TN 12th Maths — Learn, Visualize, Practice

1. Minors, cofactors and the adjoint

To invert a matrix we first need two building blocks. The minor $M_{ij}$ of an entry is the determinant of the smaller matrix left after deleting row $i$ and column $j$ . The cofactor attaches a checkerboard sign to that minor.

DefinitionCofactor and adjoint

The cofactor of entry $a_{ij}$ is $C_{ij} = (-1)^{i+j} M_{ij}$ . The adjoint (adjugate) of $A$ is the transpose of the matrix of cofactors: $\operatorname{adj}(A) = [C_{ij}]^{T}$ .

The signs follow this checkerboard pattern — start with $+$ at the top-left corner:

\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}

Note: The transpose matters: the $(i,j)$ entry of $\operatorname{adj}(A)$ is $C_{ji}$, not $C_{ij}$. For a $2\times2$ matrix this simplifies to swapping the diagonal and negating the off-diagonal: $\operatorname{adj}\!\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$.

ExampleFull 3×3 adjoint

Find

\operatorname{adj}(A)

for

A = \begin{pmatrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 3 & 2 & -1 \end{pmatrix}

.

2. Inverse of a non-singular matrix

A square matrix is called non-singular when its determinant is non-zero. Exactly these matrices can be inverted, and the adjoint gives an explicit formula.

DefinitionInverse via adjoint

If $|A| \ne 0$ , then $A$ is invertible and $A^{-1} = \dfrac{1}{|A|}\,\operatorname{adj}(A)$ , satisfying $AA^{-1} = A^{-1}A = I$ .

Key idea: If $|A| = 0$ the matrix is singular and has no inverse — you would be dividing by zero. Try it live in the Visualize tab by making two rows proportional.

Example2×2 inverse

Find the inverse of

A = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}

.

Example3×3 inverse

Find the inverse of

A = \begin{pmatrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 3 & 2 & -1 \end{pmatrix}

.

3. Determinant & adjoint identities

These identities appear constantly in objective questions — knowing them saves you from computing full matrices.

$A\,(\operatorname{adj} A) = (\operatorname{adj} A)\,A = |A|\,I$
$|\operatorname{adj}(A)| = |A|^{\,n-1}$ for an $n\times n$ matrix
$|A^{-1}| = \dfrac{1}{|A|}$
$(AB)^{-1} = B^{-1}A^{-1}$ (order reverses)
$\operatorname{adj}(AB) = \operatorname{adj}(B)\,\operatorname{adj}(A)$
$|kA| = k^{n}|A|$ for an $n\times n$ matrix

ExampleDeterminant of the adjoint

If

A

is

3\times3

with

|A| = 2

, find

|\operatorname{adj}(A)|

.

ExampleChaining the identities

For a

3\times3

matrix with

|A| = 3

, find

|A^{-1}|

and

|2A|

.

Exampleadj of adj

If

A

is

3\times3

with

|A| = 4

, find

|\operatorname{adj}(\operatorname{adj} A)|

.

4. Elementary transformations & Gauss–Jordan

Row operations — swapping two rows, scaling a row, or adding a multiple of one row to another — do not change the solution set of a system. Applying them to $[A \mid I]$ until the left block becomes $I$ turns the right block into $A^{-1}$ . This is the Gauss–Jordan method, and it is often faster than the adjoint for $3\times3$ matrices.

[\,A \mid I\,] \;\xrightarrow{\text{row operations}}\; [\,I \mid A^{-1}\,]

ExampleGauss–Jordan inverse

Find

A^{-1}

for

A = \begin{pmatrix} 0 & 5 \\ -1 & 6 \end{pmatrix}

by the Gauss–Jordan method.

5. Solving systems: the matrix-inversion method

A linear system can be packaged as a single matrix equation. If the coefficient matrix is non-singular, one multiplication solves the whole system at once.

System in matrix form

AX = B \quad\Longrightarrow\quad X = A^{-1}B

Example2-variable system

Solve

2x + 5y = 1,\ x + 3y = 2

using

X = A^{-1}B

.

