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(D) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $D$ must be $0$,and the determinants

Vedclass · Accepted Answer

$58$

Vedclass · Answer

(D) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $D$ must be $0$,and the determinants $D_1, D_2, D_3$ must also be $0$.$D = \begin{vmatrix} 1 & -2 & 1 \ 2 & \alpha & 3 \ 3 & -1 & \beta \end{vmatrix} = 1(\alpha\beta + 3) + 2(2\beta - 9) + 1(-2 - 3\alpha) = \alpha\beta - 3\alpha + 4\beta - 17 = 0 \implies \alpha\beta - 3\alpha + 4\beta = 17 \dots (1)$$D_2 = \begin{vmatrix} 1 & -4 & 1 \ 2 & 5 & 3 \ 3 & 3 & \beta \end{vmatrix} = 1(5\beta - 9) + 4(2\beta - 9) + 1(6 - 15) = 5\beta - 9 + 8\beta - 36 - 9 = 13\beta - 54 = 0 \implies \beta = \frac{54}{13}$.Substituting $\beta = \frac{54}{13}$ into equation $(1)$:$\alpha(\frac{54}{13}) - 3\alpha + 4(\frac{54}{13}) = 17$$\frac{54\alpha - 39\alpha + 216}{13} = 17$$15\alpha + 216 = 221 \implies 15\alpha = 5 \implies \alpha = \frac{1}{3}$.Now,calculate $12\alpha + 13\beta = 12(\frac{1}{3}) + 13(\frac{54}{13}) = 4 + 54 = 58$.

If the system of linear equations $x-2y+z=-4$; $2x+\alpha y+3z=5$; $3x-y+\beta z=3$ has infinitely many solutions,then $12\alpha+13\beta$ is equal to

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