Question

In the solution of the following set of linear equations by Gauss elimination using partial pivoting 5x + y + 2z = 34; 4y - 3z = 12; and 10x - 2y + z = -4; the pivots for elimination of x and y are

• A. 10 and 4
• B. 10 and 2
• C. 5 and 4
• D. 5 and -4

Answer 1

The correct option is A 10 and 4

Method I :

In the process of converting given matrix into an Echelon form, the largest absolute value in a column,that is to be used for making Echelon form, are called, PIVOT for that column.

[A : B] =

$\left[\begin{array}{ccccc}5& 1& 2& ⋮& 34\\ 0& 4& -3& ⋮& 12\\ 10& -2& 1& ⋮& -4\end{array}\right]\underset{–}{R_1\leftrightarrow R_3}\left[\begin{array}{ccccc}\underset{–}{10}& -2& 1& ⋮& -4\\ 0& 4& -3& ⋮& 12\\ 5& 1& 2& ⋮& 34\end{array}\right]$

$\underset{–}{R_3\to R_3-\frac{1}{2}{R}_1}\left[\begin{array}{ccccc}10& -2& 1& ⋮& -4\\ 0& \underset{–}{4}& -3& ⋮& 12\\ 0& 2& 3/2& ⋮& 36\end{array}\right]$

$\underset{–}{R_3\to R_3-\frac{1}{2}{R}_2}\left[\begin{array}{ccccc}10& -2& 1& ⋮& -4\\ 0& 4& -3& ⋮& 12\\ 0& 0& 3& ⋮& 30\end{array}\right]$ Echelon form

Here 10 and 4 are the largest elements that are used in making of Echelon form.

So, they can be taken as PIVOTS for $C_1$ and $C_2$ respectively

Method II :

5x + y +2z = 34

4y - 3z = 12

10x - 2y + z = -4

Elements of $C_1$ are $\left|5\right|,\left|0\right|,\left|10\right|$ so largest element can be taken as 1st pivot for $C_1$

Hence, $R_1\leftrightarrow R_3\Rightarrow$

10x - 2y + z = -4 ....(1)

4y - 3z = 12 ....(2)

5x + y + 2z = 34 ....(3)

$R_3\to R_3-\frac{R_1}{2}$

$\Rightarrow$ 10x - 2y + z = -4

0x + 4y -3z = 12

0x + 2y + $\frac{3}{2}$z = 36

Now elements of $C_2:\left|4\right|,\left|2\right|$ so can be taken as 2nd Pivot for $C_2$

Using $R_3\to R_3-\frac{1}{2}{R}_2$

$\Rightarrow$ 10x - 2y + z = -4

0x + 4y -3z = 12

0x +0y + 3z = 30

Now, we have reached at Echelon form, so pivots are :

x = 10, y = 4