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JEE Main 2025 · Mathematics · Matrices and Determinants · hard

Let A=beginbmatrix1&2&2\\2&1&-2\a&2&bendbmatrix with real a,b satisfy AA^T=9I. If det A>0, the value of det A is

Numerical answer

Let A=[12221−2a2b]A=\begin{bmatrix}1&2&2\\2&1&-2\\a&2&b\end{bmatrix} with real a,ba,b satisfy AAT=9IAA^{T}=9I. If det⁡A>0\det A>0, the value of det⁡A\det A is
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Answer: 27

AAT=9IAA^T=9I means the rows are mutually orthogonal and each has squared length 99. Orthogonality of row 3 with row 1 gives a+4+2b=0a+4+2b=0, and with row 2 gives 2a+2−2b=02a+2-2b=0. Solving, b=a+1b=a+1 and 3a+6=03a+6=0, so a=−2a=-2, b=−1b=-1; indeed the squared length is 4+4+1=94+4+1=9. Then A=[12221−2−22−1]A=\begin{bmatrix}1&2&2\\2&1&-2\\-2&2&-1\end{bmatrix} and det⁡A=1(−1+4)−2(−2−4)+2(4+2)=3+12+12=27\det A=1(-1+4)-2(-2-4)+2(4+2)=3+12+12=27. As a check, ∣A∣2=∣9I∣=729|A|^2=|9I|=729, so ∣A∣=±27|A|=\pm27, and the positive sign is the one given.
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