Chapter 3 · Determinants

3.2 Properties of Determinants

Row operations change the determinant in three predictable ways — so row reduction computes it quickly — and the determinant respects products: $\det(AB)=\det A\,\det B$, $\det A^{T}=\det A$, and $\det A\ne0$ exactly when $A$ is invertible.

Cofactor expansion is a fine way to define the determinant, but a terrible way to compute it once matrices get large: a $10\times10$ matrix has $10!=3\,628\,800$ products in its full expansion, and a $20\times20$ matrix would keep a fast computer busy for decades (Exercise 3.1 IV.7). Yet in Section 3.1 we saw one kind of matrix whose determinant costs nothing: a triangular matrix, where we just multiply the diagonal.

Row reduction turns any square matrix into a triangular one. If we know exactly how each row operation changes the determinant, we can reduce, multiply the diagonal, and undo the changes. That is the first goal of this section. Along the way the determinant reveals its real power: it tells us whether a matrix is invertible, and it turns matrix products into ordinary products of numbers.

Row Operations and the Determinant

📐 Theorem 3.3 — The effect of row operations

Let $A$ be an $n\times n$ matrix and let $B$ be obtained from $A$ by one elementary row operation.

Interchange two rows ($R_i\leftrightarrow R_j$, $i\ne j$): $\ \det B=-\det A$.
Scale one row by a number $k$ ($R_i\to kR_i$): $\ \det B=k\det A$.
Replace a row by itself plus a multiple of another row ($R_i\to R_i+kR_j$, $i\ne j$): $\ \det B=\det A$.

The same three statements hold for the corresponding column operations (Theorem 3.7 below).

Why these rules hold. Scaling: expand $\det B$ along row $i$; each term $(ka_{ij})C_{ij}$ carries the factor $k$, and the cofactors of row $i$ do not involve row $i$ at all. Interchange: for $2\times2$ matrices, $\begin{vmatrix}c&d\\a&b\end{vmatrix}=cb-da=-(ad-bc)$; for larger matrices expand along a row that was not moved — every minor has two of its rows swapped, so by the smaller case every term changes sign. Replacement: a determinant is a sum of the entries of row $i$ times cofactors that do not depend on row $i$, so splitting row $i$ as $\mathbf{r}_i+k\mathbf{r}_j$ splits the determinant as $\det A+k\det A'$, where $A'$ has row $j$ written in two places. Interchanging those two equal rows changes nothing yet must change the sign, so $\det A'=-\det A'$, i.e. $\det A'=0$.

📐 Theorem 3.4 — When a determinant is zero at sight

Let $A$ be a square matrix. If $A$ has a row of zeros, two equal rows, or one row that is a multiple of another, then $\det A=0$. (The same is true for columns.)

✏️ Example 1 — Tracking the determinant through row operations

Suppose $\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}=4$. Find (a) $\begin{vmatrix}d&e&f\\a&b&c\\g&h&i\end{vmatrix}$ (b) $\begin{vmatrix}2a&2b&2c\\d&e&f\\g&h&i\end{vmatrix}$ (c) $\begin{vmatrix}a&b&c\\d-3a&e-3b&f-3c\\g&h&i\end{vmatrix}$ (d) $\begin{vmatrix}3a&3b&3c\\3d&3e&3f\\3g&3h&3i\end{vmatrix}$ (e) $\begin{vmatrix}a+d&b+e&c+f\\d&e&f\\2g&2h&2i\end{vmatrix}$ (f) $\begin{vmatrix}g&h&i\\a&b&c\\d&e&f\end{vmatrix}$

💡 Show solution▶

(a) One interchange, $R_1\leftrightarrow R_2$: the determinant is $-4$.

(b) Row 1 scaled by $2$: $2\cdot4=8$.

(d) Every row is scaled by $3$, so the factor $3$ comes out three times: $3^3\cdot4=108$ — not $3\cdot4$.

(e) Undo the replacement $R_1\to R_1+R_2$ (no change) and take the $2$ out of row 3: $2\cdot4=8$.

(f) The rows are in the order $3,1,2$. Two interchanges restore the original order ($R_1\leftrightarrow R_2$ gives $1,3,2$; then $R_2\leftrightarrow R_3$ gives $1,2,3$), so the sign changes twice: $(-1)^2\cdot4=4$.

