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Soient les matrices suivantes : A=(10 32)A = \begin{pmatrix} -1 & 0 \ 3 & 2 \end{pmatrix} ;B=(43 02 10)B = \begin{pmatrix} 4 & -3 \ 0 & 2 \ 1 & 0 \end{pmatrix} ;C=(10 24 03)C= \begin{pmatrix} 1 & 0 \ 2 & 4 \ 0 & -3 \end{pmatrix} calcule 3B2C3B – 2 C

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Q. Soient les matrices suivantes : A=(10 32)A = \begin{pmatrix} -1 & 0 \ 3 & 2 \end{pmatrix} ;B=(43 02 10)B = \begin{pmatrix} 4 & -3 \ 0 & 2 \ 1 & 0 \end{pmatrix} ;C=(10 24 03)C= \begin{pmatrix} 1 & 0 \ 2 & 4 \ 0 & -3 \end{pmatrix} calcule 3B2C3B – 2 C
  1. Multiply Matrix B by 33: Step 11: Multiply matrix B by 33.\newlineMatrix B = (43 02 10)\begin{pmatrix} 4 & -3 \ 0 & 2 \ 1 & 0 \end{pmatrix}\newline3B=3×(43 02 10)3B = 3 \times \begin{pmatrix} 4 & -3 \ 0 & 2 \ 1 & 0 \end{pmatrix}\newline = (129 06 30)\begin{pmatrix} 12 & -9 \ 0 & 6 \ 3 & 0 \end{pmatrix}
  2. Multiply Matrix C by 22: Step 22: Multiply matrix C by 22.\newlineMatrix C = (10 24 03)\begin{pmatrix} 1 & 0 \ 2 & 4 \ 0 & -3 \end{pmatrix}\newline2C=2×(10 24 03)2C = 2 \times \begin{pmatrix} 1 & 0 \ 2 & 4 \ 0 & -3 \end{pmatrix}\newline = (20 48 06)\begin{pmatrix} 2 & 0 \ 4 & 8 \ 0 & -6 \end{pmatrix}
  3. Subtract 22C from 33B: Step 33: Subtract 2C2C from 3B3B.3B2C=129 06 3020 48 063B - 2C = \left| \begin{array}{cc} 12 -9 \ 0 6 \ 3 0 \end{array} \right| - \left| \begin{array}{cc} 2 0 \ 4 8 \ 0 -6 \end{array} \right| =12290 0468 300+6= \left| \begin{array}{cc} 12-2 -9-0 \ 0-4 6-8 \ 3-0 0+6 \end{array} \right| =109 42 36= \left| \begin{array}{cc} 10 -9 \ -4 -2 \ 3 6 \end{array} \right|

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