JEE Advanced 2023 Paper 2 · Q12 · Determinants

Question

Let $R = \left\{\begin{pmatrix}a & 3 & b\ c & 2 & d\ 0 & 5 & 0\end{pmatrix} : a, b, c, d \in \{0, 3, 5, 7, 11, 13, 17, 19\}ight\}$. Then the number of invertible matrices in $R$ is ___.

SolveKar AI · Accepted Answer

$\det = 3(0 - 0) - 5(ad - bc) + 0 = -5(ad - bc)$. Invertible iff $ad 
e bc$. Total matrices = $8^4 = 4096$. Count non-invertible (where $ad = bc$): Case 1 — both $ad$ and $bc$ are $0$: at least one of $a,d = 0$ AND at least one of $b,c = 0$. By inclusion-exclusion: number of $(a,d)$ pairs with $ad=0$ is $2\cdot 8 - 1 = 15$. Same for $(b,c)$. So $15\cdot 15 = 225$. Case 2 — $ad = bc 
e 0$: among $\{3,5,7,11,13,17,19\}$ (7 nonzero choices), count $(a,d,b,c)$ with $ad=bc$. Both products are products of two primes (or squares); enumerate: products with all four equal = $7$ (each of $a=b=c=d$); products with $a=d$ and $b=c$ (squared values $9, 25, 49, \ldots$) — these need $a^2 = b^2$, so $a = \pm b$, only $a=b$ valid in our set; $a=c$, $b=d$ symmetry; and others. By Resonance enumeration: invertible = $4096 - 316 = 3780$.

JEE Advanced 2023 Paper 2 · Q12 · Determinants

Solution