This question involves solving a system of three non-homogeneous linear equations using Cramer's Rule and/or the Matrix Inversion Method.
Given the system:
a1x+b1y+c1z=d1
a2x+b2y+c2z=d2
a3x+b3y+c3z=d3
Write the coefficient matrix A and compute ∣A∣:
A=a1a2a3b1b2b3c1c2c3
If ∣A∣=0, the unique solution is:
x=∣A∣∣Ax∣,y=∣A∣∣Ay∣,z=∣A∣∣Az∣
where Ax, Ay, Az are obtained by replacing the x-, y-, z-columns of A with the constants column d1d2d3 respectively.
Expanding along the first row:
∣A∣=a1b2b3c2c3−b1a2a3c2c3+c1a2a3b2b3
Using cofactor notation: ∣A∣=a11A11+a12A12+a13A13 where Aij=(−1)i+jMij.
Solve the system:
x+y+z=6
2x−y+z=3
x+2y−z=2
Step 1: Write the coefficient matrix and find ∣A∣.
A=1211−1211−1
Expanding along row 1:
∣A∣=1−121−1−1211−1+121−12
=1(1−2)−1(−2−1)+1(4+1)=−1+3+5=7
Step 2: Find ∣Ax∣ (replace column 1 with constants).
Ax=6321−1211−1
∣Ax∣=6(1−2)−1(−3−2)+1(6+2)=−6+5+8=7
Step 3: Find ∣Ay∣ (replace column 2 with constants).
Ay=12163211−1
∣Ay∣=1(−3−2)−6(−2−1)+1(4−3)=−5+18+1=14
Step 4: Find ∣Az∣ (replace column 3 with constants).
Az=1211−12632
∣Az∣=1(−2−6)−1(4−3)+6(4+1)=−8−1+30=21
Step 5: Apply Cramer's Rule.
x=77=1,y=714=2,z=721=3
Solution: x=1, y=2, z=3
The system AX=B has solution X=A−1B provided ∣A∣=0.
A−1=∣A∣1adj(A)
where adj(A) is the transpose of the cofactor matrix of A.
Steps:
- Compute ∣A∣. If ∣A∣=0, the method cannot be applied.
- Find the cofactor Cij=(−1)i+jMij for each element.
- Form the cofactor matrix C, then adj(A)=CT.
- Compute A−1=∣A∣1adj(A).
- Multiply: X=A−1B.
Worked Example (same system):
Using A=1211−1211−1, ∣A∣=7.
Cofactors:
C11=−121−1=1−2=−1,C12=−211−1=−(−2−1)=3
C13=21−12=4+1=5
C21=−121−1=−(−1−2)=3,C22=111−1=−1−1=−2
C23=−1112=−(2−1)=−1
C31=1−111=1+1=2,C32=−1211=−(1−2)=1
C33=121−1=−1−2=−3
adj(A)=CT=−1353−2−121−3
A−1=71−1353−2−121−3
X=A−1B=71−1353−2−121−3632=71−6+9+418−6+230−3−6=7171421=123
Solution: x=1, y=2, z=3 ✓