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If A=[2 3 -1/3 0 2], B=[1 1/2] and C=[1 -2] then prove that (AB)C = A(BC).

Mathematics

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Answer

सिद्ध हुआ\text{सिद्ध हुआ}

Step-by-step solution

5 steps
  1. ABAB की गणना करें।

    सबसे पहले, हम आव्यूह AA और BB का गुणनफल ज्ञात करेंगे।

    A=[23−1302],B=[112]A = \begin{bmatrix} 2 & 3 & -1 \\ 3 & 0 & 2 \end{bmatrix}, \quad B = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} AB=[23−1302][112]AB = \begin{bmatrix} 2 & 3 & -1 \\ 3 & 0 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} AB=[(2)(1)+(3)(1)+(−1)(2)(3)(1)+(0)(1)+(2)(2)]AB = \begin{bmatrix} (2)(1) + (3)(1) + (-1)(2) \\ (3)(1) + (0)(1) + (2)(2) \end{bmatrix} AB=[2+3−23+0+4]AB = \begin{bmatrix} 2 + 3 - 2 \\ 3 + 0 + 4 \end{bmatrix} AB=[37]AB = \begin{bmatrix} 3 \\ 7 \end{bmatrix}
  2. $(AB)C$ की गणना करें।

    अब, हम ABAB और आव्यूह CC का गुणनफल ज्ञात करेंगे।

    AB=[37],C=[1−2]AB = \begin{bmatrix} 3 \\ 7 \end{bmatrix}, \quad C = \begin{bmatrix} 1 & -2 \end{bmatrix} (AB)C=[37][1−2](AB)C = \begin{bmatrix} 3 \\ 7 \end{bmatrix} \begin{bmatrix} 1 & -2 \end{bmatrix} (AB)C=[(3)(1)(3)(−2)(7)(1)(7)(−2)](AB)C = \begin{bmatrix} (3)(1) & (3)(-2) \\ (7)(1) & (7)(-2) \end{bmatrix} (AB)C=[3−67−14](AB)C = \begin{bmatrix} 3 & -6 \\ 7 & -14 \end{bmatrix}
  3. BCBC की गणना करें।

    अब, हम आव्यूह BB और CC का गुणनफल ज्ञात करेंगे।

    B=[112],C=[1−2]B = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}, \quad C = \begin{bmatrix} 1 & -2 \end{bmatrix} BC=[112][1−2]BC = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} \begin{bmatrix} 1 & -2 \end{bmatrix} BC=[(1)(1)(1)(−2)(1)(1)(1)(−2)(2)(1)(2)(−2)]BC = \begin{bmatrix} (1)(1) & (1)(-2) \\ (1)(1) & (1)(-2) \\ (2)(1) & (2)(-2) \end{bmatrix} BC=[1−21−22−4]BC = \begin{bmatrix} 1 & -2 \\ 1 & -2 \\ 2 & -4 \end{bmatrix}
  4. A(BC)A(BC) की गणना करें।

    अंत में, हम आव्यूह AA और BCBC का गुणनफल ज्ञात करेंगे।

    A=[23−1302],BC=[1−21−22−4]A = \begin{bmatrix} 2 & 3 & -1 \\ 3 & 0 & 2 \end{bmatrix}, \quad BC = \begin{bmatrix} 1 & -2 \\ 1 & -2 \\ 2 & -4 \end{bmatrix} A(BC)=[23−1302][1−21−22−4]A(BC) = \begin{bmatrix} 2 & 3 & -1 \\ 3 & 0 & 2 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ 1 & -2 \\ 2 & -4 \end{bmatrix} A(BC)=[(2)(1)+(3)(1)+(−1)(2)(2)(−2)+(3)(−2)+(−1)(−4)(3)(1)+(0)(1)+(2)(2)(3)(−2)+(0)(−2)+(2)(−4)]A(BC) = \begin{bmatrix} (2)(1) + (3)(1) + (-1)(2) & (2)(-2) + (3)(-2) + (-1)(-4) \\ (3)(1) + (0)(1) + (2)(2) & (3)(-2) + (0)(-2) + (2)(-4) \end{bmatrix} A(BC)=[2+3−2−4−6+43+0+4−6+0−8]A(BC) = \begin{bmatrix} 2 + 3 - 2 & -4 - 6 + 4 \\ 3 + 0 + 4 & -6 + 0 - 8 \end{bmatrix} A(BC)=[3−67−14]A(BC) = \begin{bmatrix} 3 & -6 \\ 7 & -14 \end{bmatrix}
  5. तुलना करें।

    हमने पाया कि:

    (AB)C=[3−67−14](AB)C = \begin{bmatrix} 3 & -6 \\ 7 & -14 \end{bmatrix}

    और

    A(BC)=[3−67−14]A(BC) = \begin{bmatrix} 3 & -6 \\ 7 & -14 \end{bmatrix}

    चूँकि (AB)C=A(BC)(AB)C = A(BC), यह सिद्ध होता है।

    सिद्ध हुआ\text{सिद्ध हुआ}
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