LINEAR-ALGEBRA · Unit 2 · Video 1 · Interactive Practice

Gaussian Elimination & Back Substitution

IKey Formulas

Formula	Name	Description
$\ell_{ij} = \dfrac{a_{ij}}{a_{jj}}$	Multiplier	Ratio of target entry to pivot
$R_i \leftarrow R_i - \ell_{ij} \cdot R_j$	Row operation	Eliminates entry at position $(i, j)$
$\det(A) = p_1 \cdot p_2 \cdots p_n$	Determinant	Product of pivots from $U$
Back substitution	Solve $Ux = c$	Solve from bottom equation upward

IIVisualization 1 — Forward Elimination

Each row operation clears one entry below the diagonal, marching $[A \mid b]$ toward upper-triangular form.

Step 0

augmented matrix [A | b]

Step 0 · Setup

Pivot

a_{11}=1

. The entry

a_{21}=3

below it must become

0

Step 1 · Clear column 1

R_2 \leftarrow R_2 - 3R_1

, multiplier

\ell_{21}=\tfrac{3}{1}=3

. New pivot

a_{22}=2

; target

a_{32}=4

Step 2 · Clear column 2

R_3 \leftarrow R_3 - 2R_2

, multiplier

\ell_{32}=\tfrac{4}{2}=2

. Pivots

1,2,5

give

\det A = 10

IIIVisualization 2 — When Elimination Fails

Sliding the single entry $a_{33}$ slides the third pivot $p_3 = a_{33}+4$ ; when it reaches $0$ the matrix turns singular.

Entry a₃₃ 1

💡 A zero pivot, a zero determinant, and a singular matrix are one and the same event — here at $a_{33} = -4$ .

IVVisualization 3 — Back Substitution

The bottom equation has one unknown; each solved variable unlocks exactly one new unknown in the row above.

Step 0

solution vector x

VQuiz Questions

Question 1

Consider the augmented matrix from the video: $[A \mid b] = \begin{bmatrix} 1 & 2 & 1 & | & 2 \\ 3 & 8 & 1 & | & 12 \\ 0 & 4 & 1 & | & 2 \end{bmatrix}$

The first elimination step uses $R_2 \leftarrow R_2 - \ell_{21} \cdot R_1$ . What is the multiplier $\ell_{21}$ ?

✅ Correct! The multiplier is target ÷ pivot = 3/1 = 3.

❌ Not quite. The multiplier is (target entry) ÷ (pivot) = a₂₁ / a₁₁.

Show solution

Solution:

The multiplier is the ratio of the target entry to the pivot:

$\ell_{21} = \frac{a_{21}}{a_{11}} = \frac{3}{1} = 3$

We then perform $R_2 \leftarrow R_2 - 3 \cdot R_1$ to zero out position $(2,1)$ .

Common mistake: Computing $\frac{a_{11}}{a_{21}} = \frac{1}{3}$ — the multiplier is target $\div$ pivot, not the other way around.

Question 2

After forward elimination, the upper triangular matrix $U$ has pivots $1, 2, 5$ on its diagonal. What is $\det(A)$ ?

✅ Correct! det(A) = product of pivots = 1 × 2 × 5 = 10.

❌ Not quite. Remember: the determinant is the product of the pivots, not the sum.

Show solution

Solution:

The determinant equals the product of the pivots of $U$ :

$\det(A) = p_1 \times p_2 \times p_3 = 1 \times 2 \times 5 = 10$

Since $\det(A) \neq 0$ , the matrix is invertible and the system has a unique solution.

Common mistakes:

Adding pivots: $1 + 2 + 5 = 8$ ✗ (determinant is a product, not a sum)
Adding all entries on the diagonal is also wrong

Question 3

True or False: If a zero appears in a pivot position during Gaussian elimination, the matrix is always singular (not invertible).

✅ Correct! A zero pivot can sometimes be fixed by a row swap (temporary failure). Singularity only occurs when no non-zero entry exists below.

❌ Not quite. Consider: what if there's a non-zero entry below the zero pivot that we could swap?

Show solution

Answer: False

A zero in a pivot position does not always mean the matrix is singular:

Temporary failure: If a non-zero entry exists below the zero pivot, we can swap rows (a row exchange) and continue elimination. The matrix may still be invertible.
Permanent failure: If all entries below the zero pivot are also zero, no row swap can help. Only then is the matrix singular with $\det(A) = 0$ .

The key distinction: Can we find a non-zero entry below to swap with?

Question 4

During back substitution on the triangular system, after finding $z = -2$ , we substitute into Equation 2:

$2y - 2(-2) = 6$

What is the value of $y$ ?

✅ Correct! Since $-2 \times (-2) = +4$ , we get $2y + 4 = 6$ , giving $y = 1$ .

❌ Not quite. Be careful with the double negative: $-2 \times (-2) = +4$ , not $-4$ .

Show solution

Solution:

Substitute $z = -2$ into Equation 2: $2y - 2(-2) = 6$ $2y + 4 = 6 \quad \text{(double negative: } -2 \times (-2) = +4\text{)}$ $2y = 2$ $y = 1$

Common mistake: Getting the sign wrong on $-2 \times (-2)$ .

If you compute $-2 \times (-2) = -4$ instead of $+4$ , you get $2y - 4 = 6 \Rightarrow y = 5$ . ✗

Solved: 0 / 4