A system of differential equations gets written as a single matrix equation, \mathbf{x}'=A\mathbf{x}, and everything this course does with that equation — finding eigenvalues, checking whether a fundamental matrix is invertible, solving for undetermined coefficients — leans on matrix arithmetic and the two facts a determinant tells you: whether a matrix is invertible, and (through the characteristic polynomial) where its eigenvalues live.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
The lines x + y = 9 and x - y = 3 cross at one point. What is its x-coordinate?
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Why
Adding the equations eliminates y: (x+y)+(x-y) = 9+3 \Rightarrow 2x = 12 \Rightarrow x = 6. (Then y = 3, so the crossing point is (6, 3).)
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Why
Add or subtract the two equations to eliminate one variable, then solve for x.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
Two numbers have a sum of 22 and a difference of 6. What is the larger number?
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That's the smaller number — 8 = 22 - 14, and 14 - 8 = 6 checks out, but the question asks for the larger one.
16 would need a difference of 16 - 6 = 10 from the other number to sum to 22, not 6 — recheck the system x+y=22,\, x-y=6.
28 is already bigger than the sum of both numbers combined, so it can't be one of them.
Why
Let x, y be the numbers with x > y: x + y = 22 and x - y = 6. Adding gives 2x = 28, so x = 14 (and y = 8).
Let the two numbers be x and y, write the sum and difference as two equations, then add them to eliminate one variable.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
A computer reports \det A = 0.00002 for a 6\times 6 matrix A built from noisy sensor data. What's the most defensible conclusion?
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Technically true only if that number is exact — but floating-point determinants on real data are almost never exactly the algebraic value they'd have on paper, so 'nonzero' from a computer isn't the same guarantee as 'nonzero' from exact arithmetic.
A tiny nonzero float doesn't mean the true underlying matrix is singular either — it could just as easily be a well-behaved invertible matrix whose determinant is legitimately small in magnitude.
The theorem itself doesn't care where A's entries came from — it applies to any square matrix. The caution is about trusting a computed near-zero number, not about the theorem's scope.
Why
The determinant test is exact over the real numbers, but a floating-point computation on noisy data essentially never lands on exactly 0 even for a matrix that's meant to be singular — production numerical code checks the pivots elimination actually produces instead of comparing a computed determinant to zero.
Think about what a determinant extremely close to (but not exactly) zero suggests about a matrix built from noisy real-world data, versus a matrix that is exactly singular.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
A 2\times 2 matrix A has \det A = -3. Is A invertible?
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The test only cares whether \det A equals 0 or not — the sign carries no invertibility information at all, only orientation, a different question.
Symmetry never enters the determinant test. Any square A with \det A \neq 0 is invertible, symmetric or not.
The determinant already IS the row-reduction answer in disguise: a triangular echelon form's diagonal product is exactly \det A, so a nonzero determinant means that reduction already succeeded.
Why
The determinant test is a strict yes/no on one condition: \det A \neq 0. Here -3 \neq 0, so A is invertible — the negative sign says nothing about invertibility, only that the map reverses orientation.
A determinant is one number carrying two pieces of information. It says whether a matrix can be undone at all. Later, it also becomes the raw material for finding the special directions a matrix never rotates.
For a 2\times2 matrix A=\begin{pmatrix}a&b\\c&d\end{pmatrix}, the determinant \det A = ad-bc decides everything that matters here: A is invertible exactly when \det A\ne0, and the characteristic polynomial whose roots are A's eigenvalues, \det(A-\lambda I)=0, is built from this same determinant. Matrix multiplication itself — row of the left matrix against column of the right — is what turns \mathbf{x}'=A\mathbf{x} from notation into an actual pair of coupled equations once you multiply it out.
Multiply A by a vector
Step through the row-by-column multiplication that turns the matrix equation x' = Ax into two separate equations.
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Solve a 2x2 system by elimination
Row-reduce the augmented matrix to find where the two lines intersect.
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Worked example
The determinant decides invertibility
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1
Is A=\begin{pmatrix}2&4\\1&2\end{pmatrix} invertible?
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Compute \det A = (2)(2)-(4)(1)=4-4=0.
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Since \det A=0, A is not invertible. \boxed{\det A = 0,\ \text{not invertible}}
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Where this goes next
Systems of differential equations, ahead
Chapters 12 and 13 write every system as \mathbf{x}'=A\mathbf{x} and classify its long-term behavior entirely from A's eigenvalues — which are the roots of a polynomial built from \det(A-\lambda I). Nothing about that classification is new mathematics once matrix multiplication and determinants are solid.
Check your understanding (counts toward your progress)
Writing x'=2x+y and y'=x-3y as \mathbf x'=A\mathbf x, what is A?
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The rows are swapped relative to the equations' own order -- row 1 must match the x' equation.
The entries within the first row are transposed -- coefficient of x comes first, then y.
This mixes coefficients from both equations into the wrong positions.
Why
Row 1 holds x''s coefficients (2,1) and row 2 holds y''s coefficients (1,-3), read directly off the two equations in order.
A 2\times2 matrix has \det A=0. What does this say about the system A\mathbf x=\mathbf b?
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A zero determinant is exactly the condition that RULES OUT a unique solution.
The determinant is precisely the test for whether A^{-1} exists and a unique solution follows.
A zero determinant rules out uniqueness, but the system could still be inconsistent (no solution) rather than having infinitely many.
Why
\det A=0 means A has no inverse, so A\mathbf x=\mathbf b cannot be solved by \mathbf x=A^{-1}\mathbf b -- it has either no solution or infinitely many.
Recall that \det A=0 is exactly the test for whether A^{-1} fails to exist.
Multiply A=\begin{pmatrix}2&0\\1&3\end{pmatrix} by \mathbf x=\begin{pmatrix}x\\y\end{pmatrix}. Enter the first row of the resulting product as an expression in x and y.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Row 1 of A against the column \mathbf x: (2)(x)+(0)(y)=2x.
Correct answer
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Why
Multiply row 1 of A against the column vector \mathbf x=(x,y).
Find \det\begin{pmatrix}4&2\\3&1\end{pmatrix}.
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Why
\det = (4)(1)-(2)(3) = 4-6=-2.
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Why
Is B=\begin{pmatrix}3&6\\1&2\end{pmatrix} invertible? Answer 1 for yes, 0 for no.
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Why
\det B=(3)(2)-(6)(1)=6-6=0, so B is not invertible.
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Why
Compute \det B first.