2.3 Exercises

20 points · 4 extra credit

Characterizations of Invertible Matrices

  1. 1. (3 pts) Determine whether

    A=[10−231−2−5−19]A = \begin{bmatrix} 1 & 0 & -2 \\ 3 & 1 & -2 \\ -5 & -1 & 9 \end{bmatrix}

    is invertible by row reducing it. Count the pivot positions and state which statement of the Invertible Matrix Theorem you used to reach your conclusion.

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  2. 2. (3 pts) Find all values of hh for which

    A=[1−212−3h111]A = \begin{bmatrix} 1 & -2 & 1 \\ 2 & -3 & h \\ 1 & 1 & 1 \end{bmatrix}

    is not invertible.

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  3. 3. (3 pts) Let AA be an n×nn\times n matrix, and suppose the equation Ax=0A\mathbf{x} = 0 has only the trivial solution. Using the Invertible Matrix Theorem, explain why Ax=bA\mathbf{x} = \mathbf{b} is consistent for every b∈Rn\mathbf{b} \in \mathbb{R}^n and has exactly one solution.

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  4. 4. (3 pts) State whether each of the following is true or false, and justify each using the Invertible Matrix Theorem. Assume AA is n×nn\times n.

    (a) If AA has nn pivot positions, then AA is invertible.

    (b) If the columns of AA span Rn\mathbb{R}^n, then AA might still fail to be invertible.

    (c) If AA is invertible, then its reduced echelon form is InI_n.

    (d) If there is an n×nn\times n matrix CC with CA=ICA = I, then AA is invertible.

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  5. 5. (2 pts) Suppose AA is a 5×55\times5 matrix for which the equation Ax=0A\mathbf{x} = 0 has a nontrivial solution. List three statements from the Invertible Matrix Theorem that must therefore be false for AA.

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  6. 6. (2 pts) Explain why a square matrix with two identical rows cannot be invertible. Reference an appropriate statement of the Invertible Matrix Theorem.

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  7. 7. (4 pts) Let AA and BB be n×nn\times n matrices with AB=IAB = I. Prove that BB is invertible with B−1=AB^{-1} = A, and hence that AA is invertible as well. (Do not assume BA=IBA = I; deduce it.)

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  8. 8. (4 pts ••) A square matrix AA is called nilpotent if Ak=0A^k = 0 for some integer k≥1k \ge 1. Prove that if AA is nilpotent, then I−AI - A is invertible, and

    (I−A)−1=I+A+A2+⋯+Ak−1.(I - A)^{-1} = I + A + A^2 + \cdots + A^{k-1}.

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