18 & -6 \\
6 & 15 \\
-9 & 6
\end{array}\right] \]

Check if $3A - 4B$ is possible by verifying that both matrices have the same dimensions. Since both $3A$ and $4B$ will be $3 \times 2$ matrices, subtraction is possible.
First, calculate $4B$:
\[ 4B = 4 \times \left[\begin{array}{rr} 4 & 8 \\ -2 & -4 \\ 1 & 10 \end{array}\right] = \left[\begin{array}{rr}
4 \times 4 & 4 \times 8 \\
4 \times (-2) & 4 \times (-4) \\
4 \times 1 & 4 \times 10
\end{array}\right] = \left[\begin{array}{rr}
16 & 32 \\
-8 & -16 \\
4 & 40
\end{array}\right] \] Now, calculate $3A - 4B$:
\[ 3A - 4B = \left[\begin{array}{rr} 18 & -6 \\ 6 & 15 \\ -9 & 6 \end{array}\right] - \left[\begin{array}{rr}
16 & 32 \\
-8 & -16 \\
4 & 40
\end{array}\right] = \left[\begin{array}{rr}
18-16 & -6-32 \\
6-(-8) & 15-(-16) \\
-9-4 & 6-40
\end{array}\right] = \left[\begin{array}{rr}
2 & -38 \\
14 & 31 \\
-13 & -34
\end{array}\right] \] The solutions are:
(a) $A+B = \left[\begin{array}{rr} 10 & 6 \\ 0 & 1 \\ -2 & 12 \end{array}\right]$
(b) $A-B = \left[\begin{array}{rr} 2 & -10 \\ 4 & 9 \\ -4 & -8 \end{array}\right]$
(c) $3A = \left[\begin{array}{rr} 18 & -6 \\ 6 & 15 \\ -9 & 6 \end{array}\right]$
(d) $3A - 4B = \left[\begin{array}{rr} 2 & -38 \\ 14 & 31 \\ -13 & -34 \end{array}\right]$