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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].
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