For the system of linear equations2x + 4y + 2 | vedclass

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(A) The determinant of the coefficient matrix is $\Delta = \begin{vmatrix} 2 & 4 & 2a \ 1 & 2 & 3 \ 2 & -5 & 2 \end{vmatrix}$.Expanding along the fi

Vedclass · Accepted Answer

It has infinitely many solutions if $a=3, b=6$

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(A) The determinant of the coefficient matrix is $\Delta = \begin{vmatrix} 2 & 4 & 2a \ 1 & 2 & 3 \ 2 & -5 & 2 \end{vmatrix}$.Expanding along the first row: $\Delta = 2(4 + 15) - 4(2 - 6) + 2a(-5 - 4) = 2(19) - 4(-4) + 2a(-9) = 38 + 16 - 18a = 54 - 18a = 18(3 - a)$.For a unique solution,we require $\Delta 
eq 0$,which implies $18(3 - a) 
eq 0$,so $a 
eq 3$.If $a 
eq 3$,the system has a unique solution for any value of $b$.Thus,options $B$ and $C$ are correct because $a=6 
eq 3$ and $a=8 
eq 3$.For infinitely many solutions,we require $\Delta = 0$ and $\Delta_x = \Delta_y = \Delta_z = 0$.Setting $\Delta = 0$ gives $a = 3$.Now,calculate $\Delta_x = \begin{vmatrix} b & 4 & 2a \ 4 & 2 & 3 \ 8 & -5 & 2 \end{vmatrix}$.For $a = 3$,$\Delta_x = \begin{vmatrix} b & 4 & 6 \ 4 & 2 & 3 \ 8 & -5 & 2 \end{vmatrix} = b(4 + 15) - 4(8 - 24) + 6(-20 - 16) = 19b - 4(-16) + 6(-36) = 19b + 64 - 216 = 19b - 152$.For $\Delta_x = 0$,$19b = 152$,which gives $b = 8$.Therefore,the system has infinitely many solutions if $a = 3$ and $b = 8$.This makes option $D$ correct.Consequently,option $A$ is $NOT$ correct.

For the system of linear equations
$2x + 4y + 2az = b$
$x + 2y + 3z = 4$
$2x - 5y + 2z = 8$
which of the following is $NOT$ correct?

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For the system of linear equations$2x + 4y + 2az = b$$x + 2y + 3z = 4$$2x - 5y + 2z = 8$which of the following is $NOT$ correct?