The system of linear equations x + y + z = 2, | vedclass

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Question

(A) system of linear equations $AX = B$ has a unique solution if and only if the determinant of the coefficient matrix $A$ is non-zero,i.e.,$|A| 
e 0

Vedclass · Accepted Answer

$k 
e 0$

Vedclass · Answer

(A) system of linear equations $AX = B$ has a unique solution if and only if the determinant of the coefficient matrix $A$ is non-zero,i.e.,$|A| 
e 0$.The coefficient matrix $A$ is given by:$A = \begin{bmatrix} 1 & 1 & 1 \ 2 & 1 & -1 \ 3 & 2 & k \end{bmatrix}$We calculate the determinant $|A|$:$|A| = 1(1 \cdot k - (-1) \cdot 2) - 1(2 \cdot k - (-1) \cdot 3) + 1(2 \cdot 2 - 1 \cdot 3)$$|A| = 1(k + 2) - 1(2k + 3) + 1(4 - 3)$$|A| = k + 2 - 2k - 3 + 1$$|A| = -k$For a unique solution,we require $|A| 
e 0$,which implies $-k 
e 0$,or $k 
e 0$.

The system of linear equations $x + y + z = 2$,$2x + y - z = 3$,and $3x + 2y + kz = 4$ has a unique solution if

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