Problem
Let M = aᵢⱼ, i, j ∈ {1, 2, 3}, be the 3 x 3 matrix such that aᵢⱼ = 1 if j + 1 is divisible by i, otherwise aᵢⱼ = 0. Then which of the following statements is (are) true?
A. M is invertible
B. There exists a nonzero column matrix \( \begin{pmatrix} a_1 \\ a_2 \\ a_3 \\ \end{pmatrix} \) such that M \(\begin{pmatrix} a_1 \\ a_2 \\ a_3 \\ \end{pmatrix} = \begin{pmatrix} -a_1 \\ -a_2 \\ -a_3 \\ \end{pmatrix} \)
C. The set {X ∈ R³: MX = O} ≠ {O}, where O = \( \begin{pmatrix} 0 \\ 0 \\ 0 \\ \end{pmatrix} \)
D. The matrix (M – 2I) is invertible, where I is the 3 x 3 identity matrix
Solution
Let’s start by writing out the elements of M.
After applying some advanced mathematics, we’ll get M = \( \begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \\ \end{pmatrix} \).
To determine if option A is true, we’ll need to find |M|. If we expand using the last row, we’ll get |M| = 0. Sadly, the matrix M isn’t invertible, making option A false.
Option D can be figured out easily as well. M – 2I = \( \begin{pmatrix} -1 & 1 & 1 \\ 1 & -2 & 1 \\ 0 & 1 & -2 \\ \end{pmatrix} \)
To find |M – 2I|, we’ll expand using the first column, getting 0 again. |M – 2I| isn’t invertible as well, making option D false. Sad.
Now, coming to option B. Let’s call that column matrix as X. We’re looking for is a non-zero matrix X such that MX = -X.
Rearranging this equation gives (M + I)X = O. We want to check if there’s a non-zero matrix X that satisfies the equation.
In other words, we’re looking for a non-trivial solution of the equation \( \begin{pmatrix} 2 & 1 & 1 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \\ \end{pmatrix} \begin{pmatrix} a_1 \\ a_2 \\ a_3 \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ \end{pmatrix}\)
Let’s apply the Cramer’s rule! Since \( \begin{vmatrix} 2 & 1 & 1 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \\ \end{vmatrix} \) = 0, we can conclude that the given equation has infinitely many solutions.
For example, here’s one solution: a₁ = 0, a₂ = -1, a₃ = 1. So, option B is true.
The same method applies for option C.
Here, we’re looking for a non-trivial solution of the equation \( \begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \\ \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ \end{pmatrix}\)
We already know that |M| = 0. So, by the Cramer’s rule, the above equation has infinitely many solutions. Here’s one solution: x₁ = -1, x₂ = 0, x₃ = 1.
So, option C is true as well.
Comments
Weird problem overall. Firstly, not sure why the matrix was defined in such a twisted way. Secondly, options A and D are testing the same concept (and are both false 😞). Finally, options B and C are also testing the same concept. Bonus – the extra weird language used for option C. On the positive side, this was an easy problem.
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