Cramer's rule

WHAT is Cramer's Rule?

The crucial condition: it only works when $==\det A \neq 0==$ (the matrix is invertible, so a unique solution exists).

WHY does it work? (Derivation from scratch)

We derive it using one fact: the determinant is multilinear and alternating in its columns, i.e. $\det$ behaves like a "signed volume" that:

scales when you scale a column,
adds when you split a column into a sum,
vanishes if two columns are equal.

Step 1 — Build a clever matrix. Take the identity matrix $I$ and replace its $i$ -th column with the unknown vector $\mathbf{x}$ . Call it $X_i$ . For example with $n=3$ , $i=2$ : $X_2=\begin{pmatrix}1 & x_1 & 0\\ 0 & x_2 & 0\\ 0 & x_3 & 1\end{pmatrix}.$ Why this step? This matrix is upper/lower-mixed but expanding along the $i$ -th row/column gives $\det X_i = x_i$ cheaply.

Step 2 — Compute $\det X_i$ . Expanding the determinant along the special structure, every column except the $i$ -th is a standard basis vector, so $\det X_i = x_i.$ Why this step? The identity columns contribute factor $1$ ; only the entry $x_i$ on the diagonal survives.

Step 3 — Multiply by $A$ . Consider the product $A\,X_i$ . Multiplying $A$ by a matrix multiplies $A$ by each column:

A standard-basis column $\mathbf{e}_k$ picks out column $k$ of $A$ , i.e. $\mathbf{a}_k$ .
The $i$ -th column $\mathbf{x}$ gives $A\mathbf{x}=\mathbf{b}$ .

So $A X_i$ is exactly $A$ with its $i$ -th column replaced by $\mathbf{b}$ — that is $==A_i==$ : $A X_i = A_i.$ Why this step? This is the heart: matrix multiplication swaps the column for us automatically.

Step 4 — Take determinants and use multiplicativity ( $\det(MN)=\det M\det N$ ): $\det A \cdot \det X_i = \det A_i \;\Longrightarrow\; \det A \cdot x_i = \det A_i.$

Step 5 — Solve. If $\det A\neq 0$ : $\boxed{\,x_i = \dfrac{\det A_i}{\det A}\,}$

HOW to use it — worked examples

Worked example 2×2 system

Solve $\begin{cases} 2x + y = 5 \\ x - 3y = -1\end{cases}$ .

Here $A=\begin{pmatrix}2&1\\1&-3\end{pmatrix}$ , $\mathbf{b}=\begin{pmatrix}5\\-1\end{pmatrix}$ .

$\det A = (2)(-3)-(1)(1)= -7.$ Why? Need denominator first; $\neq 0$ so unique solution exists.

$x = \dfrac{\det A_1}{\det A}$ , $A_1=\begin{pmatrix}5&1\\-1&-3\end{pmatrix}$ , $\det A_1 = (5)(-3)-(1)(-1)=-14$ . So $x = \dfrac{-14}{-7}=2$ . Why? Replace column 1 (the $x$ -column) by $\mathbf{b}$ .

$y = \dfrac{\det A_2}{\det A}$ , $A_2=\begin{pmatrix}2&5\\1&-1\end{pmatrix}$ , $\det A_2 = (2)(-1)-(5)(1)=-7$ . So $y = \dfrac{-7}{-7}=1$ . Why? Replace column 2 (the $y$ -column) by $\mathbf{b}$ .

Check: $2(2)+1=5$ ✓, $2-3(1)=-1$ ✓.

Worked example 3×3 system

Solve $\begin{cases} x+y+z=6\\ 2y+5z=-4\\ 2x+5y-z=27\end{cases}$ .

$A=\begin{pmatrix}1&1&1\\0&2&5\\2&5&-1\end{pmatrix}$ , $\mathbf{b}=\begin{pmatrix}6\\-4\\27\end{pmatrix}$ .

$\det A$ : expand along row 1: $1(2\cdot(-1)-5\cdot5) -1(0\cdot(-1)-5\cdot2)+1(0\cdot5-2\cdot2)$ $= 1(-27)-1(-10)+1(-4)= -21$ . Why? Cofactor expansion is just the volume of the column-box.

$x=\det A_1/\det A$ : $A_1=\begin{pmatrix}6&1&1\\-4&2&5\\27&5&-1\end{pmatrix}$ . Expanding: $\det A_1 = -105$ , so $x = -105/-21 = 5$ .

