Math Problem Statement
matrix M= 1 k 0 k 1 0 0 0 1
find inverses and check if (M,M^-1,I) where I is unitary 3x3 if thath is linear independent
Solution
To solve this problem, we'll proceed step-by-step:
Step 1: Matrix
The given matrix is:
\begin{bmatrix} 1 & k & 0 \\ k & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ --- ### Step 2: Find the inverse $$M^{-1}$$ For a $$3 \times 3$$ matrix, the inverse exists only if the determinant is non-zero. Compute the determinant of $$M$$: $$\det(M) = 1 \cdot (1 \cdot 1 - 0 \cdot 0) - k \cdot (k \cdot 1 - 0 \cdot 0) + 0 \cdot (k \cdot 0 - 1 \cdot 0)$$ $$\det(M) = 1(1) - k(k) + 0 = 1 - k^2$$ Thus, the determinant is: $$\det(M) = 1 - k^2$$ The inverse exists if $$1 - k^2 \neq 0$$, which implies $$k \neq \pm 1$$. The inverse of $$M$$, using the formula $$M^{-1} = \frac{1}{\det(M)} \text{Adj}(M)$$, is computed as follows: 1. Find the adjugate matrix (cofactor transpose). 2. Divide by $$\det(M) = 1 - k^2$$. Adjugate of $$M$$: $$\text{Adj}(M) = \begin{bmatrix} 1 & -k & 0 \\ -k & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ Inverse $$M^{-1}$$: $$M^{-1} = \frac{1}{1 - k^2} \begin{bmatrix} 1 & -k & 0 \\ -k & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ --- ### Step 3: Check linear independence of $$(M, M^{-1}, I)$$ To check if $$M$$, $$M^{-1}$$, and $$I$$ are linearly independent, suppose there exist scalars $$a, b, c$$ such that: $$aM + bM^{-1} + cI = 0$$ Expanding this: $$a \begin{bmatrix} 1 & k & 0 \\ k & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} + \frac{b}{1 - k^2} \begin{bmatrix} 1 & -k & 0 \\ -k & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} + c \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ Combine terms element-wise: $$\begin{bmatrix} a + \frac{b}{1 - k^2} + c & ak - \frac{bk}{1 - k^2} & 0 \\ ak - \frac{bk}{1 - k^2} & a + \frac{b}{1 - k^2} + c & 0 \\ 0 & 0 & a + \frac{b}{1 - k^2} + c \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ From this equation, we extract: 1. $$a + \frac{b}{1 - k^2} + c = 0$$ 2. $$ak - \frac{bk}{1 - k^2} = 0$$ #### Solve the system From equation (2): $$ak = \frac{bk}{1 - k^2} \implies b = a(1 - k^2)$$ Substitute $$b = a(1 - k^2)$$ into equation (1): $$a + \frac{a(1 - k^2)}{1 - k^2} + c = 0$$ $$a + a + c = 0 \implies 2a + c = 0 \implies c = -2a$$ Thus, the scalars satisfy: $$b = a(1 - k^2), \quad c = -2a$$ Since $$a$$ can take any value, $$M, M^{-1}, I$$ are **linearly dependent**. --- ### Final Answer: - The inverse $$M^{-1}$$ exists if $$k \neq \pm 1$$. - $$(M, M^{-1}, I)$$ are linearly **dependent**. --- Do you have any questions or need detailed derivations? **5 Questions for Further Exploration:** 1. How is the adjugate matrix calculated step-by-step? 2. What are the geometric implications of $$\det(M) \neq 0$$? 3. Can we generalize this result to larger matrices? 4. What happens to $$M^{-1}$$ when $$k = \pm 1$$? 5. How do we test for linear independence in other contexts? **Tip:** Always verify the determinant condition before attempting to find aAsk a new question for Free
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Math Problem Analysis
Mathematical Concepts
Matrix Algebra
Determinants
Matrix Inversion
Linear Independence
Formulas
Determinant formula: det(M) = 1 - k^2
Inverse matrix formula: M^-1 = (1/det(M)) * adj(M)
Linear independence condition: aM + bM^-1 + cI = 0
Theorems
Linear Independence Theorem
Matrix Inverse Existence Theorem
Properties of Determinants
Suitable Grade Level
Undergraduate Mathematics