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Math  /  Algebra

QuestionIf possible, find A+B,A−B,3AA+B, A-B, 3 A, and 3A−4B3 A-4 B. (If not possible, enter IMPOSSIBLE in any cell of the matrix.) A=[6−225−32],B=[48−2−4110]A=\left[\begin{array}{rr} 6 & -2 \\ 2 & 5 \\ -3 & 2 \end{array}\right], \quad B=\left[\begin{array}{rr} 4 & 8 \\ -2 & -4 \\ 1 & 10 \end{array}\right] (a) A+BA+B (b) A−BA-B (c) 3A3 A

Studdy Solution

STEP 1

1. Matrices A A and B B are both 3×2 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 A + B is possible and calculate it.
2. Check if A−B A - B is possible and calculate it.
3. Calculate 3A 3A .
4. Check if 3A−4B 3A - 4B is possible and calculate it.

STEP 3

Check if A+B A + B is possible by verifying that both matrices have the same dimensions. Since both A A and B B are 3×2 3 \times 2 matrices, addition is possible.
Calculate A+B A + B by adding corresponding elements:
A+B=[6−225−32]+[48−2−4110]=[6+4−2+82+(−2)5+(−4)−3+12+10]=[10601−212] 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 A - B is possible by verifying that both matrices have the same dimensions. Since both A A and B B are 3×2 3 \times 2 matrices, subtraction is possible.
Calculate A−B A - B by subtracting corresponding elements:
A−B=[6−225−32]−[48−2−4110]=[6−4−2−82−(−2)5−(−4)−3−12−10]=[2−1049−4−8] 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 3A by multiplying each element of matrix A A by 3:
3A=3×[6−225−32]=[3×63×(−2)3×23×53×(−3)3×2]=[18−6615−96] 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]

STEP 6

Check if 3A−4B 3A - 4B is possible by verifying that both matrices have the same dimensions. Since both 3A 3A and 4B 4B will be 3×2 3 \times 2 matrices, subtraction is possible.
First, calculate 4B 4B :
4B=4×[48−2−4110]=[4×44×84×(−2)4×(−4)4×14×10]=[1632−8−16440] 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 :
3A−4B=[18−6615−96]−[1632−8−16440]=[18−16−6−326−(−8)15−(−16)−9−46−40]=[2−381431−13−34] 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=[10601−212] A+B = \left[\begin{array}{rr} 10 & 6 \\ 0 & 1 \\ -2 & 12 \end{array}\right] (b) A−B=[2−1049−4−8] A-B = \left[\begin{array}{rr} 2 & -10 \\ 4 & 9 \\ -4 & -8 \end{array}\right] (c) 3A=[18−6615−96] 3A = \left[\begin{array}{rr} 18 & -6 \\ 6 & 15 \\ -9 & 6 \end{array}\right] (d) 3A−4B=[2−381431−13−34] 3A - 4B = \left[\begin{array}{rr} 2 & -38 \\ 14 & 31 \\ -13 & -34 \end{array}\right]

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