ExampleFull 3-variable system

Solve

2x + 3y + 3z = 5,\ x - 2y + z = -4,\ 3x - y - 2z = 3

by matrix inversion.

6. Cramer's rule

When you only need the values (not $A^{-1}$ itself), Cramer's rule expresses each unknown directly as a ratio of determinants.

x_i = \dfrac{|A_i|}{|A|}, \quad |A| \ne 0

where $A_i$ is $A$ with its $i$ -th column replaced by the column $B$ .

Example2-variable Cramer

Solve

3x + 2y = 12,\ x - y = 1

by Cramer's rule.

Example3-variable Cramer

Solve

x + y + z = 6,\ 2x + 3y - z = 5,\ x - y + 2z = 5

by Cramer's rule.

7. Rank and echelon form

The rank $\rho(A)$ is the number of non-zero rows once the matrix is reduced to row-echelon form (each leading entry sits to the right of the one above it). Equivalently, it is the order of the largest non-vanishing minor.

$\rho(A) \ge 1$ if $A$ has at least one non-zero entry.
For an $m\times n$ matrix, $\rho(A) \le \min\{m, n\}$ .
A square matrix of order $n$ is invertible if and only if $\rho(A) = n$ .

ExampleRank by echelon form

Find the rank of

A = \begin{pmatrix} 3 & 1 & 2 \\ 6 & 2 & 4 \\ 3 & 1 & 2 \end{pmatrix}

.

8. Consistency of a system by rank

Not every system has a solution. Compare the rank of $A$ with that of the augmented matrix $[A \mid B]$ to decide how many solutions exist. This is the Rouché–Capelli test.

$\rho(A) = \rho([A\mid B]) = n$ → a unique solution
$\rho(A) = \rho([A\mid B]) < n$ → infinitely many solutions
$\rho(A) \ne \rho([A\mid B])$ → no solution (inconsistent)

ExampleUnique solution

Test

x + y = 3,\ x - y = 1

for consistency.

ExampleInfinitely many solutions

Test

x + y = 2,\ 2x + 2y = 4

for consistency.

ExampleNo solution

Test

x + y = 2,\ 2x + 2y = 5

for consistency.

Watch out: A homogeneous system $AX = 0$ is always consistent (it has the trivial solution $X = 0$). It has non-trivial solutions precisely when $\rho(A) < n$, i.e. when $|A| = 0$.

9. Homogeneous systems

A system $AX = 0$ (all right-hand sides zero) always has $X = 0$ . The interesting question is whether it has any other solutions.

Key idea: For a square homogeneous system, non-trivial solutions exist exactly when $|A| = 0$. If $|A| \ne 0$, the only solution is the trivial one.

ExampleCondition for non-trivial solutions

For what value of

k

does

kx + 2y = 0,\ 2x + ky = 0

have a non-trivial solution?

10. Common exam pitfalls

Watch out: Forgetting to transpose the cofactor matrix. The adjoint is $[C_{ij}]^{T}$ — mixing up rows and columns is the single most common slip in $3\times3$ inverse questions.

Watch out: Reversing the order in $(AB)^{-1}$. It is $B^{-1}A^{-1}$, not $A^{-1}B^{-1}$. The same reversal applies to $\operatorname{adj}(AB)$.

Watch out: Using $|kA| = k|A|$. For an $n\times n$ matrix the scalar is raised to the $n$-th power: $|kA| = k^{n}|A|$.

Watch out: Declaring 'no solution' when ranks are equal but less than $n$. Equal ranks always means consistent — if they are below $n$ you get infinitely many solutions, not zero.

Quick self-check before you finish

Is $|A| \ne 0$ before you write $A^{-1}$ ? A singular matrix has no inverse.
Did you divide the adjoint by $|A|$ (not multiply)?
For systems: did you compare $\rho(A)$ with $\rho([A\mid B])$ before counting solutions?
Verify with $AA^{-1} = I$ or by substituting the solution back — it takes 30 seconds and catches sign errors.