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What each row operation does to a determinant, reduction as the way one is actually computed, the four identities, and the formulas that are worth having but not worth using.
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.
By the end of this lesson you will be able to say what each of the three row operations does to a determinant and why the commonest of them is free, compute a determinant by reducing to triangular form while keeping the record that the swaps and scalings require, use $\det(AB) = \det A \det B$, $\det(A^{T}) = \det A$, $\det(cA) = c^{n}\det A$ and $\det(A^{-1}) = 1/\det A$ rather than multiplying matrices out, apply Cramer's rule where a formula is wanted instead of an algorithm, and say exactly what a zero determinant settles about $Ax = b$ and what it leaves open.
You can already compute a determinant by expanding along a well-chosen row or column, and you know what that costs: $n!$ terms, which is the end of the method somewhere around $5 \times 5$. You also know the one case that is free — a triangular matrix, whose determinant is the product of its diagonal. This lesson closes the gap between those two facts.
The three elementary row operations are a swap of two rows, a scaling of one row by a non-zero number, and a replacement that adds a multiple of one row to another. A matrix is multiplicative under the determinant in the sense that $\det(AB) = \det A \det B$. The adjugate $\operatorname{adj}(A)$ is the transpose of the matrix of cofactors. Cramer's rule names $A_i$, the matrix $A$ with its $i$-th column replaced by $b$.
Elimination turns any matrix into a triangular one. If you know what each step does to the determinant, elimination computes the determinant too — and it costs about $n^3/3$ operations instead of $n!$.
| Operation | Effect on $\det$ |
|---|---|
| swap two rows | multiply by $-1$ |
| scale a row by $c \ne 0$ | multiply by $c$ |
| add a multiple of a row to another | no change |
The third line is the one that matters. It is the operation elimination uses constantly, and it is free, so a reduction with no swaps and no scalings computes the determinant at no extra cost whatever.
Four identities follow, and between them they settle nearly every determinant question that is not a direct computation:
There is no rule for sums. $\det(A + B)$ is not $\det A + \det B$, and the two-by-two case disproves it in one line.
Another way: picture
Think of the rows as edges of a box. Scaling one edge by $c$ scales the volume by $c$ — one edge, one factor, which is exactly why scaling the whole matrix gives $c^n$ and not $c$. Adding a multiple of one edge to another slides the top of the box sideways without lifting it: a shear, which leaves the volume alone. And swapping two edges turns the box inside out, which is what the minus sign records.
Another way: steps
The determinant is linear in each row separately. Holding every other row fixed, replacing row $i$ by $u + v$ gives the sum of the two determinants you would get from $u$ and from $v$. So adding $c$ times row $j$ to row $i$ gives
$$\det(\dots, r_i + c r_j, \dots) = \det(\dots, r_i, \dots) + c \det(\dots, r_j, \dots)$$
and the second determinant has row $j$ appearing twice. A matrix with two equal rows has determinant zero — swap them and the matrix is unchanged while the determinant is negated, so it equals its own negative. The correction term vanishes, and the operation is free.
That one argument is the reason elimination is the method. It also explains the two shortcuts worth looking for before any work is done: a zero row makes the determinant zero (expand along it), and two equal or proportional rows make it zero by the same swap argument. Either fact ends the question immediately.
For an invertible $A$, the solution of $Ax = b$ is
$$x_i = \frac{\det A_i}{\det A},$$
where $A_i$ is $A$ with its $i$-th column replaced by $b$. Which column you replace is the only thing that distinguishes $x_1$ from $x_2$.
The same identity, rearranged, gives $A^{-1} = \dfrac{1}{\det A}\operatorname{adj}(A)$, where the adjugate is the transpose of the matrix of cofactors.
Both are worth knowing and neither is worth using on anything large. Cramer's rule on an $n \times n$ system asks for $n + 1$ determinants; elimination solves the whole system for less than the cost of one of them. What the formulas are for is symbolic work — a system whose coefficients are letters, an inverse that has to be differentiated, a proof that the solution depends continuously on the data. There, a formula is exactly what is wanted and an algorithm is not.
The honest summary: elimination to compute, formulas to reason with.
