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Sagot :
Certainly! Let's tackle the problem step-by-step.
Given the matrices:
[tex]\[ A = \begin{pmatrix} 1 & 4 \\ 2 & 3 \end{pmatrix} \][/tex]
[tex]\[ B = \begin{pmatrix} -4 & -1 \\ -3 & -2 \end{pmatrix} \][/tex]
### Part (a) Find the matrix [tex]\(2A + B\)[/tex]
First, we compute [tex]\( 2A \)[/tex]:
[tex]\[ 2A = 2 \begin{pmatrix} 1 & 4 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} 2 \cdot 1 & 2 \cdot 4 \\ 2 \cdot 2 & 2 \cdot 3 \end{pmatrix} = \begin{pmatrix} 2 & 8 \\ 4 & 6 \end{pmatrix} \][/tex]
Next, add matrix [tex]\( B \)[/tex] to [tex]\( 2A \)[/tex]:
[tex]\[ 2A + B = \begin{pmatrix} 2 & 8 \\ 4 & 6 \end{pmatrix} + \begin{pmatrix} -4 & -1 \\ -3 & -2 \end{pmatrix} = \begin{pmatrix} 2 + (-4) & 8 + (-1) \\ 4 + (-3) & 6 + (-2) \end{pmatrix} \][/tex]
Performing the addition element-wise:
[tex]\[ 2A + B = \begin{pmatrix} -2 & 7 \\ 1 & 4 \end{pmatrix} \][/tex]
### Part (b) Find a matrix [tex]\(C\)[/tex] such that [tex]\(C + B = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\)[/tex]
To solve for [tex]\(C\)[/tex], we start from the equation:
[tex]\[ C + B = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \][/tex]
Subtract [tex]\( B \)[/tex] from both sides to isolate [tex]\( C \)[/tex]:
[tex]\[ C = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} - B \][/tex]
Substitute [tex]\( B \)[/tex] into the equation:
[tex]\[ C = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} - \begin{pmatrix} -4 & -1 \\ -3 & -2 \end{pmatrix} \][/tex]
Subtract [tex]\( B \)[/tex] element-wise:
[tex]\[ C = \begin{pmatrix} 0 - (-4) & 0 - (-1) \\ 0 - (-3) & 0 - (-2) \end{pmatrix} = \begin{pmatrix} 4 & 1 \\ 3 & 2 \end{pmatrix} \][/tex]
Therefore, the matrix [tex]\( C \)[/tex] that satisfies [tex]\(C + B = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\)[/tex] is:
[tex]\[ C = \begin{pmatrix} 4 & 1 \\ 3 & 2 \end{pmatrix} \][/tex]
### Final Summary
(a) The matrix [tex]\(2A + B\)[/tex] is:
[tex]\[ \begin{pmatrix} -2 & 7 \\ 1 & 4 \end{pmatrix} \][/tex]
(b) The matrix [tex]\(C\)[/tex] such that [tex]\(C + B = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\)[/tex] is:
[tex]\[ \begin{pmatrix} 4 & 1 \\ 3 & 2 \end{pmatrix} \][/tex]
Given the matrices:
[tex]\[ A = \begin{pmatrix} 1 & 4 \\ 2 & 3 \end{pmatrix} \][/tex]
[tex]\[ B = \begin{pmatrix} -4 & -1 \\ -3 & -2 \end{pmatrix} \][/tex]
### Part (a) Find the matrix [tex]\(2A + B\)[/tex]
First, we compute [tex]\( 2A \)[/tex]:
[tex]\[ 2A = 2 \begin{pmatrix} 1 & 4 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} 2 \cdot 1 & 2 \cdot 4 \\ 2 \cdot 2 & 2 \cdot 3 \end{pmatrix} = \begin{pmatrix} 2 & 8 \\ 4 & 6 \end{pmatrix} \][/tex]
Next, add matrix [tex]\( B \)[/tex] to [tex]\( 2A \)[/tex]:
[tex]\[ 2A + B = \begin{pmatrix} 2 & 8 \\ 4 & 6 \end{pmatrix} + \begin{pmatrix} -4 & -1 \\ -3 & -2 \end{pmatrix} = \begin{pmatrix} 2 + (-4) & 8 + (-1) \\ 4 + (-3) & 6 + (-2) \end{pmatrix} \][/tex]
Performing the addition element-wise:
[tex]\[ 2A + B = \begin{pmatrix} -2 & 7 \\ 1 & 4 \end{pmatrix} \][/tex]
### Part (b) Find a matrix [tex]\(C\)[/tex] such that [tex]\(C + B = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\)[/tex]
To solve for [tex]\(C\)[/tex], we start from the equation:
[tex]\[ C + B = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \][/tex]
Subtract [tex]\( B \)[/tex] from both sides to isolate [tex]\( C \)[/tex]:
[tex]\[ C = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} - B \][/tex]
Substitute [tex]\( B \)[/tex] into the equation:
[tex]\[ C = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} - \begin{pmatrix} -4 & -1 \\ -3 & -2 \end{pmatrix} \][/tex]
Subtract [tex]\( B \)[/tex] element-wise:
[tex]\[ C = \begin{pmatrix} 0 - (-4) & 0 - (-1) \\ 0 - (-3) & 0 - (-2) \end{pmatrix} = \begin{pmatrix} 4 & 1 \\ 3 & 2 \end{pmatrix} \][/tex]
Therefore, the matrix [tex]\( C \)[/tex] that satisfies [tex]\(C + B = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\)[/tex] is:
[tex]\[ C = \begin{pmatrix} 4 & 1 \\ 3 & 2 \end{pmatrix} \][/tex]
### Final Summary
(a) The matrix [tex]\(2A + B\)[/tex] is:
[tex]\[ \begin{pmatrix} -2 & 7 \\ 1 & 4 \end{pmatrix} \][/tex]
(b) The matrix [tex]\(C\)[/tex] such that [tex]\(C + B = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\)[/tex] is:
[tex]\[ \begin{pmatrix} 4 & 1 \\ 3 & 2 \end{pmatrix} \][/tex]
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