Disclaimer: these are my personal solutions; they have not been reviewed, so be careful relying on them as there might be errors. Some competitions may have official solutions from the contest publishers which you may wish to refer to instead if you’re looking for verified solutions.

Problem 3

Find real numbers $c_1$ and $c_2$ so that

$I + c_1 M +c_2 M^2 = \begin{pmatrix} 0 & 0 \newline 0 & 0 \newline \end{pmatrix},$

where $M = \begin{pmatrix} 1 & 3 \newline 0 & 2 \newline \end{pmatrix}$ and $I$ is the identity matrix.

Hints

Hint #1

Can you simplify the given equation and see what you can do with it?

Hint #2

You have 2 unknowns ( $c_1$ and $c_2$ ); can you come up with equations to isolate their values, given $I$ and $M$ ?

Solution

Open to see the solution; try solving it first or see the hints above!

First, let’s compute $M^2$ :

\begin{align*} M^2 & = \begin{pmatrix} 1 & 3 \\ 0 & 2 \\ \end{pmatrix}^2 \\ & = \begin{pmatrix} 1 & 3 \\ 0 & 2 \\ \end{pmatrix} \begin{pmatrix} 1 & 3 \\ 0 & 2 \\ \end{pmatrix} \\ & = \begin{pmatrix} 1 \cdot 1 + 3 \cdot 0 & 1 \cdot 3 + 3 \cdot 2 \\ 0 \cdot 1 + 2 \cdot 0 & 0 \cdot 3 + 2 \cdot 2 \\ \end{pmatrix} \\ & = \begin{pmatrix} 1 & 9 \\ 0 & 4 \\ \end{pmatrix} \end{align*}

Now, let’s put this together into the complete equation:

\begin{align*} I + c_1 M + c_2 M^2 & = \begin{pmatrix} 0 & 0 \\ 0 & 0 \\ \end{pmatrix} \\ \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} + c_1 \begin{pmatrix} 1 & 3 \\ 0 & 2 \\ \end{pmatrix} + c_2 \begin{pmatrix} 1 & 9 \\ 0 & 4 \\ \end{pmatrix} & = \begin{pmatrix} 0 & 0 \\ 0 & 0 \\ \end{pmatrix} \\ \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} + \begin{pmatrix} c_1 & 3 c_1 \\ 0 & 2 c_1 \\ \end{pmatrix} + \begin{pmatrix} c_2 & 9 c_2 \\ 0 & 4 c_2 \\ \end{pmatrix} & = \begin{pmatrix} 0 & 0 \\ 0 & 0 \\ \end{pmatrix} \\ \begin{pmatrix} 1 + c_1 + c_2 & 3 c_1 + 9 c_2 \\ 0 & 1 + 2 c_1 + 4 c_2 \\ \end{pmatrix} & = \begin{pmatrix} 0 & 0 \\ 0 & 0 \\ \end{pmatrix} \\ \end{align*}

As a result, we have the following system of equations:

\begin{align*} 1 + c_1 + c_2 & = 0 & \qquad \text{(1)} \\ 3 c_1 + 9 c_2 & = 0 & \qquad \text{(2)} \\ 1 + 2 c_1 + 4 c_2 & = 0 & \qquad \text{(3)} \\ \end{align*}

Taking equation $(2)$ , we have:

\begin{align*} 3 c_1 + 9 c_2 & = 0 & \qquad \text{from (2) above} \\ c_1 + 3 c_2 & = 0 & \qquad \text{divide both sides by $3$} \\ c_1 & = -3 c_2 & \qquad \text{solve for $c_1$ \quad (4)} \\ \end{align*}

Now we can substitute $c_1 = -3 c_2$ into equations $(1)$ and $(3)$ to find $c_2$ and hopefully, they’re consistent!

\begin{align*} 1 + c_1 + c_2 & = 0 & \qquad \text{from (1) above} \\ 1 - 3 c_2 + c_2 & = 0 & \qquad \text{substitute from (4) above} \\ 1 - 2 c_2 & = 0 & \qquad \text{simplify} \\ 1 & = 2 c_2 & \qquad \text{bring $c_2$ to the other side} \\ \frac{1}{2} & = c_2 & \qquad \text{solve for $c_2$} \\ \end{align*}

Let’s try the same with equation $(3)$ :

\begin{align*} 1 + 2 c_1 + 4 c_2 & = 0 & \qquad \text{from (3) above} \\ 1 + 2 (-3 c_2) + 4 c_2 & = 0 & \qquad \text{substitute from (4) above} \\ 1 + -2 c_2 & = 0 & \qquad \text{simplify} \\ \end{align*}

We can stop here, as we’ve repeated the 3rd line from the previous solution, so $c_2 = \dfrac{1}{2}$ .

Now we know that $\displaystyle{c_1 = -3 c_2 = -3 \cdot \frac{1}{2} = - \frac{3}{2}}$ .

Thus, the solution is: $\boxed{ c_1 = - \frac{3}{2}, c_2 = \frac{1}{2} }. \enspace \blacksquare$

Problem 3#

Hints#

Solution#

Problem 3

Hints

Solution