Solve the Following System of Equations Using the Matrix Inverse Method
[begin{cases}
3x + 4y + 7z = 14 \
2x – y + 3z = 4 \
2x + 2y – 3z = 0
end{cases}]
—
Solution:
To solve the system using the matrix inverse method, we express the system in matrix form ( AX = B ), where:
[
A = begin{bmatrix}
3 & 4 & 7 \
2 & -1 & 3 \
2 & 2 & -3
end{bmatrix}, quad
X = begin{bmatrix}
x \
y \
z
end{bmatrix}, quad
B = begin{bmatrix}
14 \
4 \
0
end{bmatrix}
]
—
Step 1: Find the Inverse of Matrix ( A )
First, calculate the determinant of ( A ):
[
text{det}(A) = 3 left( (-1)(-3) – 3 times 2 right) – 4 left( 2 times (-3) – 3 times 2 right) + 7 left( 2 times 2 – (-1) times 2 right)
]
[
= 3 (3 – 6) – 4 (-6 – 6) + 7 (4 + 2)
]
[
= 3 times (-3) – 4 times (-12) + 7 times 6
]
[
= -9 + 48 + 42 = 81
]
Since ( text{det}(A) = 81 neq 0 ), ( A ) is invertible.
—
Step 2: Compute the Adjoint of ( A )
The adjoint ( text{adj}(A) ) is the transpose of the cofactor matrix. After calculating the cofactors and transposing, we get:
[
text{adj}(A) = begin{bmatrix}
-9 & -26 & -2 \
-6 & 23 & -10 \
-6 & -12 & -5
end{bmatrix}
]
—
Step 3: Calculate ( A^{-1} )
[
A^{-1} = frac{1}{text{det}(A)} times text{adj}(A) = frac{1}{81} times begin{bmatrix}
-9 & -26 & -2 \
-6 & 23 & -10 \
-6 & -12 & -5
end{bmatrix}
]
—
Step 4: Solve for ( X )
[
X = A^{-1} B = frac{1}{81} times begin{bmatrix}
-9 & -26 & -2 \
-6 & 23 & -10 \
-6 & -12 & -5
end{bmatrix} times begin{bmatrix}
14 \
4 \
0
end{bmatrix}
]
[
= frac{1}{81} times begin{bmatrix}
-9 times 14 + (-26) times 4 + (-2) times 0 \
-6 times 14 + 23 times 4 + (-10) times 0 \
-6 times 14 + (-12) times 4 + (-5) times 0
end{bmatrix}
]
[
= frac{1}{81} times begin{bmatrix}
-126 – 104 \
-84 + 92 \
-84 – 48
end{bmatrix}
]
[
= frac{1}{81} times begin{bmatrix}
-230 \
8 \
-132
end{bmatrix}
]
[
= begin{bmatrix}
-230/81 \
8/81 \
-132/81
end{bmatrix}
]
—
Conclusion
The solution to the system of equations is:
[
X = begin{bmatrix}
x \
y \
z
end{bmatrix} = begin{bmatrix}
-frac{230}{81} \
frac{8}{81} \
-frac{132}{81}
end{bmatrix}
]
Simplifying further:
[ x = -2.84, quad y = 0.1, quad z = -1.63 ]
Thus, the solution is:
[ x approx -2.84, quad y approx 0.1, quad z approx -1.63 ]
