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Exercise B · Q6

Q.If A=[804−236]A = \begin{bmatrix} 8 & 0 \\ 4 & -2 \\ 3 & 6 \end{bmatrix} and B=[2−242−51]B = \begin{bmatrix} 2 & -2 \\ 4 & 2 \\ -5 & 1 \end{bmatrix}, then find a matrix CC, such that 3A−2B+4C=03A - 2B + 4C = 0.

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Solve the matrix equation for CC: C=14(2B−3A)C=\tfrac14(2B-3A).

From 3A−2B+4C=O3A-2B+4C=O we isolate CC:   4C=2B−3A  ⇒  C=14(2B−3A)\;4C=2B-3A\;\Rightarrow\;C=\dfrac{1}{4}(2B-3A).

  1. Given A=[804−236], B=[2−242−51]A=\begin{bmatrix}8&0\\4&-2\\3&6\end{bmatrix},\ B=\begin{bmatrix}2&-2\\4&2\\-5&1\end{bmatrix}.
  2. 3A=[24012−6918]3A=\begin{bmatrix}24&0\\12&-6\\9&18\end{bmatrix} and 2B=[4−484−102]2B=\begin{bmatrix}4&-4\\8&4\\-10&2\end{bmatrix}.
  3. 2B−3A=[4−24−4−08−124+6−10−92−18]=[−20−4−410−19−16]2B-3A=\begin{bmatrix}4-24&-4-0\\8-12&4+6\\-10-9&2-18\end{bmatrix}=\begin{bmatrix}-20&-4\\-4&10\\-19&-16\end{bmatrix}.
  4. Divide by 44: …

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