Computing Determinants by Row Reduction

🧭 Problem-solving strategy — determinant by row reduction

Row reduce $A$ to an upper triangular matrix $U$ (an echelon form). Mainly use replacements $R_i\to R_i+kR_j$, which do not change the determinant.
Keep a record: each interchange multiplies the determinant by $-1$; if you take a common factor $k$ out of a row, write $k$ in front.
Multiply the diagonal entries of $U$ (Theorem 3.2) and apply the recorded factors.
If a zero row appears, stop: the determinant is $0$.

In symbols: if $r$ interchanges were used and the factors $k_1,\dots,k_s$ were taken out of rows, then $\det A=(-1)^{r}\,k_1k_2\cdots k_s\,u_{11}u_{22}\cdots u_{nn}$.

✏️ Example 2 — A 3×3 determinant by row reduction

Find $\det A$ for $A=\begin{bmatrix}0&2&4\\1&3&-1\\2&1&5\end{bmatrix}$.

💡 Show solution▶

Swap The top-left entry is $0$, so $R_1\leftrightarrow R_2$: $$\det A=-\begin{vmatrix}1&3&-1\\0&2&4\\2&1&5\end{vmatrix}.$$

Clear $R_3\to R_3-2R_1$ (no change), then take the common factor $2$ out of row 2: $$\det A=-\begin{vmatrix}1&3&-1\\0&2&4\\0&-5&7\end{vmatrix}=-2\begin{vmatrix}1&3&-1\\0&1&2\\0&-5&7\end{vmatrix}.$$

Triangular $R_3\to R_3+5R_2$ (no change): $$\det A=-2\begin{vmatrix}1&3&-1\\0&1&2\\0&0&17\end{vmatrix}=-2(1)(1)(17)=-34.$$

Check Cofactor expansion along row 1: $0-2\begin{vmatrix}1&-1\\2&5\end{vmatrix}+4\begin{vmatrix}1&3\\2&1\end{vmatrix}=-2(7)+4(-5)=-34$. ✓

✏️ Example 3 — A 4×4 determinant by row reduction

Find $\det C$ for $C=\begin{bmatrix}1&2&-1&3\\2&5&1&4\\-1&0&4&2\\3&6&-2&10\end{bmatrix}$.

💡 Show solution▶

Column 1 $R_2\to R_2-2R_1$, $R_3\to R_3+R_1$, $R_4\to R_4-3R_1$ — three replacements, no change: $$\det C=\begin{vmatrix}1&2&-1&3\\0&1&3&-2\\0&2&3&5\\0&0&1&1\end{vmatrix}.$$

Column 2 $R_3\to R_3-2R_2$: $$\det C=\begin{vmatrix}1&2&-1&3\\0&1&3&-2\\0&0&-3&9\\0&0&1&1\end{vmatrix}.$$

Column 3 To avoid fractions, $R_3\leftrightarrow R_4$ (sign changes), then $R_4\to R_4+3R_3$: $$\det C=-\begin{vmatrix}1&2&-1&3\\0&1&3&-2\\0&0&1&1\\0&0&-3&9\end{vmatrix}=-\begin{vmatrix}1&2&-1&3\\0&1&3&-2\\0&0&1&1\\0&0&0&12\end{vmatrix}.$$

Multiply $\det C=-(1)(1)(1)(12)=-12$. About twenty multiplications, against forty for a full cofactor expansion — and the gap grows explosively with $n$ (Exercise 3.1 IV.7).

Apply row operations of your own, or press Play to watch the matrix being reduced to triangular form. The ledger on the right keeps the equation $\det(\text{original})=c\times\det(\text{current})$ true at every step: an interchange flips the sign of $c$, scaling a row by $k$ divides $c$ by $k$, and a replacement leaves $c$ alone. Once the matrix is triangular, multiply its diagonal.

Matrix:

Operation:ijk

last operation—

factor $c$—

$\det(\text{current})$—

$\det(\text{original})$—

Determinants and Invertibility

Row reduction also explains the most important fact about determinants. Each row operation multiplies the determinant by a non-zero number ($-1$, $k\ne0$, or $1$). So the determinant of $A$ is zero exactly when the determinant of its reduced row echelon form $R$ is zero. If $A$ is invertible, $R=I$ and $\det R=1$; if not, $R$ has a zero row and $\det R=0$.