Scaling one row is not scaling the matrix. If a single row is multiplied by $c$ the determinant gains one factor of $c$; if the whole matrix is, it gains $n$ of them. Writing $\det(3A) = 3\det A$ for a $3 \times 3$ matrix is out by a factor of nine.
Forgetting the sign from the swaps. An odd number of swaps and no correction gives the right magnitude with the wrong sign, and nothing later in the calculation will object. Record each swap the moment it is made.
Dividing when you should multiply, and the reverse. If you scaled a row by $c$ on the way to triangular form, the triangular determinant is $c$ times the one you want, so you divide. Deciding this at the end from memory is how it goes wrong; write the factor down as it happens.
Inventing a rule for sums. Take $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$: both have determinant $0$, and $A + B = I$ has determinant $1$.
Reading a zero determinant as an answer about $b$. It says the solution is not unique. Whether there are none or infinitely many is decided by the augmented column, which the determinant never sees.
$\begin{pmatrix} 0 & 1 & 2 \\ 1 & 0 & 3 \\ 4 & -3 & 8 \end{pmatrix}$: the top-left entry is zero, so swap rows $1$ and $2$ and record one swap.
Record it now, not later.
Clear the first column: row $3$ minus $4$ times row $1$ gives $(0, -3, -4)$. Then row $3$ plus $3$ times row $2$ gives $(0, 0, 2)$. Both are free operations.
Two replacements, no correction.
The diagonal is $1, 1, 2$, so the triangular determinant is $2$. One swap, so the original determinant is $-2$.
Apply the record once, at the end.
$A$ is $3 \times 3$ with $\det A = 5$. Then $\det(A^{T}) = 5$ and $\det(A^{4}) = 5^{4} = 625$.
Transpose and multiplicativity.
$\det(2A) = 2^{3} \times 5 = 40$ — three rows scaled, three factors of two — and $\det(A^{-1}) = 1/5$.
The exponent is the size.
$\det(2A^{-1}A^{T}) = 8 \times \tfrac{1}{5} \times 5 = 8$. The matrix itself never appeared.
Compose the rules.
$\det(A^{3}) = (\det A)^{3} = (-2)^{3} = -8$.
Multiplicativity, three times.
Scaling the whole $2 \times 2$ matrix by $3$ contributes $3^{2} = 9$.
The exponent is the size, not the power.
So $\det(3A^{3}) = 9 \times (-8) = -72$.
Let $A = \begin{pmatrix} 3 & 6 \\ 9 & 3 \end{pmatrix}$. Write the matrix left after subtracting $3$ times row $1$ from row $2$.
This task has no paper form; do it on a device.
$\det A = 6$ and $\det B = 2$. What is $\det(A^{2}B)$?
Answer:
Match each row operation to its effect on the determinant.
| The determinant changes sign | The determinant is multiplied by $5$ | The determinant is unchanged | |
|---|---|---|---|
| Swap rows $2$ and $3$ | |||
| Multiply row $1$ by $5$ | |||
| Add $5$ times row $1$ to row $3$ |
A $3 \times 3$ determinant is being computed by reduction to triangular form. Put the steps in the order they are carried out.
Number the steps in order (write the number in the box):
A square system $Ax = b$ is given, and the coefficient matrix has determinant $6$. How many solutions does it have?
Reducing a $3 \times 3$ matrix $A$ to a triangular matrix $T$ with diagonal $3$, $3$, $10$ took three row swaps and one scaling of row $1$ by $5$; every other step added a multiple of one row to another. Fill in the table to recover $\det A$.
| Value | |
|---|---|
| The product of the diagonal of $T$ | |
| That, divided by the recorded scaling factor | |
| And corrected for the three swaps: $\det A$ |
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Let $A = \begin{pmatrix} 1 & 3 \\ 5 & 4 \end{pmatrix}$. Write the matrix left after subtracting $5$ times row $1$ from row $2$.
This task has no paper form; do it on a device.
You can compute a determinant by reduction with the bookkeeping right, answer questions about products, transposes, scalings and inverses without touching the entries, and use Cramer's rule on a small system. Say in your own words why adding a multiple of one row to another costs nothing. Next: what the number you have been computing actually measures.
10. Your turn: $\det A = -2$ for a $2 \times 2$ matrix. Find $\det(3A^{3})$, step 3
Two rules, one after the other.