If $ AX = D $ represents the system of simultaneous linear equations:
$$
\begin{aligned}
x + y + z &= 6 \\
5x - y + 2z &= 3 \\
2x + y - z &= -5
\end{aligned}
$$
Then compute: $ (\text{Adj } A) D $
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- \( \begin{bmatrix} 8 \\ -16 \\ 40 \end{bmatrix} \)
- \( \begin{bmatrix} 32 \\ 64 \\ -160 \end{bmatrix} \)
\( \begin{bmatrix} -15 \\ 30 \\ 75 \end{bmatrix} \)
- \( \begin{bmatrix} 12 \\ 24 \\ 60 \end{bmatrix} \)
The Correct Option is C
Solution and Explanation
If \( AX = D \) represents the system of simultaneous linear equations: \[ \begin{aligned} x + y + z &= 6 \quad \text{(1)} \\ 5x - y + 2z &= 3 \quad \text{(2)} \\ 2x + y - z &= -5 \quad \text{(3)} \end{aligned} \] then find \( (\text{Adj } A) D \).
Let the coefficient matrix \( A \), the variable matrix \( X \), and the constant matrix \( D \) be: \[ A = \begin{bmatrix} 1 & 1 & 1 \\ 5 & -1 & 2 \\ 2 & 1 & -1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad D = \begin{bmatrix} 6 \\ 3 \\ -5 \end{bmatrix} \]
We are to compute \( (\text{Adj } A) D \). This requires finding the adjugate of matrix \( A \) and multiplying it by matrix \( D \).
First, compute the cofactor matrix of \( A \): \[ \begin{aligned} C_{11} &= \det \begin{bmatrix} -1 & 2 \\ 1 & -1 \end{bmatrix} = -1 \\ C_{12} &= -\det \begin{bmatrix} 5 & 2 \\ 2 & -1 \end{bmatrix} = 9 \\ C_{13} &= \det \begin{bmatrix} 5 & -1 \\ 2 & 1 \end{bmatrix} = 7 \\ C_{21} &= -\det \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} = 2 \\ C_{22} &= \det \begin{bmatrix} 1 & 1 \\ 2 & -1 \end{bmatrix} = -3 \\ C_{23} &= -\det \begin{bmatrix} 1 & 1 \\ 2 & 1 \end{bmatrix} = 1 \\ C_{31} &= \det \begin{bmatrix} 1 & 1 \\ -1 & 2 \end{bmatrix} = 3 \\ C_{32} &= -\det \begin{bmatrix} 1 & 1 \\ 5 & 2 \end{bmatrix} = 3 \\ C_{33} &= \det \begin{bmatrix} 1 & 1 \\ 5 & -1 \end{bmatrix} = -6 \end{aligned} \]
The cofactor matrix of \( A \) is: \[ \begin{bmatrix} -1 & 9 & 7 \\ 2 & -3 & 1 \\ 3 & 3 & -6 \end{bmatrix} \] Taking the transpose, the adjugate of \( A \) is: \[ \text{Adj}(A) = \begin{bmatrix} -1 & 2 & 3 \\ 9 & -3 & 3 \\ 7 & 1 & -6 \end{bmatrix} \]
Now compute \( \text{Adj}(A) \cdot D \): \[ \text{Adj}(A) \cdot D = \begin{bmatrix} -1 & 2 & 3 \\ 9 & -3 & 3 \\ 7 & 1 & -6 \end{bmatrix} \cdot \begin{bmatrix} 6 \\ 3 \\ -5 \end{bmatrix} = \begin{bmatrix} -6 + 6 - 15 \\ 54 - 9 - 15 \\ 42 + 3 + 30 \end{bmatrix} = \begin{bmatrix} -15 \\ 30 \\ 75 \end{bmatrix} \]
Final Answer: \[ (\text{Adj } A) D = \begin{bmatrix} -15 \\ 30 \\ 75 \end{bmatrix} \]
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