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To find the matrix [tex]\( x \)[/tex] such that [tex]\( 2x \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 8 \\ 7 & 2 \end{bmatrix} \)[/tex], let's proceed step by step:
1. Understand the given equation:
[tex]\[ 2x \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 8 \\ 7 & 2 \end{bmatrix} \][/tex]
2. Isolate the matrix [tex]\(x\)[/tex]:
Divide both sides of the equation by 2:
[tex]\[ x \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 3 & 8 \\ 7 & 2 \end{bmatrix} = \begin{bmatrix} \frac{3}{2} & 4 \\ \frac{7}{2} & 1 \end{bmatrix} \][/tex]
3. Let's denote the [tex]\( 2 \times 2 \)[/tex] matrix [tex]\( x \)[/tex] as:
[tex]\[ x = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \][/tex]
4. Set up the matrix equation:
[tex]\[ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} \frac{3}{2} & 4 \\ \frac{7}{2} & 1 \end{bmatrix} \][/tex]
5. Multiply the matrices on the left-hand side:
[tex]\[ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} a \cdot 1 + b \cdot 3 & a \cdot 2 + b \cdot 4 \\ c \cdot 1 + d \cdot 3 & c \cdot 2 + d \cdot 4 \end{bmatrix} = \begin{bmatrix} a + 3b & 2a + 4b \\ c + 3d & 2c + 4d \end{bmatrix} \][/tex]
6. Equate this to the right-hand side:
[tex]\[ \begin{bmatrix} a + 3b & 2a + 4b \\ c + 3d & 2c + 4d \end{bmatrix} = \begin{bmatrix} \frac{3}{2} & 4 \\ \frac{7}{2} & 1 \end{bmatrix} \][/tex]
7. Set up the system of linear equations:
[tex]\[ \begin{cases} a + 3b = \frac{3}{2} \\ 2a + 4b = 4 \\ c + 3d = \frac{7}{2} \\ 2c + 4d = 1 \end{cases} \][/tex]
8. Solve the system of equations:
From the second equation:
[tex]\[ 2a + 4b = 4 \quad \Rightarrow \quad a + 2b = 2 \quad \text{(divide by 2)} \][/tex]
Now we have the system:
[tex]\[ \begin{cases} a + 3b = \frac{3}{2} \\ a + 2b = 2 \end{cases} \][/tex]
Subtract the second equation from the first:
[tex]\[ (a + 3b) - (a + 2b) = \frac{3}{2} - 2 \\ b = -\frac{1}{2} \][/tex]
Plug [tex]\( b = -\frac{1}{2} \)[/tex] back into [tex]\( a + 2b = 2 \)[/tex]:
[tex]\[ a + 2(-\frac{1}{2}) = 2 \\ a - 1 = 2 \\ a = 3 \][/tex]
Similarly, for the other two equations:
[tex]\[ \begin{cases} c + 3d = \frac{7}{2} \\ 2c + 4d = 1 \quad \Rightarrow \quad c + 2d = \frac{1}{2} \quad \text{(divide by 2)} \end{cases} \][/tex]
Subtract the second equation from the first:
[tex]\[ (c + 3d) - (c + 2d) = \frac{7}{2} - \frac{1}{2} \\ d = 3 \][/tex]
Plug [tex]\( d = 3 \)[/tex] back into [tex]\( c + 2d = \frac{1}{2} \)[/tex]:
[tex]\[ c + 2 \cdot 3 = \frac{1}{2} \\ c + 6 = \frac{1}{2} \\ c = \frac{1}{2} - 6 \\ c = -\frac{11}{2} \][/tex]
Given these calculated values, there is no set of coefficients (a, b, c, d) present, implying contradictions while calculating the original equation. Hence, there is actually no solution that would fit in all the values, thus:
[tex]\[ x \quad will \quad have \quad no \quad valid \quad solution. \][/tex]
1. Understand the given equation:
[tex]\[ 2x \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 8 \\ 7 & 2 \end{bmatrix} \][/tex]
2. Isolate the matrix [tex]\(x\)[/tex]:
Divide both sides of the equation by 2:
[tex]\[ x \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 3 & 8 \\ 7 & 2 \end{bmatrix} = \begin{bmatrix} \frac{3}{2} & 4 \\ \frac{7}{2} & 1 \end{bmatrix} \][/tex]
3. Let's denote the [tex]\( 2 \times 2 \)[/tex] matrix [tex]\( x \)[/tex] as:
[tex]\[ x = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \][/tex]
4. Set up the matrix equation:
[tex]\[ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} \frac{3}{2} & 4 \\ \frac{7}{2} & 1 \end{bmatrix} \][/tex]
5. Multiply the matrices on the left-hand side:
[tex]\[ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} a \cdot 1 + b \cdot 3 & a \cdot 2 + b \cdot 4 \\ c \cdot 1 + d \cdot 3 & c \cdot 2 + d \cdot 4 \end{bmatrix} = \begin{bmatrix} a + 3b & 2a + 4b \\ c + 3d & 2c + 4d \end{bmatrix} \][/tex]
6. Equate this to the right-hand side:
[tex]\[ \begin{bmatrix} a + 3b & 2a + 4b \\ c + 3d & 2c + 4d \end{bmatrix} = \begin{bmatrix} \frac{3}{2} & 4 \\ \frac{7}{2} & 1 \end{bmatrix} \][/tex]
7. Set up the system of linear equations:
[tex]\[ \begin{cases} a + 3b = \frac{3}{2} \\ 2a + 4b = 4 \\ c + 3d = \frac{7}{2} \\ 2c + 4d = 1 \end{cases} \][/tex]
8. Solve the system of equations:
From the second equation:
[tex]\[ 2a + 4b = 4 \quad \Rightarrow \quad a + 2b = 2 \quad \text{(divide by 2)} \][/tex]
Now we have the system:
[tex]\[ \begin{cases} a + 3b = \frac{3}{2} \\ a + 2b = 2 \end{cases} \][/tex]
Subtract the second equation from the first:
[tex]\[ (a + 3b) - (a + 2b) = \frac{3}{2} - 2 \\ b = -\frac{1}{2} \][/tex]
Plug [tex]\( b = -\frac{1}{2} \)[/tex] back into [tex]\( a + 2b = 2 \)[/tex]:
[tex]\[ a + 2(-\frac{1}{2}) = 2 \\ a - 1 = 2 \\ a = 3 \][/tex]
Similarly, for the other two equations:
[tex]\[ \begin{cases} c + 3d = \frac{7}{2} \\ 2c + 4d = 1 \quad \Rightarrow \quad c + 2d = \frac{1}{2} \quad \text{(divide by 2)} \end{cases} \][/tex]
Subtract the second equation from the first:
[tex]\[ (c + 3d) - (c + 2d) = \frac{7}{2} - \frac{1}{2} \\ d = 3 \][/tex]
Plug [tex]\( d = 3 \)[/tex] back into [tex]\( c + 2d = \frac{1}{2} \)[/tex]:
[tex]\[ c + 2 \cdot 3 = \frac{1}{2} \\ c + 6 = \frac{1}{2} \\ c = \frac{1}{2} - 6 \\ c = -\frac{11}{2} \][/tex]
Given these calculated values, there is no set of coefficients (a, b, c, d) present, implying contradictions while calculating the original equation. Hence, there is actually no solution that would fit in all the values, thus:
[tex]\[ x \quad will \quad have \quad no \quad valid \quad solution. \][/tex]
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