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3B2C3B-2C quand B=43 02 10;C=10 24 03B = \begin{vmatrix} 4 & -3 \ 0 & 2 \ 1 & 0 \end{vmatrix} ;C= \begin{vmatrix} 1 & 0 \ 2 & 4 \ 0 & -3 \end{vmatrix}

Full solution

Q. 3B2C3B-2C quand B=43 02 10;C=10 24 03B = \begin{vmatrix} 4 & -3 \ 0 & 2 \ 1 & 0 \end{vmatrix} ;C= \begin{vmatrix} 1 & 0 \ 2 & 4 \ 0 & -3 \end{vmatrix}
  1. Calculate matrix B: Step 11: Calculate the matrix B.\newlineB=(43 02 10)B = \begin{pmatrix} 4 & -3 \ 0 & 2 \ 1 & 0 \end{pmatrix}
  2. Calculate matrix C: Step 22: Calculate the matrix C.\newlineC=(10 24 03)C = \begin{pmatrix} 1 & 0 \ 2 & 4 \ 0 & -3 \end{pmatrix}
  3. Multiply B by 33: Step 33: Multiply matrix B by 33.3B=3×(43 02 10)=(129 06 30)3B = 3 \times \begin{pmatrix} 4 & -3 \ 0 & 2 \ 1 & 0 \end{pmatrix} = \begin{pmatrix} 12 & -9 \ 0 & 6 \ 3 & 0 \end{pmatrix}
  4. Multiply C by 22: Step 44: Multiply matrix C by 22.2C=2×(10 24 03)=(20 48 06)2C = 2 \times \begin{pmatrix} 1 & 0 \ 2 & 4 \ 0 & -3 \end{pmatrix} = \begin{pmatrix} 2 & 0 \ 4 & 8 \ 0 & -6 \end{pmatrix}
  5. Subtract 22C from 33B: Step 55: Subtract 2C2C from 3B3B.3B - 2C = \begin{vmatrix} 12 & -9 \end{vmatrix} - \begin{vmatrix} 2 & 0 \end{vmatrix} = \begin{vmatrix} 10 & -9 \-4 & -2 \end{vmatrix} 06 4842 36\begin{vmatrix} 0 & 6 \ 4 & 8 \end{vmatrix} \begin{vmatrix} -4 & -2 \ 3 & 6 \end{vmatrix} 30 0636\begin{vmatrix} 3 & 0 \ 0 & -6 \end{vmatrix} \begin{vmatrix} 3 & 6 \end{vmatrix}

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