Inverse of a matrix by elementary operations

MATElementary Operations & the Inverse

This subtopic finds $A^{-1}$ using elementary row (or column) operations rather than the adjoint formula. Starting from $A = IA$ , you apply the SAME row operations to the left factor and to $I$ until the left side becomes $I$ ; then the right side has become $A^{-1}$ .

It is examined as a 4-6 mark question requiring neat, sequential operations and an awareness that the inverse exists only when $A$ is non-singular.

Setup using identity (row method)

A = IA \xrightarrow{\text{row ops}} I = BA \Rightarrow A^{-1} = B

Apply each elementary row operation to BOTH the left

A

and the

I

on the right; when left becomes

I

, the right becomes

A^{-1}

Column method analogue

A = AI \xrightarrow{\text{column ops}} I = AB \Rightarrow A^{-1} = B

If using column operations, write

A=AI

and operate on columns only; never mix row and column operations in one solution.

Existence of inverse

A^{-1} \text{ exists} \iff |A| \ne 0

|A|

is the determinant; a square matrix is invertible (non-singular) precisely when its determinant is nonzero.

Elementary row operations

R_i \leftrightarrow R_j, \quad R_i \to kR_i\ (k\ne 0), \quad R_i \to R_i + kR_j

The three allowed operations: swap two rows, scale a row by nonzero

k

, add a multiple of one row to another.

Begin with $A=IA$ (for row operations) and keep the equation form throughout; both sides change together.
Use only the three elementary row operations; aim to make the left matrix into $I$ column by column.
Strategy: create a leading 1 in each pivot position, then clear all other entries in that column.
Do NOT mix row and column operations within the same method — choose one and stay consistent.
If at any stage a full row of the left matrix becomes all zeros, $A$ is singular and $A^{-1}$ does not exist.
The inverse, if it exists, is unique; you can verify by checking $AA^{-1}=I$ .
Work with exact fractions, not decimals, to keep the final inverse precise.

Where the marks go

Applying an operation to one side only — every row operation must hit both the left factor and the identity.
Mixing row and column operations in a single solution, which invalidates the method.
Continuing to compute when a zero row appears instead of concluding the matrix is non-invertible.
Rounding intermediate fractions to decimals, producing an inaccurate inverse.

How the board asks it

Numericalbegin from A = IA and reduce the left matrix to I by row operations
Using elementary row transformations, find the inverse of the matrix $A = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}$ .
Numericalcreate a leading 1 in each pivot, then clear the rest of that column
Find $A^{-1}$ by using elementary row operations, where $A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{pmatrix}$ .
Give reasonsif a full row of the left matrix becomes all zeros, the matrix is singular
Attempt to find the inverse of $A = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}$ by elementary row operations, and state with reason why $A^{-1}$ does not exist.
Diagram / graphcolumn-operation analogue using A = AI
Using elementary column transformations, obtain the inverse of $A = \begin{pmatrix} 3 & -1 \\ -4 & 2 \end{pmatrix}$ , applying each operation to both $A$ and $I$ in $A = AI$ .
Applicationsolve a linear system by writing it as AX = B and using X = A^{-1}B
Using elementary operations, find the inverse of $A = \begin{pmatrix} 2 & 1 \\ 7 & 4 \end{pmatrix}$ and hence solve the system $2x + y = 5,\ 7x + 4y = 18$ .

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.