Let A=aij3×3 be a matrix such that AAT=4I and aij+2cij=0, where cij is the cofactor of aij and I is the unit matrix of order 3 a11+4a12a13a21a22+4a23a31a32a33+4+5λa11+1a12a13a21a22+1a23a31a32a33+1=0 then the value of 10λ is ______.

Given that AAT=4I⇒|A|2=4⇒|A|=±2AT=4A−1=4 adj A|A|⇒a11a21a31a12a22a32a13a23a33=4|A|c11c21c31c12c22c32c13c23c33 Now aij=4|A|cij⇒−2cij=4|A|cijasaij+2cij=0⇒|A|=−2Now |A+4I|=A+AAT=|A|I+AT=−2I+AT=−2|I+A|⇒|A+4I|+2|A+I|=0So on comparing, we get 5λ=2⇒λ=25 Hence, 10λ=4

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$Let A = {[a_{ij}]}_{3 \times 3} be a matrix such that {AA}^{T} = 4 I and a_{ij} + 2 c_{ij} = 0, where c_{ij} is the$ $cofactor of a_{ij} and I is the unit matrix of order 3$ $|\begin{array}{ccc} a_{11} + 4 & a_{12} & a_{13} \\ a_{21} & a_{22} + 4 & a_{23} \\ a_{31} & a_{32} & a_{33} + 4 \end{array}| + 5 λ |\begin{array}{ccc} a_{11} + 1 & a_{12} & a_{13} \\ a_{21} & a_{22} + 1 & a_{23} \\ a_{31} & a_{32} & a_{33} + 1 \end{array}| = 0 then the value of 10 λ is$ ______.

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answer is 4.

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Detailed Solution

Given that ${AA}^{T} = 4 I$
$\begin{array}{l} \Rightarrow | A |^{2} = 4 \Rightarrow | A | = \pm 2 \\ A^{T} = 4 A^{- 1} = 4 \frac{adj A}{| A |} \\ \Rightarrow [\begin{array}{lll} a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32} \\ a_{13} & a_{23} & a_{33} \end{array}] = \frac{4}{| A |} [\begin{array}{lll} c_{11} & c_{21} & c_{31} \\ c_{12} & c_{22} & c_{32} \\ c_{13} & c_{23} & c_{33} \end{array}] \end{array}$
$Now a_{ij} = \frac{4}{| A |} c_{ij}$
$\begin{array}{l} \Rightarrow - 2 c_{ij} = \frac{4}{| A |} c_{ij} ({asa}_{ij} + 2 c_{ij} = 0) \\ \Rightarrow | A | = - 2 \end{array}$
Now $| A + 4 I | = |A + {AA}^{T}|$
$\begin{array}{l} = | A | |I + A^{T}| \\ = - 2 |I + A^{T}| \\ = - 2 | I + A | \end{array}$
$\Rightarrow | A + 4 I | + 2 | A + I | = 0$
So on comparing, we get $5 λ = 2 \Rightarrow λ = \frac{2}{5}$
$Hence, 10 λ = 4$