Math Insight

The determinant of a 2x2 matrix and the number of solutions

Math 2241, Spring 2023
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Due date: Feb. 1, 2023, 11:59 p.m.
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Total points: 1
  1. When we solve a matrix equation $A\vc{x}=\vc{b}$, we think of the matrix $A$ and the vector $\vc{b}$ as being known values, and our goal is to determine the value(s) of the unknown vector $\vc{x}$ for which the equation is satisfied.

    For the moment, though, let's not give numbers for the matrix $A$ or the vector $\vc{b}$. Instead, we'll represent $A$ and $\vc{b}$ as $$A = \begin{bmatrix} a & b\\ c& d \end{bmatrix}, \qquad \vc{b}=\begin{bmatrix} u \\ v \end{bmatrix},$$ where $a$, $b$, $c$, $d$, $u$, and $v$ are some numbers. (Notice that $b$ is a number while $\vc{b}$ is a completely different vector.) If the unknown vector $\vc{x}$ is $$\vc{x} = \begin{bmatrix}x\\y\end{bmatrix},$$ then we can write our matrix equation for $\vc{x}$ as $$\begin{bmatrix} a & b\\ c& d \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} u \\ v \end{bmatrix}.$$ Suppose we want to know how many solutions there are to the equation. The number of solutions will depend on the relationships between $a$, $b$, $c$, $d$, $u$, and $v$. We want to come up with conditions on these constants that will allow us to determine the number of solutions.

    1. In order to find these conditions, we will need to solve the system without replacing $a$, $b$, etc. with numbers. First, convert the matrix equation into a system of two equations. Write the two equations below.




      As the next step toward solving the system, eliminate $y$. In order to make the coefficients of $y$ in the two equations to match, multiply the first equation by $d$ and the second equation by
      . Write the resulting equations below:




      Now, subtract the second equation from the first equation. Write the resulting equation below:

      Although this equation is slightly messy, the important point is that it is just a linear equation for $x$, as $a$, $b$, $c$, $d$, $u$, and $v$ are just numbers (we call them constants). It's really just one number times $x$ equals another number; those two numbers are just combinations of the constants. Rewrite the equation in this form:
      $\big($
      $\big) x =$

      Keep in mind that the combinations of constants you entered in each of those two blanks simply represent two numbers.

    2. Let's examine the single equation in $x$ we just found. We can solve for $x$ by dividing by its coefficient, provided that coefficient is not
      . You should have gotten that the coefficient of $x$ (i.e., the “number” you entered in the first of the above two answer blanks) was $ad-bc$. Therefore, if $ad-bc \neq$
      , then there is a unique value of $x$ which will solve the system.

      To finish solving the system, we also have to determine $y$. To do so, we need to plug $x$ into one of the original equations. We'd have to do a bit of algebra to then solve for $y$, but we aren't going to worry about that algebra. We just want to know whether or not we could solve for $y$.

      It turns out that we can always solve for $y$. Here's why. Since we require that $ad-bc \neq$
      , we know that both $b$ and $d$ cannot be
      . This means that $y$ must appear in at least one of the two original equations. Plug in the value of $x$ into that equation and solve for $y$.

      We have determined the condition that guarantees the existence of a unique solution to the system:

      $\neq 0$

    3. What happens if $ad-bc=0$? If we assume that $ad-bc=0$, this simplifies the equation we found at the end of part a. Write the simplified equation here:


      This equation contains only $b$, $d$, $u$, and $v$, which are all constants. In particular, the equation does not depend on the variables $x$ and $y$. Since the equation simply states that two numbers are equal, there are two possibilities. The first possibility is that the number $du$ is the same as the number $bv$ in which case the equation is
      . The second possibility is that the number $du$ is not the same number as $bv$, in which case the equation is
      .

    4. Something strange is going on when $ad-bc=0$. A geometric perspective might give us some insight what is happening. The original equations $ax+by=u$ and $cx+dy=v$ describe two lines in the $xy$-plane. When trying to find the solution to both equations, we are trying to find points $(x,y)$ where the two lines intersect.

      For simplicity, let's assume that both $b$ and $d$ are not zero. (Our conclusions are still true in the case where they are zero, but then we have vertical lines with infinite slope. Let's avoid this case so we can talk about slopes and $y$-intercepts.)

