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If A=[[7,5,-4],[-2,5,1],[10,-4,-7]] and B=[[2,-2,3],[8,1,4],[-4,2,-8]]. Evaluate the following. a) A.B b) Solve for X if 3X-4A=2B

33. If A=[75−4−25110−4−7] A=\left[\begin{array}{ccc}7 & 5 & -4 \\ -2 & 5 & 1 \\ 10 & -4 & -7\end{array}\right] and B=[2−23814−42−8] B=\left[\begin{array}{ccc}2 & -2 & 3 \\ 8 & 1 & 4 \\ -4 & 2 & -8\end{array}\right] . Evaluate the following.\newlinea) A.B\newlineb) Solve for X X if 3X−4A=2B 3 X-4 A=2 B

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Q. 33. If A=[75−4−25110−4−7] A=\left[\begin{array}{ccc}7 & 5 & -4 \\ -2 & 5 & 1 \\ 10 & -4 & -7\end{array}\right] and B=[2−23814−42−8] B=\left[\begin{array}{ccc}2 & -2 & 3 \\ 8 & 1 & 4 \\ -4 & 2 & -8\end{array}\right] . Evaluate the following.\newlinea) A.B\newlineb) Solve for X X if 3X−4A=2B 3 X-4 A=2 B
  1. Calculate Matrix Product: Calculate the matrix product A.BA.B. To multiply two matrices, we take the dot product of the rows of the first matrix with the columns of the second matrix. Let's calculate A.BA.B. A.B=[[(7×2+5×(−2)+(−4)×(−4)),(7×(−2)+5×1+(−4)×2),(7×3+5×4+(−4)×(−8))],[(−2×2+5×(−2)+1×(−4)),(−2×(−2)+5×1+1×2),(−2×3+5×4+1×(−8))],[(10×2+(−4)×(−2)+(−7)×(−4)),(10×(−2)+(−4)×1+(−7)×2),(10×3+(−4)×4+(−7)×(−8))]]A.B = [[(7\times2 + 5\times(-2) + (-4)\times(-4)), (7\times(-2) + 5\times1 + (-4)\times2), (7\times3 + 5\times4 + (-4)\times(-8))], [(-2\times2 + 5\times(-2) + 1\times(-4)), (-2\times(-2) + 5\times1 + 1\times2), (-2\times3 + 5\times4 + 1\times(-8))], [(10\times2 + (-4)\times(-2) + (-7)\times(-4)), (10\times(-2) + (-4)\times1 + (-7)\times2), (10\times3 + (-4)\times4 + (-7)\times(-8))]] A.B=[[(14−10+16),(−14+5−8),(21+20+32)],[(−4−10−4),(4+5+2),(−6+20−8)],[(20+8+28),(−20−4−14),(30−16+56)]]A.B = [[(14 - 10 + 16), (-14 + 5 - 8), (21 + 20 + 32)], [(-4 - 10 - 4), (4 + 5 + 2), (-6 + 20 - 8)], [(20 + 8 + 28), (-20 - 4 - 14), (30 - 16 + 56)]] A.B=[[20,−17,73],[−18,11,6],[56,−38,70]]A.B = [[20, -17, 73], [-18, 11, 6], [56, -38, 70]]
  2. Solve for XX: Solve for XX in the equation 3X−4A=2B3X-4A=2B. First, we need to express the equation in terms of XX. 3X=2B+4A3X = 2B + 4A Now, we divide both sides by 33 to solve for XX. X=2B+4A3X = \frac{2B + 4A}{3} Let's calculate 2B2B and 4A4A first. XX00 XX11 XX22 XX33 Now, we add 2B2B and 4A4A. XX66 XX77 Now, we divide each element by 33 to get XX. 3X−4A=2B3X-4A=2B00

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