📐 Theorem 3.5 — The determinant test for invertibility

A square matrix $A$ is invertible if and only if $\det A\ne0$.

📐 Theorem 3.6 — The Invertible Matrix Theorem (continued)

Let $A$ be an $n\times n$ matrix. The following statements are equivalent — either all are true or all are false:

$A$ is invertible.
$A$ is row equivalent to $I_n$.
$A$ has $n$ pivot positions.
The equation $A\mathbf{x}=\mathbf{0}$ has only the trivial solution $\mathbf{x}=\mathbf{0}$.
The equation $A\mathbf{x}=\mathbf{b}$ has at least one solution for every $\mathbf{b}$ in $\mathbb{R}^n$.
The equation $A\mathbf{x}=\mathbf{b}$ has exactly one solution for every $\mathbf{b}$ in $\mathbb{R}^n$.
$A$ is a product of elementary matrices.
There is an $n\times n$ matrix $C$ with $CA=I$.
There is an $n\times n$ matrix $D$ with $AD=I$.
$A^{T}$ is invertible.
$\det A\ne0$.

Statements (a)–(j) are the Invertible Matrix Theorem of Section 2.2, with the same labels; (k) is new, and it is equivalent to (a) by Theorem 3.5. More equivalent statements will join the list in later chapters.

✏️ Example 4 — For which values is a matrix invertible?

For which values of $k$ is $M=\begin{bmatrix}1&1&1\\1&k&3\\1&3&k\end{bmatrix}$ invertible?

💡 Show solution▶

Reduce $R_2\to R_2-R_1$, $R_3\to R_3-R_1$ (no change): $$\det M=\begin{vmatrix}1&1&1\\0&k-1&2\\0&2&k-1\end{vmatrix}=1\cdot\begin{vmatrix}k-1&2\\2&k-1\end{vmatrix}=(k-1)^2-4,$$ expanding along column 1.

Factor $(k-1)^2-4=(k-1-2)(k-1+2)=(k-3)(k+1)$.

Answer By Theorem 3.5, $M$ is invertible for every $k$ except $k=3$ and $k=-1$.

Check At $k=3$, rows 2 and 3 are both $(1,3,3)$ — equal rows, determinant $0$. At $k=-1$, $R_2+R_3=(2,2,2)=2R_1$, so the rows are dependent. ✓

Products, Transposes, Inverses and Scalar Multiples

📐 Theorem 3.7 — Determinant of a product and of a transpose

Let $A$ and $B$ be $n\times n$ matrices. Then $$\det(AB)=(\det A)(\det B)\qquad\text{and}\qquad\det A^{T}=\det A .$$ Because rows of $A^{T}$ are columns of $A$, every statement about rows (Theorems 3.3 and 3.4) is also true for columns.

Why. For an elementary matrix $E$, Theorem 3.3 says exactly that $\det(EB)=(\det E)(\det B)$ (with $\det E=-1$, $k$ or $1$). If $A$ is invertible it is a product of elementary matrices $A=E_1E_2\cdots E_m$, and peeling them off one at a time gives $$\det(AB)=\det E_1\cdots\det E_m\,\det B=(\det A)(\det B).$$ If $A$ is not invertible, neither is $AB$, and both sides are $0$. For the transpose, expanding $\det A^{T}$ along its first row is expanding $\det A$ down its first column, which gives the same number by Theorem 3.1.

📐 Theorem 3.8 — Inverses, powers and scalar multiples

Let $A$ be an $n\times n$ matrix and $k$ a number.

If $A$ is invertible, $\det(A^{-1})=\dfrac{1}{\det A}$. (Take determinants of $AA^{-1}=I$.)
$\det(A^{m})=(\det A)^{m}$ for every positive integer $m$.
$\det(kA)=k^{n}\det A$ — the factor $k$ comes out of each of the $n$ rows.

⚠️ Common mistakes with determinant rules

1. $\det(A+B)\ne\det A+\det B$ in general. With $A=\begin{bmatrix}2&1\\1&3\end{bmatrix}$ and $B=\begin{bmatrix}1&-2\\4&0\end{bmatrix}$: $\det A=5$, $\det B=8$, but $\det(A+B)=\begin{vmatrix}3&-1\\5&3\end{vmatrix}=14\ne13$. Products work; sums do not.

