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The Timpson Lyceum · Linear Algebra (2026)

2.1 Exercises

Problem sheet · 20 points · 4 extra credit

Matrix Operations

  1. 1. (2 pts) Let

    A=[1−2340−1],B=[210−352].A = \begin{bmatrix} 1 & -2 & 3 \\ 4 & 0 & -1 \end{bmatrix}, \qquad B = \begin{bmatrix} 2 & 1 & 0 \\ -3 & 5 & 2 \end{bmatrix}.

    Compute 2A−B2A - B.

  2. 2. (3 pts) Let A=[2−103]A = \begin{bmatrix} 2 & -1 \\ 0 & 3 \end{bmatrix} and B=[14−25]B = \begin{bmatrix} 1 & 4 \\ -2 & 5 \end{bmatrix}. Compute both ABAB and BABA, and state what the results show about matrix multiplication.

  3. 3. (3 pts) Compute the product ABAB, where

    A=[102−131],B=[210−142].A = \begin{bmatrix} 1 & 0 & 2 \\ -1 & 3 & 1 \end{bmatrix}, \qquad B = \begin{bmatrix} 2 & 1 \\ 0 & -1 \\ 4 & 2 \end{bmatrix}.

  4. 4. (2 pts) Write down the transpose ATA^{\mathsf T} of the matrix AA from Problem 1.

  5. 5. (3 pts) Let A=[1101]A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}. Compute A2A^2 and A3A^3, then conjecture a formula for AnA^n (n≥1n \ge 1).

  6. 6. (3 pts) State whether each of the following is true or false, and justify each answer.

    (a) For all n×nn\times n matrices AA and BB, (A+B)2=A2+2AB+B2(A+B)^2 = A^2 + 2AB + B^2.

    (b) If AA and BB are matrices with AB=0AB = 0, then A=0A = 0 or B=0B = 0.

    (c) The (i,j)(i,j) entry of ABAB equals the sum of products of the entries of row ii of AA with the corresponding entries of column jj of BB.

  7. 7. (2 pts) Let u=[123]\mathbf{u} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} and v=[2−1]\mathbf{v} = \begin{bmatrix} 2 \\ -1 \end{bmatrix}. Compute the outer product uvT\mathbf{u}\mathbf{v}^{\mathsf T} and state its size.

  8. 8. (2 pts) Let AA be any square matrix. Show that A+ATA + A^{\mathsf T} is symmetric and that A−ATA - A^{\mathsf T} is skew-symmetric (i.e., equal to the negative of its own transpose).

  9. 9. (4 pts ••) The trace of an n×nn\times n matrix AA, written tr⁡(A)\operatorname{tr}(A), is the sum of its diagonal entries a11+a22+⋯+anna_{11} + a_{22} + \cdots + a_{nn}. Prove that for any n×nn\times n matrices AA and BB,

    tr⁡(AB)=tr⁡(BA),\operatorname{tr}(AB) = \operatorname{tr}(BA),

    even though AB≠BAAB \ne BA in general.