This question involves solving a system of 3×3 non-homogeneous linear equations using the matrix inversion method and/or Cramer's Rule.
For a system AX=B, if ∣A∣=0, the unique solution is:
X=A−1B
Steps:
- Write the system in matrix form AX=B.
- Find ∣A∣ (determinant of A) using cofactor expansion.
- If ∣A∣=0, find A−1 using row operations on the augmented matrix [A∣I].
- Multiply: X=A−1B to get the solution.
For the system:
a1x+b1y+c1z=d1
a2x+b2y+c2z=d2
a3x+b3y+c3z=d3
Let D=∣A∣. Then:
x=DDx,y=DDy,z=DDz
where:
- Dx = determinant of A with the first column replaced by B
- Dy = determinant of A with the second column replaced by B
- Dz = determinant of A with the third column replaced by B
Condition: Cramer's Rule applies only when D=∣A∣=0.
To find A−1, form the augmented matrix [A∣I3] and apply elementary row operations until the left side becomes I3. The right side then becomes A−1.
Key row operations:
- Ri↔Rj (swap rows)
- Ri→kRi (multiply row by scalar k=0)
- Ri→Ri+kRj (add multiple of one row to another)
For matrix A=a1a2a3b1b2b3c1c2c3, expanding along Row 1:
∣A∣=a1C11+b1C12+c1C13
where Cij=(−1)i+jMij and Mij is the minor obtained by deleting row i and column j.