2. $\det(kA)=k^n\det A$, not $k\det A$. Doubling a $3\times3$ matrix multiplies its determinant by $8$.

3. $\det(AB)=\det(BA)$ even when $AB\ne BA$. For the matrices above, $AB=\begin{bmatrix}6&-4\\13&-2\end{bmatrix}$ and $BA=\begin{bmatrix}0&-5\\8&4\end{bmatrix}$ are different, yet both have determinant $40=5\cdot8$.

4. $\det A=0$ does not mean $A$ has a zero row or two equal rows — only that its rows (and columns) are linearly dependent, as in Example 4 with $k=-1$.

✏️ Example 5 — Using the rules without the matrices

$A$ and $B$ are $4\times4$ matrices with $\det A=3$ and $\det B=-2$. Find (a) $\det(AB)$; (b) $\det(2A)$; (c) $\det(A^{T}B^{-1})$; (d) $\det(A^{3})$; (e) $\det(-A)$; (f) $\det(BAB^{-1})$.

💡 Show solution▶

(a), (b) $\det(AB)=3(-2)=-6$. $\ \det(2A)=2^4\cdot3=48$ (since $n=4$).

(c) $\det(A^{T}B^{-1})=\det A^{T}\cdot\det B^{-1}=3\cdot\dfrac{1}{-2}=-\dfrac32$. ($B$ is invertible because $\det B\ne0$.)

(d), (e) $\det(A^3)=3^3=27$. $\ \det(-A)=(-1)^4\det A=3$ — for an even size, $-A$ has the same determinant as $A$.

(f) $\det(BAB^{-1})=\det B\cdot\det A\cdot\dfrac{1}{\det B}=\det A=3$. Numbers commute even when matrices do not.

First $B$:then $A$:

$\det B$—

$\det A$—

$\det(AB)$ from the product matrix—

$\det A\cdot\det B$—

✏️ Example 6 — A determinant that factors: the Vandermonde matrix

Show that $\begin{vmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{vmatrix}=(b-a)(c-a)(c-b)$, and deduce that through any three points $(a,y_1)$, $(b,y_2)$, $(c,y_3)$ with distinct $a,b,c$ there passes exactly one parabola (or line) $y=p+qx+rx^2$.

💡 Show solution▶

Reduce $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$ (no change), using $b^2-a^2=(b-a)(b+a)$: $$\begin{vmatrix}1&a&a^2\\0&b-a&(b-a)(b+a)\\0&c-a&(c-a)(c+a)\end{vmatrix}.$$

Factor Take $(b-a)$ out of row 2 and $(c-a)$ out of row 3: $$(b-a)(c-a)\begin{vmatrix}1&a&a^2\\0&1&b+a\\0&1&c+a\end{vmatrix}.$$

Finish $R_3\to R_3-R_2$ gives the triangular matrix with diagonal $1,1,c-b$. So the determinant is $(b-a)(c-a)(c-b)$.

Parabola The conditions $p+qa+ra^2=y_1$, $p+qb+rb^2=y_2$, $p+qc+rc^2=y_3$ form a system whose coefficient matrix is this Vandermonde matrix. If $a,b,c$ are distinct, its determinant is non-zero, so by Theorem 3.6 there is exactly one solution $(p,q,r)$.

Exercises

Given $\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}=-6$, find each determinant without expanding.

1 $\begin{vmatrix}d&e&f\\a&b&c\\g&h&i\end{vmatrix}$
2 $\begin{vmatrix}a&b&c\\d&e&f\\4g&4h&4i\end{vmatrix}$
3 $\begin{vmatrix}a&b&c\\d+2a&e+2b&f+2c\\g&h&i\end{vmatrix}$
4 $\begin{vmatrix}-a&-b&-c\\-d&-e&-f\\-g&-h&-i\end{vmatrix}$
5 $\begin{vmatrix}a&d&g\\b&e&h\\c&f&i\end{vmatrix}$
6 $\begin{vmatrix}g&h&i\\d&e&f\\a&b&c\end{vmatrix}$
7 $\begin{vmatrix}a&b&c\\2d&2e&2f\\3g&3h&3i\end{vmatrix}$
8 $\begin{vmatrix}a+g&b+h&c+i\\d&e&f\\g-a&h-b&i-c\end{vmatrix}$

