LINEAR-ALGEBRA · Unit 3 · Video 3 · Interactive Practice

Finding Matrix Inverses Using Gauss-Jordan Elimination

IKey Formulas

Formula	Name	Description
$[A \mid I] \xrightarrow{\text{row ops}} [I \mid A^{-1}]$	Gauss-Jordan Method	Append identity, row-reduce, read off inverse
$E_k \cdots E_1 A = I$	Elementary Matrices	Each row operation equals left-multiplication by an elementary matrix
$A^{-1} = E_k \cdots E_1$	Inverse as Product	The inverse accumulates all row operations applied to $I$
$AA^{-1} = A^{-1}A = I$	Verification	Multiply to confirm the result is correct

IIVisualization 1 — Building A⁻¹ by Row Reduction

As the left block reduces to $I$ , every row operation is mirrored on the right block — which records them and becomes $A^{-1}$ .

Stage 0

augmented matrix [A | I]

Stage 0 · Setup [A | I]

Pivot

a_{11} = 1

. Eliminate the

2

below it:

R_2 \leftarrow R_2 - 2R_1

Stage 1 · Gauss (forward)

R_2 \leftarrow R_2 - 2R_1

done. Pivot

a_{22} = 1

; eliminate the

3

above it:

R_1 \leftarrow R_1 - 3R_2

Stage 2 · Jordan (backward)

Left half

= I

. Right half

= A^{-1} = \begin{bmatrix} 7 & -3 \\ -2 & 1 \end{bmatrix}

IIIVisualization 2 — The Inverse as Determinant → 0

How does $A^{-1}$ behave as the determinant $ad - bc$ shrinks toward zero?

a 1 b 3 c 2 d 7

💡 When one row is a scalar multiple of the other, the rows are linearly dependent, forcing $\det = 0$ — and no inverse exists.

IVVisualization 3 — Verifying A·A⁻¹ = I

Each entry of $A\,A^{-1}$ is a row of $A$ dotted with a column of $A^{-1}$ : the diagonal gives $1$ , off-diagonal gives $0$ .

Product entry

VQuiz Questions

Question 1

You are performing Gauss-Jordan elimination on $[A \mid I]$ where $A = \begin{bmatrix} 1 & 3 \\ 2 & 7 \end{bmatrix}$ .

After the Gauss phase (forward elimination), the augmented matrix is: $\left[\begin{array}{cc|cc} 1 & 3 & 1 & 0 \\ 0 & 1 & -2 & 1 \end{array}\right]$

What row operation completes the Jordan phase (backward elimination)?

✅ Correct! We subtract 3 times Row 2 from Row 1 to eliminate the 3 above the second pivot.

❌ Not quite. The Jordan phase creates zeros above pivots. The entry to eliminate is the 3 in position (1,2). Which operation zeros it out?

Show solution

Solution:

The Jordan phase eliminates entries above each pivot. The second pivot is at position $(2,2)$ with value $1$ . The entry above it at position $(1,2)$ is $3$ .

To eliminate it: $R_1 \leftarrow R_1 - 3 \cdot R_2$

Check each entry of the new Row 1:

Column 1: $1 - 3(0) = 1$
Column 2: $3 - 3(1) = 0$ ✓ (eliminated!)
Column 3: $1 - 3(-2) = 7$
Column 4: $0 - 3(1) = -3$

Result: $\left[\begin{array}{cc|cc} 1 & 0 & 7 & -3 \\ 0 & 1 & -2 & 1 \end{array}\right]$

The answer is R1 ← R1 − 3·R2.

Question 2

True or False: If you encounter a zero in the pivot position during Gauss-Jordan elimination, the matrix must be singular (non-invertible).

✅ Correct! A zero pivot can often be fixed by swapping rows. The method fails only when no nonzero entry exists below the pivot in that column.

❌ Not quite. Think about what happens when you swap rows — can you sometimes fix a zero pivot?

Show solution

Solution: False.

A zero in the pivot position does not mean the matrix is singular. You can swap rows to bring a nonzero entry into the pivot position — row swapping is a valid elementary row operation.

The matrix is singular only when no nonzero entry exists at or below the current pivot position in that column.

Example: $A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ has a zero at position $(1,1)$ , but swapping rows gives $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ . This matrix is invertible — it is its own inverse!

Question 3

Find the inverse of $A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}$ using Gauss-Jordan elimination. What is $A^{-1}$ ?

✅ Correct! $A^{-1} = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}$ . You can verify by multiplying $AA^{-1} = I$ .

❌ Not quite. Try setting up $[A \mid I]$ and carefully performing each row operation. Watch the signs closely!

Show solution

Solution:

Set up $[A \mid I]$ : $\left[\begin{array}{cc|cc} 2 & 5 & 1 & 0 \\ 1 & 3 & 0 & 1 \end{array}\right]$

Swap rows for a leading 1 pivot: $\left[\begin{array}{cc|cc} 1 & 3 & 0 & 1 \\ 2 & 5 & 1 & 0 \end{array}\right]$

Gauss phase — $R_2 \leftarrow R_2 - 2R_1$ : $\left[\begin{array}{cc|cc} 1 & 3 & 0 & 1 \\ 0 & -1 & 1 & -2 \end{array}\right]$

Scale — $R_2 \leftarrow -R_2$ : $\left[\begin{array}{cc|cc} 1 & 3 & 0 & 1 \\ 0 & 1 & -1 & 2 \end{array}\right]$

Jordan phase — $R_1 \leftarrow R_1 - 3R_2$ : $\left[\begin{array}{cc|cc} 1 & 0 & 3 & -5 \\ 0 & 1 & -1 & 2 \end{array}\right]$

So $A^{-1} = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}$ .

Verify: $AA^{-1} = \begin{bmatrix} 2(3)+5(-1) & 2(-5)+5(2) \\ 1(3)+3(-1) & 1(-5)+3(2) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ ✓

Question 4

In Gauss-Jordan elimination on $[A \mid I]$ , suppose you stop after completing only the Gauss phase (forward elimination), leaving the left half in upper-triangular form. What does the right half represent at this point?

✅ Correct! The right half records all operations applied so far. After only forward elimination, it holds $E_k \cdots E_1$ — not the full inverse.

❌ Not quite. The right half starts as $I$ and accumulates every row operation. After only the Gauss phase, the left half is upper-triangular, not $I$ — so the process isn't complete yet.

Show solution

Solution:

After the Gauss phase, we have applied elementary matrices $E_1, E_2, \ldots, E_k$ (the forward elimination steps) to both sides:

$E_k \cdots E_1 \, [A \mid I] = [\,E_k \cdots E_1 A \;\mid\; E_k \cdots E_1 I\,]$

Left half: $E_k \cdots E_1 A = U$ (upper triangular — not yet $I$ )
Right half: $E_k \cdots E_1 \cdot I = E_k \cdots E_1$ (product of forward-elimination elementary matrices)

This is not the full inverse. The Jordan phase applies additional elementary matrices $E_{k+1}, \ldots, E_m$ to finish reducing $U$ to $I$ . Only when the left half equals $I$ does the right half equal $A^{-1} = E_m \cdots E_1$ .

The correct answer is the product of the forward-elimination elementary matrices.

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