Get detailed and accurate responses to your questions with IDNLearn.com. Ask any question and receive timely, accurate responses from our dedicated community of experts.
Sagot :
To solve the given matrix operations, we need to perform addition, subtraction, scalar multiplication, and a combination of these on the matrices provided. Here are the matrices:
[tex]\[ A = \left[\begin{array}{cc} 3 & 1 \\ 1 & 3 \end{array}\right], \quad B = \left[\begin{array}{rr} -2 & -2 \\ 1 & 2 \end{array}\right] \][/tex]
### (a) [tex]\(A + B\)[/tex]
To find [tex]\(A + B\)[/tex], we add the corresponding elements of matrices [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
[tex]\[ A + B = \left[\begin{array}{cc} 3 & 1 \\ 1 & 3 \end{array}\right] + \left[\begin{array}{rr} -2 & -2 \\ 1 & 2 \end{array}\right] = \left[\begin{array}{cc} 3 + (-2) & 1 + (-2) \\ 1 + 1 & 3 + 2 \end{array}\right] = \left[\begin{array}{cc} 1 & -1 \\ 2 & 5 \end{array}\right] \][/tex]
So, [tex]\(A + B = \left[\begin{array}{cc} 1 & -1 \\ 2 & 5 \end{array}\right]\)[/tex].
### (b) [tex]\(A - B\)[/tex]
To find [tex]\(A - B\)[/tex], we subtract the corresponding elements of matrix [tex]\(B\)[/tex] from matrix [tex]\(A\)[/tex]:
[tex]\[ A - B = \left[\begin{array}{cc} 3 & 1 \\ 1 & 3 \end{array}\right] - \left[\begin{array}{rr} -2 & -2 \\ 1 & 2 \end{array}\right] = \left[\begin{array}{cc} 3 - (-2) & 1 - (-2) \\ 1 - 1 & 3 - 2 \end{array}\right] = \left[\begin{array}{cc} 5 & 3 \\ 0 & 1 \end{array}\right] \][/tex]
So, [tex]\(A - B = \left[\begin{array}{cc} 5 & 3 \\ 0 & 1 \end{array}\right]\)[/tex].
### (c) [tex]\(5A\)[/tex]
To find [tex]\(5A\)[/tex], we multiply each element of matrix [tex]\(A\)[/tex] by the scalar [tex]\(5\)[/tex]:
[tex]\[ 5A = 5 \cdot \left[\begin{array}{cc} 3 & 1 \\ 1 & 3 \end{array}\right] = \left[\begin{array}{cc} 5 \cdot 3 & 5 \cdot 1 \\ 5 \cdot 1 & 5 \cdot 3 \end{array}\right] = \left[\begin{array}{cc} 15 & 5 \\ 5 & 15 \end{array}\right] \][/tex]
So, [tex]\(5A = \left[\begin{array}{cc} 15 & 5 \\ 5 & 15 \end{array}\right]\)[/tex].
### (d) [tex]\(5A - 2B\)[/tex]
To find [tex]\(5A - 2B\)[/tex], we first find [tex]\(2B\)[/tex] by multiplying each element of matrix [tex]\(B\)[/tex] by the scalar [tex]\(2\)[/tex], and then subtract it from [tex]\(5A\)[/tex].
[tex]\[ 2B = 2 \cdot \left[\begin{array}{rr} -2 & -2 \\ 1 & 2 \end{array}\right] = \left[\begin{array}{rr} 2 \cdot (-2) & 2 \cdot (-2) \\ 2 \cdot 1 & 2 \cdot 2 \end{array}\right] = \left[\begin{array}{rr} -4 & -4 \\ 2 & 4 \end{array}\right] \][/tex]
Now, subtract [tex]\(2B\)[/tex] from [tex]\(5A\)[/tex]:
[tex]\[ 5A - 2B = \left[\begin{array}{cc} 15 & 5 \\ 5 & 15 \end{array}\right] - \left[\begin{array}{rr} -4 & -4 \\ 2 & 4 \end{array}\right] = \left[\begin{array}{cc} 15 - (-4) & 5 - (-4) \\ 5 - 2 & 15 - 4 \end{array}\right] = \left[\begin{array}{cc} 19 & 9 \\ 3 & 11 \end{array}\right] \][/tex]
So, [tex]\(5A - 2B = \left[\begin{array}{cc} 19 & 9 \\ 3 & 11 \end{array}\right]\)[/tex].
