If the system of equations x+ky+3z=-2, 4x+3y+ | vedclass

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Question

(A) The given system of equations is:$x + ky + 3z = -2$$4x + 3y + kz = 14$$2x + y + 2z = 3$For the system to be solved by the matrix inversion method,

Vedclass · Accepted Answer

$k 
eq 0$ and $\frac{9}{2}$

Vedclass · Answer

(A) The given system of equations is:$x + ky + 3z = -2$$4x + 3y + kz = 14$$2x + y + 2z = 3$For the system to be solved by the matrix inversion method, the coefficient matrix $A$ must be invertible, which means the determinant $|A|$ must be non-zero $(|A| 
eq 0)$.$|A| = \begin{vmatrix} 1 & k & 3 \ 4 & 3 & k \ 2 & 1 & 2 \end{vmatrix}$Expanding along the first row:$|A| = 1(3 	imes 2 - k 	imes 1) - k(4 	imes 2 - k 	imes 2) + 3(4 	imes 1 - 3 	imes 2)$$|A| = 1(6 - k) - k(8 - 2k) + 3(4 - 6)$$|A| = 6 - k - 8k + 2k^2 - 6$$|A| = 2k^2 - 9k$Since $|A| 
eq 0$:$2k^2 - 9k 
eq 0$$k(2k - 9) 
eq 0$Therefore, $k 
eq 0$ and $k 
eq \frac{9}{2}$.

If the system of equations $x+ky+3z=-2$, $4x+3y+kz=14$, and $2x+y+2z=3$ can be solved by the matrix inversion method, then:

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