$y$ : $A_2=\begin{pmatrix}1&6&1\\0&-4&5\\2&27&-1\end{pmatrix}$ , $\det A_2 = -63$ , $y = -63/-21 = 3$ .

$z$ : $A_3=\begin{pmatrix}1&1&6\\0&2&-4\\2&5&27\end{pmatrix}$ , $\det A_3 = 42$ , $z = 42/-21 = -2$ .

Check: $5+3+(-2)=6$ ✓, $2(3)+5(-2)=6-10=-4$ ✓, $2(5)+5(3)-(-2)=10+15+2=27$ ✓. Solution $(x,y,z)=(5,3,-2)$ .

Forecast-then-Verify

Common mistakes

Recall Feynman: explain to a 12-year-old

Imagine a recipe: to make a smoothie ( $\mathbf{b}$ ) you mix some banana, apple and milk (the columns of $A$ ). Cramer's rule tells you exactly how many bananas you need. The trick: measure the "size" of the original ingredient-box (that's $\det A$ ), then make a new box where you swap the banana ingredient for the whole smoothie and measure that box ( $\det A_1$ ). Divide the two sizes — that ratio is exactly the amount of banana! Do the same for each ingredient. If the original box is flat (size $0$ ), the ingredients are "stuck together" and there's no single recipe.

Flashcards

What does Cramer's rule compute, and when is it valid?

x_i=\det A_i/\det A

, valid only when

\det A\neq 0

(unique solution).

What is

A_i

in Cramer's rule?

A

with its

i

-th column replaced by the vector

\mathbf{b}

Key identity that proves Cramer's rule

A X_i = A_i

and

\det X_i = x_i

, so

\det A\cdot x_i=\det A_i

Why must

\det A\neq 0

Otherwise you divide by zero; the system has no unique solution.

What property of

\det

gives

\det X_i = x_i

Multilinearity + the identity columns being basis vectors.

Doubling

\mathbf{b}

does what to the solution?

Doubles every

x_i

(columns of

A_i

are linear in

\mathbf{b}

Why is Cramer's rule bad for large numerical systems?

Cost grows like

(n+1)

determinants and dividing by tiny

\det A

is unstable; Gaussian elimination is faster/stabler.

Common error to avoid

Replacing a row instead of a column with

\mathbf{b}

Connections

Determinants — the engine; Cramer is "determinant as volume."
Matrix Inverse — Cramer ⇔ $A^{-1}=\frac{1}{\det A}\operatorname{adj}(A)$ applied to $\mathbf{b}$ .
Cofactor Expansion — how the $\det A_i$ are computed.
Gaussian Elimination — the practical alternative for big systems.
Multilinear and Alternating Maps — properties that make the proof work.
Invertible Matrix Theorem — $\det A\neq 0 \iff$ unique solution.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Cramer's rule ka core idea bahut simple hai: jab tumhe $A\mathbf{x}=\mathbf{b}$ solve karna ho aur $\det A \neq 0$ ho, to har ek unknown $x_i$ nikaalne ke liye sirf do determinant chahiye. Pehle original matrix $A$ ka determinant nikaalo (yeh denominator hai). Phir $A$ ki $i$ -th column ko $\mathbf{b}$ se replace kar do, us naye matrix $A_i$ ka determinant nikaalo (yeh numerator hai). Bas, $x_i = \det A_i / \det A$ .

Iske peeche ka magic yeh hai ki determinant ek tarah ka "volume" (ya 2D me area) measure karta hai. Jab tum ek column ko $\mathbf{b}$ se badalte ho, to volume exactly utne factor se change hota hai jitna woh unknown ki value hai. Proof me hum ek special matrix $X_i$ banate hain (identity ki $i$ -th column ki jagah $\mathbf{x}$ ), aur $A X_i = A_i$ ban jaata hai — bas determinant le lo dono taraf, ho gaya derive.

Dhyaan rakhna: yeh sirf tab kaam karta hai jab $\det A \neq 0$ . Agar zero hai to ya to koi solution nahi ya infinitely many — tab Gaussian elimination use karo. Aur exam me sabse common galti: row replace mat karna, hamesha column replace karni hai $\mathbf{b}$ se. Bade systems ke liye Cramer slow hota hai, lekin chote $2\times2$ , $3\times3$ aur theory/proofs ke liye yeh bahut elegant aur clean tareeka hai.

WHAT is Cramer's Rule?

WHY does it work? (Derivation from scratch)

HOW to use it — worked examples

Forecast-then-Verify

Common mistakes

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Linear Algebra (Full)

Connections