Question

12, The system of equations x-4y+7z=12 3x+8y-2z=0, 26z-8y=6.
a)unique solution
b) no solution
c) Infinite number of solution​

Answer 1

Given:

A system of equations

$x-4y+7z=12$

$3x+8y-2z=0$

$26z-8y=6$

To find:

Whether the system of equations has a unique solution, no solution or an infinite no. of solutions.

Solution:

Consider a system of equations:

$a_{1}x+b_{1}y+c_{1}z=d_{1}$,

$a_{2}x+b_{2}y+c_{2}z=d_{2}$,

$a_{3}x+b_{3}y+c_{3}z=d_{3}$

Then, they can be written in a matrix form:

$A=\left[\begin{array}{ccc}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{array}\right]$

where, $a_{1},a_{2},a_{3}$ are the coefficients of $x$, $b_{1},b_{2},b_{3}$ are the coefficients of $y$ and $c_{1},c_{2},c_{3}$ are the coefficients of $z$.

$A_{1}=\left[\begin{array}{ccc}d_{1}&b_{1}&c_{1}\\d_{2}&b_{2}&c_{2}\\d_{3}&b_{3}&c_{3}\end{array}\right]$  , $A_{2}=\left[\begin{array}{ccc}a_{1}&d_{1}&c_{1}\\a_{2}&d_{2}&c_{2}\\a_{3}&d_{3}&c_{3}\end{array}\right]$   ,   $A_{3}=\left[\begin{array}{ccc}a_{1}&b_{1}&d_{1}\\a_{2}&b_{2}&d_{2}\\a_{3}&b_{3}&d_{3}\end{array}\right]$

If $\det A\neq 0$ then the system of equations has a unique solution.

If $\det A=\det A_{1}=\det A_{2}=\det A_{3}=0$, then the system of equations has infinite no. of solutions.

If $\det A=0$ and one of the $\det A_{1}, \det A_{2}, \det A_{3}$ is not equal to zero, then the system of equations has no solution.

Here, the equations given are:

$x-4y+7z=12$

$3x+8y-2z=0$

$26z-8y=6$

Thus,

$A=\left[\begin{array}{ccc}1&-4&7\\3&8&-2\\0&-8&26\end{array}\right]$

$\det A=1[(8*26)-(-2)*(-8)]-(-4)[(3*26)-0*(-2)]+7[3*(-8)-0*8]$

$\det A=1(208-16)+4(78+0)+7(-24-0)$

$\det A=192+312-168=336\neq 0$

Hence, the system of equations has a unique solution because $\det A\neq 0$. Thus, option (a) is the correct answer.

The system of equations: $x-4y+7z=12$, $3x+8y-2z=0$, $26z-8y=6$ has a unique solution. Option (a) is the correct answer.

Answer 2

a)unique solution

Step-by-step explanation:

                 $x-4y+7z=12$

                $3x+8y-2z=0$

$26z-8y=6$  ⇒$-8y+26z=6$.

Given equations can be written in the matrix form as

$\left[\begin{array}{ccc}1&-4&7\\3&-8&-2\\0&-8&26\end{array}\right]$ $\left[\begin{array}{ccc}x&\\y\\z\end{array}\right]$  $=\left[\begin{array}{ccc}12&\\0&\\6&\end{array}\right]$

                    $A$      $X$      $= B$

$(A , B) ~ \left[\begin{array}{cccc}1&-4&7&12\\3&-8&-2&0\\0&-8&26&6\end{array}\right]$

order of the Matric is 3 × 4

           ∴ 8(A) ≤ 3

Consider the third order minor.

$\left[\begin{array}{ccc}1&-4&7\\3&-8&-2\\0&-8&26\end{array}\right]$

$= 1(-208-16) + 4(78-0)+7(-24-0)}$

$= 1(-224) + 312 - 168\\=-224 + 312 - 168\\=-392+312\\=80 \neq 0$

• There is a minor of 3 which is not zero
• Therefore the rank of (A₂B) = 3 = 4
• No. of in knowing (n) = 3

To Find the system has unique , infinity or no solutions.

• some of The conditions is to satisfied.
• (i) If the rank of (A , B) = A = n

Then the system has unique solution.

• (ii) (If the rank of (A, B) = A ∠n
• Then the system has infinitely many solutions.

• (iii) If the rank (A , B) ≠ A
• has NO solution.

Here the rank of matrix (A , B) = A = 1 so the given system has unique solution