IMO Practice Test — Determinants
6 Questions • 20 min • Olympiad level
20:00
Question 1 of 6
hard
$\begin{vmatrix}1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z\end{vmatrix}=$
$xyz+xy+yz+zx$
$xyz$
$1+xyz$
$x+y+z+xyz$
Explanation: $=xyz+xy+yz+zx$ (proved by row operations: $R_1-R_3$, $R_2-R_3$, then expand).
Question 2 of 6
hard
If $\omega$ is a cube root of unity ($\omega^3=1$, $1+\omega+\omega^2=0$), then $\begin{vmatrix}1&\omega&\omega^2\\\omega&\omega^2&1\\\omega^2&1&\omega\end{vmatrix}=$
$0$
$1$
$-1$
$3$
Explanation: $C_1+C_2+C_3$ gives $1+\omega+\omega^2=0$ in each row. So det$=0$.
Question 3 of 6
medium
For what value of $k$ is $\begin{vmatrix}k & 2 \\ 3 & k\end{vmatrix} = \begin{vmatrix}1 & 0 \\ 2 & 1\end{vmatrix}$?
$\pm\sqrt{7}$
$\pm\sqrt{5}$
$\pm\sqrt{6}$
$\pm\sqrt{8}$
Explanation: LHS $=k^2-6$; RHS $=1$. So $k^2-6=1$, $k^2=7$, $k=\pm\sqrt{7}$.
Question 4 of 6
hard
If $f(x)=\begin{vmatrix}1+\sin^2x & \cos^2x & 4\sin 2x \\ \sin^2x & 1+\cos^2x & 4\sin 2x \\ \sin^2x & \cos^2x & 1+4\sin 2x\end{vmatrix}$, then the maximum value of $f(x)$ is:
6
4
8
2
Explanation: $R_1-R_2$ and $R_1-R_3$, then expand. Result = $2+4\sin 2x$. Max when $\sin 2x=1$: $2+4=6$.
Question 5 of 6
hard
If $A$ is a skew-symmetric matrix of odd order, then $|A|=$
$0$
$1$
$-1$
Cannot determine
Explanation: For a skew-symmetric matrix, $|A|=|A'|=|-A|=(-1)^n|A|$. For odd $n$: $|A|=-|A|$, so $|A|=0$.
Question 6 of 6
hard
The system $x+2y+3z=1$, $2x+y+3z=2$, $5x+5y+9z=4$ has:
Unique solution
No solution
Infinitely many solutions
Two solutions
Explanation: Compute $|A|$: it equals 0. Check (adj$A)B$: equals $O$. So infinitely many solutions.