💡 Show solutions▶

1 One interchange: $6$.

2 Row 3 scaled by $4$: $4(-6)=-24$.

3 Replacement $R_2\to R_2+2R_1$: unchanged, $-6$.

4 This is $\det(-A)$ with $n=3$: $(-1)^3(-6)=6$.

5 This is $\det A^{T}=\det A=-6$.

6 One interchange $R_1\leftrightarrow R_3$: $6$.

7 Factors $2$ and $3$ from rows 2 and 3: $6(-6)=-36$.

8 $R_3\to R_3+R_1$ (no change) turns row 3 into $(2g,2h,2i)$; take out the $2$; then $R_1\to R_1-R_3$ (no change) leaves $(a,b,c)$. Result: $2(-6)=-12$.

Evaluate by row reduction to triangular form. Record every interchange and every factor you take out.

1 $\begin{vmatrix}1&2&3\\2&5&8\\3&7&12\end{vmatrix}$
2 $\begin{vmatrix}0&3&1\\1&2&-2\\4&1&6\end{vmatrix}$
3 $\begin{vmatrix}2&-4&6\\1&3&5\\-1&2&4\end{vmatrix}$
4 $\begin{vmatrix}3&6&9\\1&4&2\\2&1&5\end{vmatrix}$
5 $\begin{vmatrix}1&0&2&1\\2&1&3&0\\-1&2&0&4\\0&1&1&3\end{vmatrix}$
6 $\begin{vmatrix}2&4&-2&6\\1&3&0&2\\3&5&1&7\\1&1&1&4\end{vmatrix}$
7 $\begin{vmatrix}1&2&3&4\\0&1&5&2\\2&4&6&8\\3&1&0&7\end{vmatrix}$
8 $\begin{vmatrix}1&1&1\\1&2&4\\1&4&16\end{vmatrix}$ (then check with Example 6)

Check: type a square matrix (rows separated by ";") to see one possible reduction and the determinant

Matrix:

interchanges—

diagonal of the triangular form—

$\det$—

💡 Show solutions▶

1 $R_2\to R_2-2R_1$, $R_3\to R_3-3R_1$: rows $(0,1,2)$, $(0,1,3)$; then $R_3\to R_3-R_2$: $(0,0,1)$. Diagonal $1,1,1$: $\det=1$.

2 $R_1\leftrightarrow R_2$ (sign $-$). $R_3\to R_3-4R_1$: $(0,-7,14)=7(0,-1,2)$ (factor $7$). $R_2\leftrightarrow R_3$ (sign $-$) gives rows $(1,2,-2)$, $(0,-1,2)$, $(0,3,1)$; $R_3\to R_3+3R_2$: $(0,0,7)$. $\det=(-1)(7)(-1)\cdot\big[(1)(-1)(7)\big]=-49$.

3 Take $2$ out of row 1: $(1,-2,3)$. $R_2\to R_2-R_1$: $(0,5,2)$; $R_3\to R_3+R_1$: $(0,0,7)$. $\det=2(1)(5)(7)=70$.

4 Take $3$ out of row 1: $(1,2,3)$. $R_2\to R_2-R_1$: $(0,2,-1)$; $R_3\to R_3-2R_1$: $(0,-3,-1)$; $R_3\to R_3+\tfrac32R_2$: $\left(0,0,-\tfrac52\right)$. $\det=3(1)(2)\left(-\tfrac52\right)=-15$.

5 $R_2\to R_2-2R_1$: $(0,1,-1,-2)$; $R_3\to R_3+R_1$: $(0,2,2,5)$. $R_3\to R_3-2R_2$: $(0,0,4,9)$; $R_4\to R_4-R_2$: $(0,0,2,5)$. $R_4\to R_4-\tfrac12R_3$: $\left(0,0,0,\tfrac12\right)$. $\det=(1)(1)(4)\left(\tfrac12\right)=2$.