[tex]\[ A = \left[\begin{array}{cc} 3 & 1 \\ 1 & 3 \end{array}\right], \quad B = \left[\begin{array}{rr} -2 & -2 \\ 1 & 2 \end{array}\right] \][/tex]
### (a) [tex]\(A + B\)[/tex]
To find [tex]\(A + B\)[/tex], we add the corresponding elements of matrices [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
[tex]\[ A + B = \left[\begin{array}{cc} 3 & 1 \\ 1 & 3 \end{array}\right] + \left[\begin{array}{rr} -2 & -2 \\ 1 & 2 \end{array}\right] = \left[\begin{array}{cc} 3 + (-2) & 1 + (-2) \\ 1 + 1 & 3 + 2 \end{array}\right] = \left[\begin{array}{cc} 1 & -1 \\ 2 & 5 \end{array}\right] \][/tex]
So, [tex]\(A + B = \left[\begin{array}{cc} 1 & -1 \\ 2 & 5 \end{array}\right]\)[/tex].
### (b) [tex]\(A - B\)[/tex]
To find [tex]\(A - B\)[/tex], we subtract the corresponding elements of matrix [tex]\(B\)[/tex] from matrix [tex]\(A\)[/tex]:
[tex]\[ A - B = \left[\begin{array}{cc} 3 & 1 \\ 1 & 3 \end{array}\right] - \left[\begin{array}{rr} -2 & -2 \\ 1 & 2 \end{array}\right] = \left[\begin{array}{cc} 3 - (-2) & 1 - (-2) \\ 1 - 1 & 3 - 2 \end{array}\right] = \left[\begin{array}{cc} 5 & 3 \\ 0 & 1 \end{array}\right] \][/tex]
So, [tex]\(A - B = \left[\begin{array}{cc} 5 & 3 \\ 0 & 1 \end{array}\right]\)[/tex].
### (c) [tex]\(5A\)[/tex]
To find [tex]\(5A\)[/tex], we multiply each element of matrix [tex]\(A\)[/tex] by the scalar [tex]\(5\)[/tex]:
[tex]\[ 5A = 5 \cdot \left[\begin{array}{cc} 3 & 1 \\ 1 & 3 \end{array}\right] = \left[\begin{array}{cc} 5 \cdot 3 & 5 \cdot 1 \\ 5 \cdot 1 & 5 \cdot 3 \end{array}\right] = \left[\begin{array}{cc} 15 & 5 \\ 5 & 15 \end{array}\right] \][/tex]
So, [tex]\(5A = \left[\begin{array}{cc} 15 & 5 \\ 5 & 15 \end{array}\right]\)[/tex].
### (d) [tex]\(5A - 2B\)[/tex]
To find [tex]\(5A - 2B\)[/tex], we first find [tex]\(2B\)[/tex] by multiplying each element of matrix [tex]\(B\)[/tex] by the scalar [tex]\(2\)[/tex], and then subtract it from [tex]\(5A\)[/tex].
[tex]\[ 2B = 2 \cdot \left[\begin{array}{rr} -2 & -2 \\ 1 & 2 \end{array}\right] = \left[\begin{array}{rr} 2 \cdot (-2) & 2 \cdot (-2) \\ 2 \cdot 1 & 2 \cdot 2 \end{array}\right] = \left[\begin{array}{rr} -4 & -4 \\ 2 & 4 \end{array}\right] \][/tex]
Now, subtract [tex]\(2B\)[/tex] from [tex]\(5A\)[/tex]:
[tex]\[ 5A - 2B = \left[\begin{array}{cc} 15 & 5 \\ 5 & 15 \end{array}\right] - \left[\begin{array}{rr} -4 & -4 \\ 2 & 4 \end{array}\right] = \left[\begin{array}{cc} 15 - (-4) & 5 - (-4) \\ 5 - 2 & 15 - 4 \end{array}\right] = \left[\begin{array}{cc} 19 & 9 \\ 3 & 11 \end{array}\right] \][/tex]
So, [tex]\(5A - 2B = \left[\begin{array}{cc} 19 & 9 \\ 3 & 11 \end{array}\right]\)[/tex].
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. IDNLearn.com is committed to providing accurate answers. Thanks for stopping by, and see you next time for more solutions.
