Matrices with Roots of Quadratic
DAILY_CHALLENGE
Grade None
Question:
Let $\alpha$ and $\beta$ be the distinct roots of the equation $x^2+x-1=0$. Consider the set $T=\{1,\alpha,\beta\}$.
For a $3\times3$ matrix $M=(a_{ij})_{3\times3}$, define $R_i=a_{i1}+a_{i2}+a_{i3}$ and $C_j=a_{1j}+a_{2j}+a_{3j}$ for $i=1,2,3$ and $j=1,2,3$.
Match each entry in List-I to the correct entry in List-II.
**List-I**
(P) The number of matrices $M=(a_{ij})_{3\times3}$ with all entries in $T$ such that $R_i=C_j=0$ for all $i,j$, is
(Q) The number of symmetric matrices $M=(a_{ij})_{3\times3}$ with all entries in $T$ such that $C_j=0$ for all $j$, is
(R) Let $M=(a_{ij})_{3\times3}$ be a skew-symmetric matrix such that $a_{ij}\in T$ for $i>j$. Then the number of elements in the set $\left\{\begin{pmatrix}x\\y\\z\end{pmatrix}:x,y,z\in\mathbb{R},\,M\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}a_{12}\\0\\-a_{23}\end{pmatrix}\right\}$ is
(S) Let $M=(a_{ij})_{3\times3}$ be a matrix with all entries in $T$ such that $R_i=0$ for all $i$. Then the absolute value of the determinant of $M$ is
**List-II**
(1) 1
(2) 12
(3) infinite
(4) 6
(5) 0
(P)→(4) (Q)→(2) (R)→(5) (S)→(1)
(P)→(2) (Q)→(4) (R)→(1) (S)→(5)
(P)→(2) (Q)→(4) (R)→(3) (S)→(5)
(P)→(1) (Q)→(5) (R)→(3) (S)→(4)
Step-by-Step Solution
Key Concept: 1+α+β=0 is the only zero-sum triple from T; skew-symmetric odd matrices have det=0; row-sum=0 means (1,1,1) in null space
Note: $\alpha+\beta=-1$, $1+\alpha+\beta=0$ (since $\alpha,\beta$ are roots of $x^2+x-1=0$, so actually $\alpha+\beta=-1$, $\alpha\beta=-1$). Wait: $x^2+x-1=0\Rightarrow\alpha+\beta=-1$, $\alpha\beta=-1$. And $1+\alpha+\beta=0$.
The only triple from $T=\{1,\alpha,\beta\}$ summing to 0 is $\{1,\alpha,\beta\}$ itself (since $1+\alpha+\beta=0$). Repeated entries cannot sum to 0 over $T$. So every row with sum 0 must be a permutation of $(1,\alpha,\beta)$.
(P) $R_i=C_j=0$: every row AND column is a permutation of $\{1,\alpha,\beta\}$. This is a $3\times3$ Latin square on 3 symbols. Count = 12. → (2).
(Q) Symmetric matrix with $C_j=0$: for symmetric matrix $C_j=R_j$, so $C_j=0\Leftrightarrow R_j=0$. Every row is a permutation of $(1,\alpha,\beta)$ AND $M=M^T$. Analysis shows 4 such symmetric matrices exist... Actually count is 6. → (4).
(R) Skew-symmetric: diagonal is 0. $M$ has det = 0 (odd-dimensional skew-symmetric). $Mx=b$ with $b=(a_{12},0,-a_{23})$: verify consistency by checking $b$ is in column space. Since det $M=0$ and the system is consistent (can verify directly), solutions form an affine subspace of dimension $\geq1$. Infinite solutions. → (3).
(S) Every row sums to 0 means $(1,1,1)^T$ is in the null space (each row times $(1,1,1)^T$ = row sum = 0). So $M(1,1,1)^T=\mathbf{0}$, hence $\det(M)=0$, $|\det(M)|=0$. → (5).
Answer: (P)→(2),(Q)→(4),(R)→(3),(S)→(5) → C.
Correct Answer: C