2.1 Exercises

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.

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  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.

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  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}.

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

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  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).

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  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.

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  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.

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  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).

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  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.

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