Question

solve by inversion method iii) x - y + z = 4, 2x + y - 3z = 0 and x + y + z = 2​

Answer 1

Answer:

above image is right answer your questions

Answer 2

The solution of equations are:

x=2, y=-1 and z= 1

Given:

• $x - y + z = 4 \\ 2x + y - 3z = 0 \\ x + y + z = 2$

To find:

• Solution of equations using matrix inversion method.

Solution:

Concept/Method to be used:

1. Write the equations in matrix form as AX= B where A: Coefficients matrix, X: Unknown variables matrix, B: Constant matrix.
2. Now calculate $A^{-1}$ and solution is $X= A^{-1}B$

Step 1:

Write equations in matrix form.

AX= B

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

Step 2:

Find $A^{-1}$.

$A^{-1} = \frac{Adj. A }{ |A| }$

Find determinant of A.

$|A| = \left |\begin{array}{ccc}1&-1&1\\2&1&-3\\1&1&1\end{array}\right |$

Expand the determinant along R1

=1(1+3)+(2+3)+1(2-1)

=4+5+1

|A|= 10

|A|≠0, Thus inverse is exists.

Find Minor matrix for A:

Minor of each element of matrix can be find by hiding the elements of that row and column and cross multiply the elements and subtract.

Just like we are doing while calculating determinant.

Thus,

M11=1×1-(-3×1)= 4

by this way find all Minors.

M12=5

M13=1

M21=-2

M22=0

M23=2

M31=2

M32=-5

M33=3

Arrange all these in 3×3 matrix.

$M_{ij} = \left[\begin{array}{ccc}4&5&1\\ - 2&0&2\\2& - 5&3\end{array}\right]$

Now find Co-factor matrix:

$C_{ij}=(-1)^{1+j}M_{ij}$

$C_{ij} = \left[\begin{array}{ccc} 4& - 5&1\\ 2&0& - 2\\2& 5&3\end{array}\right]$

Find Adj.(A):

Now take transpose of Co-factor matrix

$Adj.(A)=[C_{ij}]^T$

$Adj(A) = \left[\begin{array}{ccc} 4& 2&2\\ - 5&0& 5\\1& - 2&3\end{array}\right]$

Thus,

$A^{-1} = \frac{1}{10} \left[\begin{array}{ccc} 4& 2&2\\ - 5&0& 5\\1& - 2&3\end{array}\right]$

Step 3:

Multiply $A^{-1}$ with B.

$\left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\frac{1}{10} \left[\begin{array}{ccc}4&2&2\\- 5&0& 5\\1&- 2&3\end{array}\right] \left[\begin{array}{ccc}4\\0\\2\end{array}\right]$

or

$\left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\frac{1}{10} \left[\begin{array}{ccc} 16 + 0 + 4\\ - 20 + 0 + 10\\4 + 0 + 6\end{array}\right]$

or

$\left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\frac{1}{10} \left[\begin{array}{ccc} 20\\ - 10\\10\end{array}\right]$

or

$\left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc} 2\\ - 1\\1\end{array}\right]$

Thus,

$\bf x = 2 \\\bf y = - 1 \\ \bf z = 1$

_______________________________

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