Solve the following equations, using inverse of a matrix :
`{:(3x+4y+7z=4),(2x-y+3z=-3),(x+2y-3z=8):}`

Answer

To solve the given system of equations using the inverse of a matrix, we can follow these steps:

### Step 1: Write the equations in matrix form
The given equations are:
1. \(3x + 4y + 7z = 4\)
2. \(2x - y + 3z = -3\)
3. \(x + 2y - 3z = 8\)

We can express this in the form \(AX = B\), where:
\[ A = \begin{pmatrix} 3 & 4 & 7 \\ 2 & -1 & 3 \\ 1 & 2 & -3 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 4 \\ -3 \\ 8 \end{pmatrix} \]

### Step 2: Find the determinant of matrix \(A\)
To find the inverse of matrix \(A\), we first need to calculate its determinant. The determinant of a \(3 \times 3\) matrix is given by:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix \(A\):
\[ \text{det}(A) = 3((-1)(-3) - (3)(2)) - 4((2)(-3) - (3)(1)) + 7((2)(2) - (-1)(1)) \]
Calculating this step by step:
- \(3(3 - 6) = 3(-3) = -9\)
- \(-4(-6 - 3) = -4(-9) = 36\)
- \(7(4 + 1) = 7(5) = 35\)

Now, summing these values:
\[ \text{det}(A) = -9 + 36 + 35 = 62 \]

### Step 3: Find the adjoint of matrix \(A\)
The adjoint of a matrix is the transpose of its cofactor matrix. We calculate the cofactor matrix \(C\) of \(A\):
\[ C = \begin{pmatrix} \text{Cofactor}_{11} & \text{Cofactor}_{12} & \text{Cofactor}_{13} \\ \text{Cofactor}_{21} & \text{Cofactor}_{22} & \text{Cofactor}_{23} \\ \text{Cofactor}_{31} & \text{Cofactor}_{32} & \text{Cofactor}_{33} \end{pmatrix} \]
Calculating each cofactor:
- \(\text{Cofactor}_{11} = (-1)^{1+1} \cdot \text{det}\begin{pmatrix}-1 & 3 \\ 2 & -3\end{pmatrix} = 3 + 6 = 9\)
- \(\text{Cofactor}_{12} = (-1)^{1+2} \cdot \text{det}\begin{pmatrix}2 & 3 \\ 1 & -3\end{pmatrix} = -(-6 - 3) = 9\)
- \(\text{Cofactor}_{13} = (-1)^{1+3} \cdot \text{det}\begin{pmatrix}2 & -1 \\ 1 & 2\end{pmatrix} = 4 + 1 = 5\)

Continuing this for all elements, we get:
\[ C = \begin{pmatrix} 9 & 9 & 5 \\ -6 & -9 & -3 \\ -7 & -5 & 4 \end{pmatrix} \]

Now, the adjoint \(A^*\) is the transpose of \(C\):
\[ A^* = \begin{pmatrix} 9 & -6 & -7 \\ 9 & -9 & -5 \\ 5 & -3 & 4 \end{pmatrix} \]

### Step 4: Find the inverse of matrix \(A\)
The inverse of matrix \(A\) is given by:
\[ A^{-1} = \frac{1}{\text{det}(A)} A^* \]
Thus,
\[ A^{-1} = \frac{1}{62} \begin{pmatrix} 9 & -6 & -7 \\ 9 & -9 & -5 \\ 5 & -3 & 4 \end{pmatrix} \]

### Step 5: Multiply \(A^{-1}\) with \(B\)
Now we can find \(X\) by multiplying \(A^{-1}\) with \(B\):
\[ X = A^{-1}B = \frac{1}{62} \begin{pmatrix} 9 & -6 & -7 \\ 9 & -9 & -5 \\ 5 & -3 & 4 \end{pmatrix} \begin{pmatrix} 4 \\ -3 \\ 8 \end{pmatrix} \]

Calculating this product:
- First row: \(9 \cdot 4 + (-6) \cdot (-3) + (-7) \cdot 8 = 36 + 18 - 56 = -2\)
- Second row: \(9 \cdot 4 + (-9) \cdot (-3) + (-5) \cdot 8 = 36 + 27 - 40 = 23\)
- Third row: \(5 \cdot 4 + (-3) \cdot (-3) + 4 \cdot 8 = 20 + 9 + 32 = 61\)

Thus,
\[ X = \frac{1}{62} \begin{pmatrix} -2 \\ 23 \\ 61 \end{pmatrix} \]

### Step 6: Final solution
Calculating the final values:
\[ x = \frac{-2}{62}, \quad y = \frac{23}{62}, \quad z = \frac{61}{62} \]
This gives us:
\[ x = -\frac{1}{31}, \quad y = \frac{23}{62}, \quad z = \frac{61}{62} \]