6 Take $2$ out of row 1: $(1,2,-1,3)$. $R_2-R_1$: $(0,1,1,-1)$; $R_3-3R_1$: $(0,-1,4,-2)$; $R_4-R_1$: $(0,-1,2,1)$. $R_3+R_2$: $(0,0,5,-3)$; $R_4+R_2$: $(0,0,3,0)$. $R_3\leftrightarrow R_4$ (sign $-$), then $R_4\to R_4-\tfrac53R_3$: $(0,0,0,-3)$. $\det=2\cdot(-1)\cdot\big[(1)(1)(3)(-3)\big]=18$.

7 Row 3 is $2\times$ row 1 (Theorem 3.4), so $\det=0$ — no reduction needed.

8 $R_2-R_1$: $(0,1,3)$; $R_3-R_1$: $(0,3,15)$; $R_3-3R_2$: $(0,0,6)$. $\det=6$. Example 6 with $a=1$, $b=2$, $c=4$: $(2-1)(4-1)(4-2)=6$. ✓

III

$A$ and $B$ are $3\times3$ matrices with $\det A=2$ and $\det B=-3$. Find:

1 $\det(AB)$
2 $\det(A^{T})$
3 $\det(A^{-1})$
4 $\det(3A)$
5 $\det(B^{2}A^{-1})$
6 $\det(B^{T}A^{2})$
7 $\det(-B)$
8 $\det(A^{-1}BA)$
9 $\det(2A^{-1})$

💡 Show solutions▶

1 $2(-3)=-6$.

2 $\det A^{T}=\det A=2$.

3 $\dfrac12$.

4 $3^3\cdot2=54$.

5 $(-3)^2\cdot\dfrac12=\dfrac92$.

6 $(-3)\cdot2^2=-12$.

7 $(-1)^3(-3)=3$.

8 $\dfrac12\cdot(-3)\cdot2=-3$.

9 $2A^{-1}$ is the $3\times3$ matrix $A^{-1}$ scaled by $2$: $2^3\cdot\dfrac12=4$.

Invertibility, proofs and an application.

1 For which $k$ is $\begin{bmatrix}k&1&0\\1&k&1\\0&1&k\end{bmatrix}$ not invertible?
2 For which $t$ is $\begin{bmatrix}1&t&0\\2&1&t\\0&2&1\end{bmatrix}$ invertible?
3 A matrix is skew-symmetric if $A^{T}=-A$. Prove that every $3\times3$ skew-symmetric matrix has $\det A=0$. Does the same argument work for $2\times2$?
4 Prove that if $Q^{T}Q=I$ then $\det Q=1$ or $\det Q=-1$.
5 A matrix with $A^{2}=A$ is called idempotent. Show that $\det A$ is $0$ or $1$.
6 A matrix with $A^{k}=O$ for some positive integer $k$ is called nilpotent. Show that a nilpotent matrix is never invertible.
7 Nutrition. Three foods contain, per serving, these grams of (protein, fat, carbohydrate): food 1 $(10,2,5)$, food 2 $(4,6,12)$, food 3 $(6,4,8)$. (a) Use a determinant to decide whether every target (protein, fat, carbohydrate) can be met by exactly one combination of servings $(x_1,x_2,x_3)$. (b) Food 3 is replaced by a blend that is half food 1 and half food 2. Repeat (a) and interpret.
8 True or false? Give a reason or a counterexample. (a) $\det(A+B)=\det A+\det B$. (b) $\det(AB)=\det(BA)$ for square $A$, $B$ of the same size. (c) If $\det A=0$ then $A$ has a row of zeros. (d) If $A$ is $n\times n$ and invertible, $\det(A^{-1}A^{T})=1$.

💡 Show solutions▶

1 Expanding along row 1: $k(k^2-1)-1(k-0)=k^3-2k=k\left(k^2-2\right)$. Not invertible exactly when $k=0$ or $k=\pm\sqrt2$.

2 Along row 1: $1(1-2t)-t(2-0)+0=1-4t$. Invertible exactly when $t\ne\tfrac14$.

3 $\det A=\det A^{T}=\det(-A)=(-1)^3\det A=-\det A$, so $2\det A=0$ and $\det A=0$. For $2\times2$ the factor is $(-1)^2=+1$ and the argument proves nothing; indeed $\begin{bmatrix}0&1\\-1&0\end{bmatrix}$ is skew-symmetric with determinant $1$.

