Linear Equations and Determinants - Class XI

$\displaystyle \begin{bmatrix} 3x+7 & 5

\ y+1 & 2-3x \end{bmatrix}=\begin{bmatrix} 0 & y-2 \ 8 & 4

\end{bmatrix}$
Equating the corresponding elements, we get,
$\displaystyle 3x+7=0\Rightarrow x=-\frac { 7 }{ 3 } $
$\displaystyle 5=y-2\Rightarrow y=7$
$\displaystyle y+1=8\Rightarrow y=7$
$\displaystyle 2-3x=4\Rightarrow x=-\frac { 2 }{ 3 } $
We find that on comparing the corresponding elements of the two matrices, we get two different values of $x$, which is not possible.
Hence, it is not possible to find the values of $x$ and $y$ for which the given matrices are equal.

$kx-ay=66-----(1)$

From x+z=2, z=2-x. Substitute into the first two: 3x-4y+2(2-x)=-1 -> x-4y=-5. 2x+3y+5(2-x)=7 -> -3x+3y=-3 -> -x+y=-1 -> y=x-1. Substitute y into x-4y=-5: x-4(x-1)=-5 -> -3x+4=-5 -> -3x=-9 -> x=3.

For infinitely many solutions,
$\displaystyle \frac { k+1 }{ k } =\frac { 8 }{ k+3 } =\frac { 4 }{ 3k-1 } $

$\Rightarrow k^{2}-4k+3=0 $ and $24k-8=4k+12$
$\Rightarrow k=3,k=1$ and $k=1$

Hence, $k=1$

For the system to have infinitely many solutions, the augmented matrix must lead to a row of zeros equal to zero (0 = 0). This occurs when a = -1 and b = 9.

$\Delta =\begin{vmatrix} k & 1 & 1 \ 1 & k & 1 \ 1 & 1 & k \end{vmatrix}$

$\Rightarrow \Delta =(k-1)^{2}(k+2)$

Taking $\Delta=0$
$\Rightarrow (k-1)^{2}(k+2)=0$
$\Rightarrow k=1, k=-2$
At these values of k, system can have either no solution or infinitely many solution.

For k=1, equations takes the form $x+y+z=1$
Hence, infinitely many solution for k=1.

For k=-2,
$D _{1}=\begin{vmatrix} 1 & 1 & 1 \ -2 & -2 & 1 \ 4 & 1 & -2 \end{vmatrix}$
$D _{1}=9 \ne 0$

So, $D=0$, at least one $D _{1}\ne 0$
Hence, system has no solution for k=-2

Given system of equations can be written as
$AX=B$

$\begin{bmatrix} 1 & 1 & 1 \ 2 & -1 & 3 \ 1 & -2 & -1 \end{bmatrix}: \begin{bmatrix} x \ y \ z \end{bmatrix}=\begin{bmatrix} 2 \ 5 \ -1 \end{bmatrix}$

Here, $D =\begin{vmatrix} \alpha & 1 & 1 \ 1 & \alpha & 1 \ 1 & 1 & \alpha \end{vmatrix}$

$\Rightarrow D =(\alpha-1)^{2}(\alpha+2)$

Taking $D=0$
$\Rightarrow (\alpha-1)^{2}(\alpha+2)=0$
$\Rightarrow \alpha=1, \alpha=-2$
At these values of $\alpha$, system can have either no solution or infinitely many solution.

For $\alpha=1$, equations takes the form $x+y+z=0$
Hence, infinitely many solution for $\alpha=1$.

For $\alpha=-2$,
$D _{1}=\begin{vmatrix} -3 & 1 & 1 \ -3 & -2 & 1 \ -3 & 1 & -2 \end{vmatrix}$
$D _{1}=27 \ne 0$

So, $D=0$, at least one $D _{1}\ne 0$
Hence, system has no solution for $\alpha=-2$

If the given matrices are equal then we have the following equations,

$(b+c)(y+z)-ax=b-c\$       ..............(1)

Given system of equations can be written as
$AX=B$
where $A=\begin{bmatrix} { 1 } & { 1 } & { 1 } \ { 1 } & { 2 } & 3\ { 2 } & 3 & 4 \end{bmatrix}$ 
$X=\begin{bmatrix} x \ y \ z \end{bmatrix}$ ;$B=\begin{bmatrix} 3 \ 4 \ 7 \end{bmatrix}$

