Inverse Matrix & Systems of Equations
SRMJEEE – Mathematics
1. Simple Explanation
-
Inverse Matrix
For a square matrix (A) (size (n \times n)), its inverse (A^{-1}) satisfies
[ A,A^{-1} = A^{-1},A = I_{n} ] where (I_{n}) is the identity matrix.
Only invertible (non‑singular) matrices have an inverse.
A matrix is invertible ⇔ its determinant is non‑zero. -
Systems of Linear Equations
A system
[ \begin{cases} a_{11}x_{1}+a_{12}x_{2}+ \dots + a_{1n}x_{n}=b_{1}\ \vdots\ a_{n1}x_{1}+a_{n2}x_{2}+ \dots + a_{nn}x_{n}=b_{n} \end{cases} ] can be written in matrix form as (A\mathbf{x}=\mathbf{b}).
If (A) is invertible, the unique solution is
[ \mathbf{x}=A^{-1}\mathbf{b}. ]
2. Key Points
| Topic | Important Facts |
|---|---|
| Existence of Inverse | A square matrix (A) has an inverse iff (\det(A)\neq 0). |
| Determinant of 2×2 and 3×3 | (\det\begin{bmatrix}a&b\c&d\end{bmatrix}=ad-bc). (\det\begin{bmatrix}a&b&c\d&e&f\g&h&i\end{bmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)). |
| Adjugate (Adjoint) Method | (A^{-1} = \dfrac{1}{\det(A)},\operatorname{adj}(A)). The adjugate is the transpose of the cofactor matrix. |
| Gauss‑Jordan Elimination | Augment (A) with (I) and row‑reduce to (I |
| Properties of Inverse | ((A^{-1})^{-1}=A). ((AB)^{-1}=B^{-1}A^{-1}). ((A^T)^{-1}=(A^{-1})^T). |
| Solving (A\mathbf{x}=\mathbf{b}) | 1. Compute (A^{-1}). 2. Multiply (A^{-1}\mathbf{b}). Alternative: Use Gaussian elimination or Cramer’s Rule. |
| Cramer’s Rule | For each variable (x_i), replace column (i) of (A) by (\mathbf{b}) and compute determinant: (x_i = \dfrac{\det(A_i)}{\det(A)}). |
| Singular Matrix | (\det(A)=0) → no inverse → system may have no solution or infinitely many solutions. |
3. Worked Example
Problem
Find the inverse of
[
A=\begin{bmatrix}
2 & 1 & 3\
0 & 4 & 1\
-1 & 2 & 5
\end{bmatrix}
]
and use it to solve the system
[
A\mathbf{x}=\mathbf{b},\quad \mathbf{b}=\begin{bmatrix} 5\ 6\ 4 \end{bmatrix}.
]
Step 1: Compute (\det(A))
[ \det(A)= 2\begin{vmatrix}4&1\2&5\end{vmatrix} -1\begin{vmatrix}0&1\-1&5\end{vmatrix} +3\begin{vmatrix}0&4\-1&2\end{vmatrix} ]
[ =2(4\cdot5-1\cdot2)-1(0\cdot5-1\cdot-1)+3(0\cdot2-4\cdot-1) ] [ =2(20-2)-1(0+1)+3(0+4)=2(18)-1+12=36-1+12=47 ]
Since (\det(A)=47\neq 0), (A) is invertible.
Step 2: Find Cofactor Matrix
Compute cofactors (C_{ij}) (apply ((-1)^{i+j}) times minor determinant).
| (i,j) | Minor | Cofactor |
|---|---|---|
| (1,1) | (\begin{vmatrix}4&1\2&5\end{vmatrix}=18) | (+18) |
| (1,2) | (\begin{vmatrix}0&1\-1&5\end{vmatrix}=1) | (-1) |
| (1,3) | (\begin{vmatrix}0&4\-1&2\end{vmatrix}=4) | (+4) |
| (2,1) | (\begin{vmatrix}1&3\2&5\end{vmatrix}=1) | (-1) |
| (2,2) | (\begin{vmatrix}2&3\-1&5\end{vmatrix}=13) | (+13) |
| (2,3) | (\begin{vmatrix}2&1\-1&2\end{vmatrix}=3) | (-3) |
| (3,1) | (\begin{vmatrix}1&3\4&1\end{vmatrix}=-11) | (-11) |
| (3,2) | \ |