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Question 16 of 44

Q.If A=(11−12−343−23)A = \begin{pmatrix} 1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{pmatrix} and BT=(1−25−2413−6−1)B^T = \begin{pmatrix} 1 & -2 & 5 \\ -2 & 4 & 1 \\ 3 & -6 & -1 \end{pmatrix}, then find the rank of ABAB.

Tamil Nadu DgeTamil Nadu HSC (DGE) Commerce Board 2020Subjective· 3mImportance★★★★★
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Transpose BTB^T to get BB, compute ABAB, find det⁡(AB)=0\det(AB)=0 but a 2×22\times2 minor e0 e 0; hence the rank is 22.

Step 1 — recover BB. Since BT=(1−25−2413−6−1)B^T = \begin{pmatrix} 1 & -2 & 5 \\ -2 & 4 & 1 \\ 3 & -6 & -1 \end{pmatrix},

B=(BT)T=(1−23−24−651−1).B = (B^T)^T = \begin{pmatrix} 1 & -2 & 3 \\ -2 & 4 & -6 \\ 5 & 1 & -1 \end{pmatrix}.

Step 2 — compute ABAB with A=(11−12−343−23)A = \begin{pmatrix} 1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{pmatrix}:

AB=(−61−228−122022−1118).AB = \begin{pmatrix} -6 & 1 & -2 \\ 28 & -12 & 20 \\ 22 & -11 & 18 \end{pmatrix}.

(For example, row 1: 1⋅1+1⋅(−2)+(−1)⋅5=−61\cdot1+1\cdot(-2)+(-1)\cdot5 = -6; 1⋅(−2)+1⋅4+(−1)⋅1=11\cdot(-2)+1\cdot4+(-1)\cdot1 = 1; 1⋅3+1⋅(−6)+(−1)⋅(−1)=−21\cdot3+1\cdot(-6)+(-1)\cdot(-1) = -2.)

Step 3 — determinant. …

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