Here, $|A|=0$
Now, we will find $(adj A)B$

$adj A=C^{T}={\begin{bmatrix} { -1 } & { 2 } & { -1 } \ { -1 } & { 2 } & -1 \ { 1 } & -2 & 1 \end{bmatrix}}^T$

$\Rightarrow adj A=\begin{bmatrix} { -1 } & { -1 } & { 1 } \ { 2 } & { 2 } & -2 \ { -1 } & -1 & 1 \end{bmatrix}$

Now, $(adj A)B=\begin{bmatrix} { -1 } & { -1 } & { 1 } \ { 2 } & { 2 } & -2 \ { -1 } & -1 & 1 \end{bmatrix}\begin{bmatrix} 3 \ 4 \ 7 \end{bmatrix}$

$\Rightarrow (adj A)B=\begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix}$
$\Rightarrow (adj A)B=O$

Hence,the system of equations has infinitely many solutions.
Let $z=k$ where $k\in R$
Then 
$x+y=3-k$
$x+2y=4-3k$
Solving these eqns, we get 
$y=1-2k ; x=2+k$

$\Delta =\begin{vmatrix} 1 & 1 & 1 \ 1 & 2 & 3 \ 1 & 2 & \lambda  \end{vmatrix}=1\left( 2\lambda -6 \right) -1\left( \lambda -3 \right) +1\left( 2-2 \right) =2\lambda -6-\lambda +3+0=\lambda -3\ \Delta _{ 1 }=\begin{vmatrix} 6 & 1 & 1 \ 10 & 2 & 3 \ \mu  & 2 & \lambda  \end{vmatrix}=6\left( 2\lambda -6 \right) -1\left( 10\lambda -3\mu  \right) +1\left( 20-2\mu  \right) \ =12\lambda -36-10\lambda +3\mu +20-2\mu =2\lambda +\mu -16$

$\Delta=\begin{vmatrix} 1 &4  &-2  \3  &1  &5  \  2&3  &1  \end{vmatrix}=1(1-15)-4(3-10)-2(9-2)=0$
and $\Delta _1$ is not zero
Therefore, it has no solution

$2x+3y+5z=9$,

${2}x+{3}y+z={5}$

Given $\omega$ is a cube root of unity and $x+ y + z = a, x + \omega y + \omega^2 z = b, x + \omega^2 y + \omega z = c$.

Given $x+y+z=a $...(i)

For the solution of the system of equations
$x - 2y + 3z = -1$
$-x + y - 2z = 2$
$kx - 3y + 4z = 1$, 
$\begin{bmatrix}
1 &-2  &3 \ 
-1 &1  &-2 \ 
k &-3  &4 
\end{bmatrix} 
$
If determinant of the matrix is zero than the given lines are coplanar and if the determinant is non zero the they are non co-planar.
The solution for the system of equations exist when the lines are coplanar.
Hence, $1(-2 - 4k) - 3(-1 - k) -1(-4 + 2) \neq 0$
$\Rightarrow -2 - 4k + 3 + 3k + 2 \neq 0$
i.e. $k \neq 3$

$D=\begin{vmatrix}1 &1  &1 \ 2 & 2 & 2\ 3 & 3 & 3\end{vmatrix}=0$
$D _1=0,$     $D _2=0$,      $D _3=0$
Let $z=t$
$x+y=1-t$
$2x+2y=3-2t$
Since both the lines are parallel hence no value of x and y
Hence there is no solution of the given equations.

We find $D = 0$, where $D$ is the determinant formed by the coefficients of $x,\ y,\ z$ in the three equations and since no pair of planes are parallel, so there is an infinite number of solutions.
Let $\alpha P _{1} + \lambda P _{2} = P _{3}$
$\Rightarrow P _{1} + 7P _{2} = 13P _{3}$
$\Rightarrow b _{1} + 7b _{2} = 13b _{3}$
(A) $D\neq 0\Rightarrow$ unique solution for any $b _{1}, b _{2}, b _{3}$
(B) $D = 0$ but $P _{1} + 7P _{2} \neq 13P _{3}$
(C) $D = 0$ Also $b _{2} = -2b _{1}, b _{3} = -b _{1}$
Satisfied $b _{1} + 7b _{2} = 13b _{3}$ (Actually all three planes are co-incident)
(D) $D\neq 0$.

Let u=e^x and v=e^y. The system is au+bv=c and pu+qv=d. By Cramer's Rule, u = Delta2/Delta1 and v = Delta3/Delta1. Since u=e^x and v=e^y, x=log(Delta2/Delta1) and y=log(Delta3/Delta1).