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Worked Examples · Example 6

Q.Find the values of aa, bb, cc and dd from the following equation: [424−311]=[2a+b4c+3da−2b5c−d]\begin{bmatrix} 4 & 24 \\ -3 & 11 \end{bmatrix} = \begin{bmatrix} 2a+b & 4c+3d \\ a-2b & 5c-d \end{bmatrix}.

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Equating corresponding entries gives two pairs of simultaneous equations, one in a,ba,b and one in c,dc,d.

Equal matrices ⇒\Rightarrow corresponding entries equal, giving:

2a+b=4,a−2b=−3,4c+3d=24,5c−d=11.2a+b=4,\quad a-2b=-3,\quad 4c+3d=24,\quad 5c-d=11.

  1. Solve for a,ba,b. From 2a+b=42a+b=4 and a−2b=−3a-2b=-3. Multiply the first by 2: 4a+2b=84a+2b=8. Add to a−2b=−3a-2b=-3:

5a=5 ⇒ a=1.5a=5\ \Rightarrow\ a=1.

  1. Then b=4−2a=4−2(1)=2b=4-2a=4-2(1)=2. Check a−2b=1−4=−3a-2b=1-4=-3 ✓. …

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