TN 12th MathsLearn · Visualize · Practice

Chapter 1

Applications of Matrices and Determinants

Inverse of a matrix, elementary transformations, and solving systems of linear equations.

1. Minors, cofactors and the adjoint

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

DefinitionCofactor and adjoint

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

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: $\operatorname{adj}\!\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$.

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 A0|A| \ne 0, then AA is invertible and A1=1Aadj(A)A^{-1} = \dfrac{1}{|A|}\,\operatorname{adj}(A), satisfying AA1=A1A=IAA^{-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.
Example
Find the inverse of A=(2513)A = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}.

3. Determinant & adjoint identities

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

  • A(adjA)=(adjA)A=AIA\,(\operatorname{adj} A) = (\operatorname{adj} A)\,A = |A|\,I
  • adj(A)=An1|\operatorname{adj}(A)| = |A|^{\,n-1} for an n×nn\times n matrix
  • A1=1A|A^{-1}| = \dfrac{1}{|A|}
  • (AB)1=B1A1(AB)^{-1} = B^{-1}A^{-1} (order reverses)
  • adj(AB)=adj(B)adj(A)\operatorname{adj}(AB) = \operatorname{adj}(B)\,\operatorname{adj}(A)
Example
If AA is 3×33\times3 with A=2|A| = 2, find adj(A)|\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 [AI][A \mid I] until the left block becomes II turns the right block into A1A^{-1}. This is the Gauss–Jordan method, and it is often faster than the adjoint for 3×33\times3 matrices.

[AI]  row operations  [IA1][\,A \mid I\,] \;\xrightarrow{\text{row operations}}\; [\,I \mid A^{-1}\,]

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=BX=A1BAX = B \quad\Longrightarrow\quad X = A^{-1}B
Example
Solve 2x+5y=1, x+3y=22x + 5y = 1,\ x + 3y = 2 using X=A1BX = A^{-1}B.

6. Cramer's rule

When you only need the values (not A1A^{-1} itself), Cramer's rule expresses each unknown directly as a ratio of determinants.

xi=AiA,A0x_i = \dfrac{|A_i|}{|A|}, \quad |A| \ne 0

where AiA_i is AA with its ii-th column replaced by the column BB.

7. Consistency of a system by rank

Not every system has a solution. The rank of a matrix — the number of non-zero rows after reducing to echelon form — decides everything. Compare the rank of AA with that of the augmented matrix [AB][A \mid B].

  • ρ(A)=ρ([AB])=n\rho(A) = \rho([A\mid B]) = n → a unique solution
  • ρ(A)=ρ([AB])<n\rho(A) = \rho([A\mid B]) < n → infinitely many solutions
  • ρ(A)ρ([AB])\rho(A) \ne \rho([A\mid B]) → no solution (inconsistent)
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 $|A| = 0$.