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A pair of simultaneous equations written as one matrix equation, solved with the inverse of a 2x2 matrix, and what a zero determinant says about the system.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
In this lesson a pair of simultaneous equations becomes a single object. Written as $AX = B$, the whole system is one multiplication, and solving it is undoing that multiplication with the inverse matrix — a rule you can write down once and then apply to any system of the same size. You will build the inverse of a two-by-two matrix, use it to solve, and read the determinant for what it really says: not a step in the arithmetic, but the answer to whether the system has one solution at all.
You can solve a pair of simultaneous equations by elimination or substitution, and you have just learned to multiply matrices and to compute a determinant. This lesson connects the two: the same pair of equations, written once as a single matrix equation, and solved by undoing a multiplication rather than by juggling the lines.
Coefficient matrix: the grid of numbers multiplying the unknowns.
Matrix equation: $AX = B$, with $X$ the column of unknowns and $B$ the column of constants.
Inverse matrix $A^{-1}$: the matrix with $A^{-1}A = AA^{-1} = I$.
Singular: having determinant zero, and therefore no inverse.
Consistent: having at least one solution.
Unique solution: exactly one — what an invertible coefficient matrix guarantees.
The system $$\begin{aligned} ax + by &= e\\ cx + dy &= f \end{aligned}$$ is the single matrix equation $AX = B$ with $A = \begin{pmatrix} a & b \\ c & d\end{pmatrix}$, $X = \begin{pmatrix} x \\ y\end{pmatrix}$ and $B = \begin{pmatrix} e \\ f\end{pmatrix}$: multiplying row by column reproduces the two lines exactly. If $A$ has an inverse, multiply on the left by it — the side $A$ is on — to get $X = A^{-1}B$, and that is the whole solution. For a $2\times2$ matrix the inverse is $$A^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a\end{pmatrix}:$$ swap the main diagonal, negate the other, divide by the determinant. So $\begin{pmatrix} 2 & 1 \\ 1 & 1\end{pmatrix}$ has determinant $1$ and inverse $\begin{pmatrix} 1 & -1 \\ -1 & 2\end{pmatrix}$. When the determinant is zero there is no inverse and no unique solution: the rows are proportional, the two lines are parallel, and the system has either none (parallel and apart) or infinitely many (the same line twice). The determinant is therefore not a step in the calculation — it is the answer to whether the calculation is possible at all.
Another way: picture
Two pairs of axes. On the left, two lines crossing at a single marked point, with $\det A \ne 0$ written beneath. On the right, two parallel lines that never meet, with $\det A = 0$ beneath — the same picture the algebra is describing.
Another way: steps
"$X = B/A$." There is no division of matrices. You multiply by the inverse, and you must do it on the correct side: from $AX = B$ you get $X = A^{-1}B$, not $BA^{-1}$.
"Determinant zero means no solution." It means no unique solution. The two equations may describe the very same line, in which case there are infinitely many. Deciding which needs a look at the constants, not just at the determinant.
Forgetting the $\frac{1}{ad-bc}$. The inverse of $\begin{pmatrix} a & b \\ c & d\end{pmatrix}$ is $\frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a\end{pmatrix}$. Swapping and negating without dividing gives a matrix whose product with the original is $(ad-bc)I$, not $I$.
$A = \begin{pmatrix} 1 & 1 \\ 1 & -1\end{pmatrix}$, $B = \begin{pmatrix} 5 \\ 1\end{pmatrix}$, and $\det A = -1 - 1 = -2 \ne 0$.
A non-zero determinant promises exactly one solution.
$A^{-1} = -\tfrac{1}{2}\begin{pmatrix} -1 & -1 \\ -1 & 1\end{pmatrix} = \begin{pmatrix} \tfrac12 & \tfrac12 \\ \tfrac12 & -\tfrac12\end{pmatrix}$.
Swap, negate, divide.
$X = A^{-1}B = \begin{pmatrix} \tfrac52 + \tfrac12 \\ \tfrac52 - \tfrac12\end{pmatrix} = \begin{pmatrix} 3 \\ 2\end{pmatrix}$, so $x = 3$, $y = 2$ — and both equations check out.
$\det\begin{pmatrix} 2 & 4 \\ 3 & 6\end{pmatrix} = 12 - 12 = 0$, so there is no unique solution.
Singular, so decide between the two remaining cases.
The coefficients are in the ratio $2:3$ but the constants are $6:10$, which is not $2:3$. The lines are parallel and apart: no solution at all.
The determinant is $6 - 5 = 1$, so there is nothing to divide by.
Swap the main diagonal and negate the other: $\begin{pmatrix} 2 & -1 \\ -5 & 3\end{pmatrix}$.
Solve the system $5x + 4y = 28$ and $x + 4y = 12$.
x = x, y = y
Enter the inverse of $\begin{pmatrix} 7 & 3 \\ 2 & 1 \end{pmatrix}$.
This task has no paper form; do it on a device.
For which value of $k$ does $\begin{pmatrix} 3 & 7 \\ 6 & k \end{pmatrix}$ fail to have an inverse?
Answer:
The coefficient matrix of a system of two equations in two unknowns has determinant zero. What follows about the system?
$A$ is invertible and $AX = B$. Which of these is $X$?
A café sells only tea and coffee. One receipt shows $6$ teas and $5$ coffees for $22$ pounds. A second shows $12$ teas and $10$ coffees for $44$ pounds. Can the two prices be worked out from these receipts?
A blender takes $x$ scoops of oats and $y$ scoops of nuts. Every scoop of oats adds $4$ g of fibre and $2$ g of fat; every scoop of nuts adds $1$ g of fibre and $3$ g of fat. The finished mix is measured at $16$ g of fibre and $28$ g of fat. How many scoops of each went in?
oats = x scoops, nuts = y scoops
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Solve the system $2x + 4y = 2$ and $3x + 4y = -3$.
x = x, y = y
You can solve a two-by-two system with matrices and say what a zero determinant means. From memory: why is the solution of $AX = B$ written $A^{-1}B$ and not $BA^{-1}$, and what are the two things a zero determinant leaves open?
8. Your turn: the inverse of $\begin{pmatrix} 3 & 1 \\ 5 & 2\end{pmatrix}$, step 2