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Sagot :
To find matrix [tex]\(Y\)[/tex] given the equation [tex]\(X - 2Y = Z\)[/tex], we follow these steps:
1. Write down the matrices [tex]\(X\)[/tex], [tex]\(Y\)[/tex], and [tex]\(Z\)[/tex]:
[tex]\[ X = \begin{pmatrix} b & a \\ 4 & a \end{pmatrix}, Y = \begin{pmatrix} c & d \\ a & b \end{pmatrix}, Z = \begin{pmatrix} a & c \\ 16 & b \end{pmatrix} \][/tex]
2. Substitute these matrices into the equation [tex]\(X - 2Y = Z\)[/tex]:
[tex]\[ \begin{pmatrix} b & a \\ 4 & a \end{pmatrix} - 2 \begin{pmatrix} c & d \\ a & b \end{pmatrix} = \begin{pmatrix} a & c \\ 16 & b \end{pmatrix} \][/tex]
3. Multiply matrix [tex]\(Y\)[/tex] by 2:
[tex]\[ 2Y = 2 \begin{pmatrix} c & d \\ a & b \end{pmatrix} = \begin{pmatrix} 2c & 2d \\ 2a & 2b \end{pmatrix} \][/tex]
4. Subtract [tex]\(2Y\)[/tex] from [tex]\(X\)[/tex]:
[tex]\[ \begin{pmatrix} b & a \\ 4 & a \end{pmatrix} - \begin{pmatrix} 2c & 2d \\ 2a & 2b \end{pmatrix} = \begin{pmatrix} a & c \\ 16 & b \end{pmatrix} \][/tex]
5. Calculate the elements of the resulting matrix:
- For the first element [tex]\((1,1)\)[/tex], we have [tex]\(b - 2c = a\)[/tex]. Thus, [tex]\(2c = b - a\)[/tex].
- For the second element [tex]\((1,2)\)[/tex], we have [tex]\(a - 2d = c\)[/tex]. Thus, [tex]\(2d = a - c\)[/tex].
- For the third element [tex]\((2,1)\)[/tex], we have [tex]\(4 - 2a = 16\)[/tex]. Thus, [tex]\(2a = 4 - 16 = -12\)[/tex] and [tex]\(a = -6\)[/tex].
- For the fourth element [tex]\((2,2)\)[/tex], we have [tex]\(a - 2b = b\)[/tex]. Thus, [tex]\(2b = a - b\)[/tex].
Since we need the elements of matrix [tex]\(Y\)[/tex]:
- [tex]\(c = \frac{b - a}{2}\)[/tex]
- [tex]\(d = \frac{a - c}{2}\)[/tex]
- [tex]\(a = -6\)[/tex]
- [tex]\(b = x\)[/tex] (undetermined in the provided information)
After simplifying:
- Let’s use specific values to solve further since the equation can be ambiguous with undetermined variables. From further simplification, there can be specific solutions for distinct [tex]\(a\)[/tex] and [tex]\(b\)[/tex].
Given correct calculations:
[tex]\[ Y = \begin{pmatrix} \text{(relate c to b and a specific value )} & \text{new d (solved constant) } \\ -6 & \text{ solved b content }\end{pmatrix} \][/tex]
1. Write down the matrices [tex]\(X\)[/tex], [tex]\(Y\)[/tex], and [tex]\(Z\)[/tex]:
[tex]\[ X = \begin{pmatrix} b & a \\ 4 & a \end{pmatrix}, Y = \begin{pmatrix} c & d \\ a & b \end{pmatrix}, Z = \begin{pmatrix} a & c \\ 16 & b \end{pmatrix} \][/tex]
2. Substitute these matrices into the equation [tex]\(X - 2Y = Z\)[/tex]:
[tex]\[ \begin{pmatrix} b & a \\ 4 & a \end{pmatrix} - 2 \begin{pmatrix} c & d \\ a & b \end{pmatrix} = \begin{pmatrix} a & c \\ 16 & b \end{pmatrix} \][/tex]
3. Multiply matrix [tex]\(Y\)[/tex] by 2:
[tex]\[ 2Y = 2 \begin{pmatrix} c & d \\ a & b \end{pmatrix} = \begin{pmatrix} 2c & 2d \\ 2a & 2b \end{pmatrix} \][/tex]
4. Subtract [tex]\(2Y\)[/tex] from [tex]\(X\)[/tex]:
[tex]\[ \begin{pmatrix} b & a \\ 4 & a \end{pmatrix} - \begin{pmatrix} 2c & 2d \\ 2a & 2b \end{pmatrix} = \begin{pmatrix} a & c \\ 16 & b \end{pmatrix} \][/tex]
5. Calculate the elements of the resulting matrix:
- For the first element [tex]\((1,1)\)[/tex], we have [tex]\(b - 2c = a\)[/tex]. Thus, [tex]\(2c = b - a\)[/tex].
- For the second element [tex]\((1,2)\)[/tex], we have [tex]\(a - 2d = c\)[/tex]. Thus, [tex]\(2d = a - c\)[/tex].
- For the third element [tex]\((2,1)\)[/tex], we have [tex]\(4 - 2a = 16\)[/tex]. Thus, [tex]\(2a = 4 - 16 = -12\)[/tex] and [tex]\(a = -6\)[/tex].
- For the fourth element [tex]\((2,2)\)[/tex], we have [tex]\(a - 2b = b\)[/tex]. Thus, [tex]\(2b = a - b\)[/tex].
Since we need the elements of matrix [tex]\(Y\)[/tex]:
- [tex]\(c = \frac{b - a}{2}\)[/tex]
- [tex]\(d = \frac{a - c}{2}\)[/tex]
- [tex]\(a = -6\)[/tex]
- [tex]\(b = x\)[/tex] (undetermined in the provided information)
After simplifying:
- Let’s use specific values to solve further since the equation can be ambiguous with undetermined variables. From further simplification, there can be specific solutions for distinct [tex]\(a\)[/tex] and [tex]\(b\)[/tex].
Given correct calculations:
[tex]\[ Y = \begin{pmatrix} \text{(relate c to b and a specific value )} & \text{new d (solved constant) } \\ -6 & \text{ solved b content }\end{pmatrix} \][/tex]
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