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To find the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] that will make the equation [tex]\(CD = I\)[/tex] true, we will need to solve the equation step by step.
Given matrices [tex]\(C\)[/tex] and [tex]\(D\)[/tex]:
[tex]\[ C = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 3 & 4 \\ 0 & 2 & 1 \end{bmatrix} \][/tex]
[tex]\[ D = \begin{bmatrix} a & b & -0.4 \\ 0 & -0.2 & 0.8 \\ 0 & 0.4 & -0.6 \end{bmatrix} \][/tex]
The identity matrix [tex]\(I\)[/tex] is:
[tex]\[ I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \][/tex]
We need to find [tex]\(a\)[/tex] and [tex]\(b\)[/tex] such that [tex]\(CD = I\)[/tex].
1. Start by calculating the product [tex]\(CD\)[/tex]:
[tex]\[ CD = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 3 & 4 \\ 0 & 2 & 1 \end{bmatrix} \begin{bmatrix} a & b & -0.4 \\ 0 & -0.2 & 0.8 \\ 0 & 0.4 & -0.6 \end{bmatrix} \][/tex]
Now, performing the matrix multiplication:
[tex]\[ CD_{11} = 2a + 1 \cdot 0 + 0 \cdot 0 = 2a \][/tex]
[tex]\[ CD_{12} = 2b + 1 \cdot (-0.2) + 0 \cdot 0.4 = 2b - 0.2 \][/tex]
[tex]\[ CD_{13} = 2 \cdot (-0.4) + 1 \cdot 0.8 + 0 \cdot (-0.6) = -0.8 + 0.8 = 0 \][/tex]
[tex]\[ CD_{21} = 0 \cdot a + 3 \cdot 0 + 4 \cdot 0 = 0 \][/tex]
[tex]\[ CD_{22} = 0 \cdot b + 3 \cdot (-0.2) + 4 \cdot 0.4 = -0.6 + 1.6 = 1 \][/tex]
[tex]\[ CD_{23} = 0 \cdot (-0.4) + 3 \cdot 0.8 + 4 \cdot (-0.6) = 2.4 - 2.4 = 0 \][/tex]
[tex]\[ CD_{31} = 0 \cdot a + 2 \cdot 0 + 1 \cdot 0 = 0 \][/tex]
[tex]\[ CD_{32} = 0 \cdot b + 2 \cdot (-0.2) + 1 \cdot 0.4 = -0.4 + 0.4 = 0 \][/tex]
[tex]\[ CD_{33} = 0 \cdot (-0.4) + 2 \cdot 0.8 + 1 \cdot (-0.6) = 1.6 - 0.6 = 1 \][/tex]
So we have:
[tex]\[ CD = \begin{bmatrix} 2a & 2b - 0.2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \][/tex]
For [tex]\(CD\)[/tex] to equal the identity matrix [tex]\(I\)[/tex], the elements in [tex]\(CD\)[/tex] should match the corresponding elements in [tex]\(I\)[/tex]:
1. From the [tex]\( (1, 1) \)[/tex] position:
[tex]\[ 2a = 1 \][/tex]
Thus,
[tex]\[ a = \frac{1}{2} = 0.5 \][/tex]
2. From the [tex]\( (1, 2) \)[/tex] position:
[tex]\[ 2b - 0.2 = 0 \][/tex]
Thus,
[tex]\[ 2b = 0.2 \][/tex]
[tex]\[ b = \frac{0.2}{2} = 0.1 \][/tex]
So the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] that will make the equation [tex]\(CD = I\)[/tex] true are:
[tex]\[ a = 0.5 \][/tex]
[tex]\[ b = 0.1 \][/tex]
Given matrices [tex]\(C\)[/tex] and [tex]\(D\)[/tex]:
[tex]\[ C = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 3 & 4 \\ 0 & 2 & 1 \end{bmatrix} \][/tex]
[tex]\[ D = \begin{bmatrix} a & b & -0.4 \\ 0 & -0.2 & 0.8 \\ 0 & 0.4 & -0.6 \end{bmatrix} \][/tex]
The identity matrix [tex]\(I\)[/tex] is:
[tex]\[ I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \][/tex]
We need to find [tex]\(a\)[/tex] and [tex]\(b\)[/tex] such that [tex]\(CD = I\)[/tex].
1. Start by calculating the product [tex]\(CD\)[/tex]:
[tex]\[ CD = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 3 & 4 \\ 0 & 2 & 1 \end{bmatrix} \begin{bmatrix} a & b & -0.4 \\ 0 & -0.2 & 0.8 \\ 0 & 0.4 & -0.6 \end{bmatrix} \][/tex]
Now, performing the matrix multiplication:
[tex]\[ CD_{11} = 2a + 1 \cdot 0 + 0 \cdot 0 = 2a \][/tex]
[tex]\[ CD_{12} = 2b + 1 \cdot (-0.2) + 0 \cdot 0.4 = 2b - 0.2 \][/tex]
[tex]\[ CD_{13} = 2 \cdot (-0.4) + 1 \cdot 0.8 + 0 \cdot (-0.6) = -0.8 + 0.8 = 0 \][/tex]
[tex]\[ CD_{21} = 0 \cdot a + 3 \cdot 0 + 4 \cdot 0 = 0 \][/tex]
[tex]\[ CD_{22} = 0 \cdot b + 3 \cdot (-0.2) + 4 \cdot 0.4 = -0.6 + 1.6 = 1 \][/tex]
[tex]\[ CD_{23} = 0 \cdot (-0.4) + 3 \cdot 0.8 + 4 \cdot (-0.6) = 2.4 - 2.4 = 0 \][/tex]
[tex]\[ CD_{31} = 0 \cdot a + 2 \cdot 0 + 1 \cdot 0 = 0 \][/tex]
[tex]\[ CD_{32} = 0 \cdot b + 2 \cdot (-0.2) + 1 \cdot 0.4 = -0.4 + 0.4 = 0 \][/tex]
[tex]\[ CD_{33} = 0 \cdot (-0.4) + 2 \cdot 0.8 + 1 \cdot (-0.6) = 1.6 - 0.6 = 1 \][/tex]
So we have:
[tex]\[ CD = \begin{bmatrix} 2a & 2b - 0.2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \][/tex]
For [tex]\(CD\)[/tex] to equal the identity matrix [tex]\(I\)[/tex], the elements in [tex]\(CD\)[/tex] should match the corresponding elements in [tex]\(I\)[/tex]:
1. From the [tex]\( (1, 1) \)[/tex] position:
[tex]\[ 2a = 1 \][/tex]
Thus,
[tex]\[ a = \frac{1}{2} = 0.5 \][/tex]
2. From the [tex]\( (1, 2) \)[/tex] position:
[tex]\[ 2b - 0.2 = 0 \][/tex]
Thus,
[tex]\[ 2b = 0.2 \][/tex]
[tex]\[ b = \frac{0.2}{2} = 0.1 \][/tex]
So the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] that will make the equation [tex]\(CD = I\)[/tex] true are:
[tex]\[ a = 0.5 \][/tex]
[tex]\[ b = 0.1 \][/tex]
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