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Show that (A+B)(A+B) != A2 + 2AB + B2 for the given matrices A and B.

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Answer

b

Shown in steps above.\text{Shown in steps above.}

Step-by-step solution

14 steps

Here are the solutions to the matrix problems.

Given matrices:

A=[−1201]A = \begin{bmatrix} -1 & 2 \\ 0 & 1 \end{bmatrix} B=[10−12]B = \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix}

Part a· 8 steps

Show that (A+B)(A+B)≠A2+2AB+B2(A+B)(A+B) \neq A^2 + 2AB + B^2

  1. Calculate A+BA+B.

    A+B=[−1201]+[10−12]=[−1+12+00−11+2]=[02−13]A+B = \begin{bmatrix} -1 & 2 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} -1+1 & 2+0 \\ 0-1 & 1+2 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -1 & 3 \end{bmatrix}

  2. Calculate (A+B)(A+B)(A+B)(A+B).

    (A+B)(A+B)=[02−13][02−13](A+B)(A+B) = \begin{bmatrix} 0 & 2 \\ -1 & 3 \end{bmatrix} \begin{bmatrix} 0 & 2 \\ -1 & 3 \end{bmatrix} =[(0)(0)+(2)(−1)(0)(2)+(2)(3)(−1)(0)+(3)(−1)(−1)(2)+(3)(3)]= \begin{bmatrix} (0)(0)+(2)(-1) & (0)(2)+(2)(3) \\ (-1)(0)+(3)(-1) & (-1)(2)+(3)(3) \end{bmatrix} =[−26−37]= \begin{bmatrix} -2 & 6 \\ -3 & 7 \end{bmatrix}

  3. Calculate A2A^2.

    A2=A⋅A=[−1201][−1201]A^2 = A \cdot A = \begin{bmatrix} -1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 2 \\ 0 & 1 \end{bmatrix} =[(−1)(−1)+(2)(0)(−1)(2)+(2)(1)(0)(−1)+(1)(0)(0)(2)+(1)(1)]= \begin{bmatrix} (-1)(-1)+(2)(0) & (-1)(2)+(2)(1) \\ (0)(-1)+(1)(0) & (0)(2)+(1)(1) \end{bmatrix} =[1001]= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

  4. Calculate B2B^2.

    B2=B⋅B=[10−12][10−12]B^2 = B \cdot B = \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix} =[(1)(1)+(0)(−1)(1)(0)+(0)(2)(−1)(1)+(2)(−1)(−1)(0)+(2)(2)]= \begin{bmatrix} (1)(1)+(0)(-1) & (1)(0)+(0)(2) \\ (-1)(1)+(2)(-1) & (-1)(0)+(2)(2) \end{bmatrix} =[10−34]= \begin{bmatrix} 1 & 0 \\ -3 & 4 \end{bmatrix}

  5. Calculate ABAB.

    AB=[−1201][10−12]AB = \begin{bmatrix} -1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix} =[(−1)(1)+(2)(−1)(−1)(0)+(2)(2)(0)(1)+(1)(−1)(0)(0)+(1)(2)]= \begin{bmatrix} (-1)(1)+(2)(-1) & (-1)(0)+(2)(2) \\ (0)(1)+(1)(-1) & (0)(0)+(1)(2) \end{bmatrix} =[−34−12]= \begin{bmatrix} -3 & 4 \\ -1 & 2 \end{bmatrix}

  6. Calculate $2AB$.

    2AB=2[−34−12]=[−68−24]2AB = 2 \begin{bmatrix} -3 & 4 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} -6 & 8 \\ -2 & 4 \end{bmatrix}

  7. Calculate A2+2AB+B2A^2 + 2AB + B^2.

    A2+2AB+B2=[1001]+[−68−24]+[10−34]A^2 + 2AB + B^2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} -6 & 8 \\ -2 & 4 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ -3 & 4 \end{bmatrix} =[1−6+10+8+00−2−31+4+4]= \begin{bmatrix} 1-6+1 & 0+8+0 \\ 0-2-3 & 1+4+4 \end{bmatrix} =[−48−59]= \begin{bmatrix} -4 & 8 \\ -5 & 9 \end{bmatrix}

  8. Compare (A+B)(A+B)(A+B)(A+B) and A2+2AB+B2A^2 + 2AB + B^2.

    Since [−26−37]≠[−48−59]\begin{bmatrix} -2 & 6 \\ -3 & 7 \end{bmatrix} \neq \begin{bmatrix} -4 & 8 \\ -5 & 9 \end{bmatrix}, it is shown that (A+B)(A+B)≠A2+2AB+B2(A+B)(A+B) \neq A^2 + 2AB + B^2.

Part b

Shown in steps above.\text{Shown in steps above.}

Show that (A+B)(A−B)≠A2−B2(A+B)(A-B) \neq A^2 - B^2

  1. We already have A+B=[02−13]A+B = \begin{bmatrix} 0 & 2 \\ -1 & 3 \end{bmatrix} from part (a).

  2. Calculate A−BA-B.

    A−B=[−1201]−[10−12]=[−1−12−00−(−1)1−2]=[−221−1]A-B = \begin{bmatrix} -1 & 2 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} -1-1 & 2-0 \\ 0-(-1) & 1-2 \end{bmatrix} = \begin{bmatrix} -2 & 2 \\ 1 & -1 \end{bmatrix}

  3. Calculate (A+B)(A−B)(A+B)(A-B).

    (A+B)(A−B)=[02−13][−221−1](A+B)(A-B) = \begin{bmatrix} 0 & 2 \\ -1 & 3 \end{bmatrix} \begin{bmatrix} -2 & 2 \\ 1 & -1 \end{bmatrix} =[(0)(−2)+(2)(1)(0)(2)+(2)(−1)(−1)(−2)+(3)(1)(−1)(2)+(3)(−1)]= \begin{bmatrix} (0)(-2)+(2)(1) & (0)(2)+(2)(-1) \\ (-1)(-2)+(3)(1) & (-1)(2)+(3)(-1) \end{bmatrix} =[2−25−5]= \begin{bmatrix} 2 & -2 \\ 5 & -5 \end{bmatrix}

  4. We already have A2=[1001]A^2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} and B2=[10−34]B^2 = \begin{bmatrix} 1 & 0 \\ -3 & 4 \end{bmatrix} from part (a).

  5. Calculate A2−B2A^2 - B^2.

    A2−B2=[1001]−[10−34]A^2 - B^2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ -3 & 4 \end{bmatrix} =[1−10−00−(−3)1−4]= \begin{bmatrix} 1-1 & 0-0 \\ 0-(-3) & 1-4 \end{bmatrix} =[003−3]= \begin{bmatrix} 0 & 0 \\ 3 & -3 \end{bmatrix}

  6. Compare (A+B)(A−B)(A+B)(A-B) and A2−B2A^2 - B^2.

    Since [2−25−5]≠[003−3]\begin{bmatrix} 2 & -2 \\ 5 & -5 \end{bmatrix} \neq \begin{bmatrix} 0 & 0 \\ 3 & -3 \end{bmatrix}, it is shown that (A+B)(A−B)≠A2−B2(A+B)(A-B) \neq A^2 - B^2.

    The final answer is Shown in steps above.\boxed{\text{Shown in steps above.}}

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