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Q.Given A=[201345023]A = \begin{bmatrix} 2 & 0 & 1 \\ 3 & 4 & 5 \\ 0 & 2 & 3 \end{bmatrix} and B=[11−5−51−51−24]B = \begin{bmatrix} 1 & 1 & -5 \\ -5 & 1 & -5 \\ 1 & -2 & 4 \end{bmatrix}, find BABA.

CBSECBSE Class XII Board 2025Subjective· 2mImportance★★★★★
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Multiplying BB (rows) into AA (columns) entry-by-entry gives BA=[5−6−9−7−6−15−403]BA = \begin{bmatrix} 5 & -6 & -9 \\ -7 & -6 & -15 \\ -4 & 0 & 3 \end{bmatrix}.

(BA)ij=∑kBik Akj(BA)_{ij} = \sum_{k} B_{ik}\,A_{kj} — the (i,j)(i,j) entry is the dot product of row ii of BB with column jj of AA.

With A=[201345023]A = \begin{bmatrix} 2 & 0 & 1 \\ 3 & 4 & 5 \\ 0 & 2 & 3 \end{bmatrix} and B=[11−5−51−51−24]B = \begin{bmatrix} 1 & 1 & -5 \\ -5 & 1 & -5 \\ 1 & -2 & 4 \end{bmatrix}:

  1. Row 1 of B=(1, 1, −5)B = (1,\,1,\,-5):
    • col 1: 1(2)+1(3)+(−5)(0)=51(2)+1(3)+(-5)(0) = 5
    • col 2: 1(0)+1(4)+(−5)(2)=−61(0)+1(4)+(-5)(2) = -6
    • col 3: 1(1)+1(5)+(−5)(3)=−91(1)+1(5)+(-5)(3) = -9
  2. Row 2 of B=(−5, 1, −5)B = (-5,\,1,\,-5):
    • col 1: −5(2)+1(3)+(−5)(0)=−7-5(2)+1(3)+(-5)(0) = -7 …

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