4 $1=\det I=\det(Q^{T}Q)=\det Q^{T}\det Q=(\det Q)^2$, so $\det Q=\pm1$. (Such matrices are rotations and reflections — Chapter 8.)

5 $\det A=\det(A^2)=(\det A)^2$, so $\det A\left(\det A-1\right)=0$: $\det A=0$ or $1$.

6 $(\det A)^k=\det(A^k)=\det O=0$, so $\det A=0$ and $A$ is not invertible (Theorem 3.5).

7 The servings must solve $N\mathbf{x}=\mathbf{t}$, where the columns of $N$ are the foods. (a) $\det\begin{bmatrix}10&4&6\\2&6&4\\5&12&8\end{bmatrix}=-20\ne0$, so every target corresponds to exactly one $(x_1,x_2,x_3)$ (though some targets may need negative servings, which are not practical). (b) The blend is $\tfrac12(10,2,5)+\tfrac12(4,6,12)=(7,4,8.5)$, and $\det\begin{bmatrix}10&4&7\\2&6&4\\5&12&8.5\end{bmatrix}=0$ because column 3 is $\tfrac12$(column 1 + column 2). The third food adds nothing new: some targets cannot be met at all, and those that can be met have infinitely many combinations.

8 (a) False $A=B=I_2$: $4\ne2$. (b) True both equal $\det A\det B$. (c) False $\begin{bmatrix}1&2\\2&4\end{bmatrix}$ has determinant $0$ and no zero row. (d) True $\dfrac{1}{\det A}\cdot\det A=1$.

🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. Is a small determinant a reliable warning that a linear system will be hard to solve accurately?

Why it matters. Surveyors fixing a position from GPS ranges, engineers computing forces in a bridge truss and economists fitting models all solve systems $A\mathbf{x}=\mathbf{b}$ whose data contain measurement errors. A matrix that is "nearly singular" can turn a tiny error in $\mathbf{b}$ into a huge error in $\mathbf{x}$. It is tempting to measure "nearly singular" by $\det A$ — but is that right?

What to do.

Let $A=\begin{bmatrix}1&1\\1&1.0001\end{bmatrix}$. Find $\det A$. Solve $A\mathbf{x}=\begin{bmatrix}2\\2.0001\end{bmatrix}$ and then $A\mathbf{x}=\begin{bmatrix}2\\2.0002\end{bmatrix}$. How much did $\mathbf{x}$ change when $\mathbf{b}$ changed by $0.0001$?
Now let $D=0.1\,I_{10}$ (the $10\times10$ identity times $0.1$). Use Theorem 3.8 to find $\det D$. Solve $D\mathbf{x}=\mathbf{b}$ by hand for any $\mathbf{b}$ — is it hard or sensitive?
Multiply every entry of $A$ from step 1 by $1000$. What happens to the determinant, and to the sensitivity of the solution?
Look up the condition number of a matrix. Compute it for $A$ and $D$ with an online calculator or Python (numpy.linalg.cond) and compare with the determinants.

Tools: Calculator, spreadsheet or Python with NumPy.

Hand in: A half-page note answering the research question, using your two examples as evidence.

Going further. Find out how a GPS receiver's "dilution of precision" (DOP) number is computed and how it relates to the geometry of the satellites in the sky.

Instructor note: $\det A=0.0001$; the solutions are $(1,1)$ and $(0,2)$ — a change of $0.0001$ in $\mathbf{b}$ moves $\mathbf{x}$ by about $1.4$. $\det D=10^{-10}$, yet $\mathbf{x}=10\mathbf{b}$ is perfectly stable (condition number $1$). Scaling $A$ by $1000$ multiplies $\det A$ by $10^6$ (to $100$) without changing the sensitivity at all (condition number $\approx4\times10^{4}$ in both cases). Conclusion: the size of $\det A$ depends on scale and is not a measure of sensitivity; the condition number is.

🧪 Self-check quiz

0/5 correct

Q1. Interchanging two rows of a square matrix multiplies its determinant by

Q2. $A$ is $3\times3$ with $\det A=5$. Then $\det(2A)=$

Q3. If $\det A=4$, then $\det(A^{-1})=$

Q4. Which row operation leaves the determinant unchanged?

Q5. $A$ is $n\times n$ with $\det A=0$. Which statement must be true?

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