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The Timpson Lyceum · Linear Algebra (2026)

2.3 Exercises

Problem sheet · 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.

  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.

  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.

  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.

  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.

  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.

  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.)

  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}.