      The condition $ad-bc=0$ is the same as $ad=bc$. Divide both sides of this equation by $bd$. The condition becomes:
      . This condition will tell us something about the slopes of the lines.

      What is the slope of the original first equation $ax+by=u$?
      What is the slope of the original second equation $cx+dy=v$?
      If the condition you wrote, $_$, is true, then what must be true of the slopes?

      In other words, when $ad-bc=0$, the two lines $ax+by=u$ and $cx+dy=v$ are
      .

    5. No wonder, we ran into problems. Parallel lines don't typically intersect. The only way for two parallel lines to intersect is when they are identical, in which case they intersect at a infinite number of points rather than just one.

      How do we know if the lines are identical? Since, for simplicity, we are assuming the lines are not vertical, we can check their $y$-intercepts. What is the $y$-intercept of the line $ax+by=u$?
      What is the $y$-intercept of the line $cx+dy=v$?
      The two parallel lines are identical if their $y$-intercepts are the same, i.e., if
      . Multiply both sides of this equation by $bd$ to rewrite this equation as
      . Does this condition look familiar? It's exactly the same condition we encountered above when trying to solve for $x$ when $ad-bc=0$.

      In summary, if $ad-bc=0$ and $bv \ne ud$, then the lines $ax+by=u$ and $cx+dy=v$ are
      . This means that the system of equations has
      solutions.

      On the other hand, if $ad-bc=0$ and $bv = ud$, then the lines $ax+by=u$ and $cx+dy=v$ are
      . This means that the system of equations has
      solutions. Pick any point on that one line and it will be a solution to the system of equations.

      For the easy case, $ad-bc \ne 0$, the lines $ax+by=u$ and $cx+dy=v$ are
      . This means that the system of equations has
      solution. In this case, the number of solutions doesn't depend on the values of $u$ and $v$.

    6. The value of $ad-bc$ determines whether or not the system has a unique solution. In fact, this number has other properties that help in determining the behavior of a system, and so we give it a name. The determinant of the $2\times 2$ matrix $$\begin{bmatrix} a & b\\ c& d \end{bmatrix} $$ is $$\det \left( \begin{bmatrix} a & b\\ c& d \end{bmatrix} \right) = ad-bc.$$ The determinant is defined for all square matrices, but the formula for it is quite complicated for larger matrices. In all cases, a matrix equation has a unique solution if and only if the determinant is nonzero.

      What is the determinant of the matrix $\begin{bmatrix} 3 & 1\\ -1& 2 \end{bmatrix}$?
      How many solutions does the following matrix equation have?
      $$\begin{bmatrix} 3 & 1\\ -1& 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 3 \\ 2 \end{bmatrix}$$ Does the number of solutions depend on the vector on the right-hand side?

  2. Calculate the determinants of the coefficient matrices $A$ in the following matrix equations of the form $A\vc{x}=\vc{b}$. Determine how many solutions each equation has.
    1. $\left[\begin{matrix}3 & -6\\-2 & 4\end{matrix}\right]\left[\begin{matrix}x\\y\end{matrix}\right] = \left[\begin{matrix}3\\1\end{matrix}\right]$
      $\det A =$

      number of solutions:
    2. $\left[\begin{matrix}8 & 3\\3 & 1\end{matrix}\right]\left[\begin{matrix}x\\y\end{matrix}\right] = \left[\begin{matrix}4\\-1\end{matrix}\right]$
      $\det A =$

      number of solutions:
    3. $\left[\begin{matrix}1 & 1\\-4 & -4\end{matrix}\right]\left[\begin{matrix}x\\y\end{matrix}\right] = \left[\begin{matrix}3\\-12\end{matrix}\right]$
      $\det A =$

      number of solutions:

  3. Consider the equation $A\vc{x}=\vc{b}$ where $$A = \left[\begin{matrix}-5 & 4\\10 & -8\end{matrix}\right], \quad \vc{x} = \begin{bmatrix}x\\y\end{bmatrix}, \quad \text{and} \quad \vc{b}=\left[\begin{matrix}7\\-14\end{matrix}\right].$$

    How many solutions does the system of equation have?
    .

    To find a solution, pick any value for $x$:
    . Then, calculate $y$:
    . Therefore, one solution is $\vc{x} = $
    .

    To find a solution, pick any different value for $x$:
    . Then, calculate $y$:
    . A second solution is therefore $\vc{x} = $
    .