Linear-Algebra-Done-Right · Unit 29 · Video 3 · Interactive Practice

det Aᵀ = det A: Row Operations and Gaussian Elimination

IKey Formulas

Formula	Name	Consequence
$\det A^{\mathrm{t}} = \det A$	Transpose ( $A$ square)	$\det$ is alternating and $n$ -linear in the rows as well as the columns
$\det T' = \det T, \quad \det T^* = \overline{\det T}$	Dual map; adjoint (inner product space)	Matrices $A^{\mathrm{t}}$ and $\overline{A}^{\mathrm{t}}$ in suitable bases
swap $\to \times(-1)$ , scale a row by $c \to \times c$ , add a multiple of a row $\to \times 1$	Row operations	Two equal rows give $\det = 0$

Key Insight: Elimination reaches an upper-triangular matrix, whose determinant is the product of its diagonal; keeping a ledger of the swaps and scalings recovers $\det A$ .

IIElimination with a Ledger

Each row operation multiplies the determinant by a known factor, so the triangular end state gives $\det M$ .

Step 0

IIIRows and Columns Span Equal Areas

The parallelogram of the columns and the parallelogram of the rows of a $2$ -by- $2$ matrix always have the same signed area.

IVQuiz Questions

Problem 1 · Two Operations in a Row

Given: a $3$ -by- $3$ matrix $A$ with $\det A = 4$ . The matrix $B$ is obtained by swapping rows $1$ and $3$ of $A$ and then multiplying row $2$ by $5$ . Find $\det B$ .

✅ Correct! The swap contributes

-1

and the scaling

5

4 \cdot (-1) \cdot 5 = -20

❌ Not quite. Swapping two rows flips the sign of the determinant.

❌ Not quite. Multiplying one row by

5

multiplies the determinant by

5

❌ Not quite. Only one row was multiplied by

5

, so the factor is

5

, not

5^3

Show solution

Because $\det$ is alternating and multilinear in the rows, a swap multiplies it by $-1$ and scaling one row by $5$ multiplies it by $5$ :

\det B = 5 \cdot (-1) \cdot \det A = -20.

Problem 2 · A Row Replacement That Is Not Just Adding

Given: a $3$ -by- $3$ matrix $A$ with $\det A = 6$ . The matrix $B$ is obtained by replacing row $1$ of $A$ with $2 \cdot (\text{row } 1) + (\text{row } 2)$ ; rows $2$ and $3$ are unchanged. Find $\det B$ .

✅ Correct! Linearity in row

1

gives

2\det A + \det(\text{matrix with two equal rows}) = 12 + 0

❌ Not quite. Adding a multiple of row

2

is harmless, but row

1

itself was also doubled.

❌ Not quite. The added copy of row

2

contributes a determinant with two equal rows, which is

0

, not

6

❌ Not quite. Doubling a row multiplies the determinant by

2

; it does not divide it.

Show solution

Write the rows of $A$ as $r_1, r_2, r_3$ . The determinant is linear in the first row:

\det(2r_1 + r_2,\ r_2,\ r_3) = 2\det(r_1, r_2, r_3) + \det(r_2, r_2, r_3) = 2 \cdot 6 + 0 = 12.

The second determinant is $0$ because it has two equal rows.

Problem 3 · Eliminate a 4-by-4

Find $\det \begin{pmatrix} 0 & 1 & 2 & 1 \\ 1 & 2 & 1 & 0 \\ 2 & 4 & 3 & 1 \\ 1 & 3 & 3 & 4 \end{pmatrix}$ by row operations.

✅ Correct! One swap, then only row additions, ending at the diagonal

1, 1, 1, 3

❌ Almost. The first step must swap rows

1

and

2

to get a nonzero pivot, and that swap flips the sign.

❌ Not quite. Track each operation: one swap, then only additions of multiples of rows.

Show solution

Swap rows $1$ and $2$ (factor $-1$ ). Then subtract $2\cdot$ row $1$ from row $3$ and row $1$ from row $4$ , and finally subtract row $2$ from row $4$ (no change):

\begin{pmatrix} 1 & 2 & 1 & 0 \\ 0 & 1 & 2 & 1 \\ 2 & 4 & 3 & 1 \\ 1 & 3 & 3 & 4 \end{pmatrix} \to \begin{pmatrix} 1 & 2 & 1 & 0 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 1 & 2 & 4 \end{pmatrix} \to \begin{pmatrix} 1 & 2 & 1 & 0 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 3 \end{pmatrix}.

The last matrix is upper triangular with determinant $1 \cdot 1 \cdot 1 \cdot 3 = 3$ , so the original determinant is $(-1) \cdot 3 = -3$ .

Problem 4 · The Adjoint Conjugates

Given: $T \in \mathcal{L}(\mathbf{C}^3)$ , with the standard inner product, and $\det T = 1 - i$ . Find $\det(T^*T)$ .

✅ Correct!

\det(T^*T) = \overline{\det T}\,\det T = (1 + i)(1 - i) = 2

❌ Not quite.

(1 - i)^2 = -2i

treats

\det T^*

\det T

; the adjoint conjugates the determinant.

❌ Not quite.

\sqrt{2} = |\det T|

; the product

\overline{\det T}\,\det T

|\det T|^2

❌ Not quite.

1 + i

\det T^*

alone; multiply by

\det T

as well.

Show solution

With respect to the standard (orthonormal) basis, $\mathcal{M}(T^*)$ is the conjugate transpose of $\mathcal{M}(T)$ , so $\det T^* = \overline{\det T} = 1 + i$ . By the product rule,

\det(T^*T) = \det T^* \det T = (1 + i)(1 - i) = 1 - i^2 = 2.

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