If the system of equations 2x + 3y - z = 5,x | vedclass

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(B) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix must be zero,and the augmented matrix

Vedclass · Accepted Answer

$1120$

Vedclass · Answer

(B) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix must be zero,and the augmented matrix must have a rank less than $3$.The coefficient matrix is $A = \begin{bmatrix} 2 & 3 & -1 \ 1 & \alpha & 3 \ 3 & -1 & \beta \end{bmatrix}$.Setting $|A| = 0$:$2(\alpha\beta + 3) - 3(\beta - 9) - 1(-1 - 3\alpha) = 0$$2\alpha\beta + 6 - 3\beta + 27 + 1 + 3\alpha = 0$$2\alpha\beta + 3\alpha - 3\beta + 34 = 0$  (Equation $1$)Since the system has infinitely many solutions,the planes must be linearly dependent. We can express the third equation as a linear combination of the first two: $L_3 = c_1 L_1 + c_2 L_2$.Comparing coefficients:$2c_1 + c_2 = 3$$3c_1 + c_2\alpha = -1$$-c_1 + 3c_2 = \beta$$5c_1 - 4c_2 = 7$Solving the system of equations for $c_1$ and $c_2$ using the first and last equations:$c_2 = 3 - 2c_1$$5c_1 - 4(3 - 2c_1) = 7 \implies 5c_1 - 12 + 8c_1 = 7 \implies 13c_1 = 19 \implies c_1 = \frac{19}{13}$$c_2 = 3 - 2(\frac{19}{13}) = \frac{39 - 38}{13} = \frac{1}{13}$Substituting $c_1$ and $c_2$ into the other equations:$3(\frac{19}{13}) + \alpha(\frac{1}{13}) = -1 \implies 57 + \alpha = -13 \implies \alpha = -70$$-(\frac{19}{13}) + 3(\frac{1}{13}) = \beta \implies \beta = \frac{-16}{13}$Finally,$13\alpha\beta = 13(-70)(\frac{-16}{13}) = 70 	imes 16 = 1120$.

If the system of equations $2x + 3y - z = 5$,$x + \alpha y + 3z = -4$,and $3x - y + \beta z = 7$ has infinitely many solutions,then $13\alpha\beta$ is equal to

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