Question

Solve the following system of equations using matrices: 3x+2y+z=6, 4x-y+2z=5, 7x+3y-3z=7​

Answer 1

Answer:

x+jjd sj und= jsji +isii

Answer 2

The solution for the following systems of Linear Equation is x=1, y=1, z=1

Given

3x+2y+z=6

4x-y+2z=5

7x+3y-3z=7​

To Find

Values of x, y, and z

Solution

Writing the systems of equations in matrix form we get

$\begin{pmatrix}3 & 2 & 1\\4 & -1 & 2\\7 & 3 & -3\\\end{pmatrix}$ $\begin{pmatrix}x\\y\\z\\\end{pmatrix}$$= \begin{pmatrix}6\\5\\7\\\end{pmatrix}$

Let

$A = \begin{pmatrix}3 & 2 & 1\\4 & -1 & 2\\7 & 3 & -3\\\end{pmatrix}$

$X = \begin{pmatrix}x\\y\\z\\\end{pmatrix}$

$B = \begin{pmatrix}6\\5\\7\\\end{pmatrix}$

Now, we can write this as

AX = B

Multiplying the equation with the inverse of A i.e A⁻¹ we get,

A⁻¹AX= A⁻¹B

We know that the product of any matrix and its inverse is the Identity Matrix, i.e I

Therefore

IX= A⁻¹B

The product of any matric with the identity matrix is the matrix itself.

Therefore,

X= A⁻¹B

A⁻¹ = Adj(A)/ |A|

|A| = 3X[(-1 X -3) - (3X2)] - 2[(4X-3)-(7X2)] + 1[(4X3)-(-1X7)]

= 3[3-6] - 2[-12 - 14] + [12 + 7]

= 3X(-3) - 2X(-26) + 19

= -9 +  52 + 19

= 62

Therefore |A|=62

Adj(A) = transpose of  [a]ₙₓₙ Where [a]ₙₓₙ  is cofactor of A

The cofactors are

$A = \begin{pmatrix}3 & 2 & 1\\4 & -1 & 2\\7 & 3 & -3\\\end{pmatrix}$

C₁₁= (-1X-3 - 3X2)(-1)¹⁺¹= 3-6 = -3

C₁₂ = (-12-14)(-1)¹⁺²= 26

C₁₃= (12 + 7)(-1)⁴= 19

C₂₁ = (-6-3)(-1)³= 9

C₂₂ = (-9 -7)(-1)⁴ = -16

C₂₃= (9 - 14)(-1)⁵ = 5

C₃₁ = (4 + 1)(-1)⁴ = 5

C₃₂ = (6 - 4)(-1)⁵ = -2

C₃₃ = (-3 - 8)(-1)⁶ = -11

$Adj(A) = \begin{pmatrix}-3 & 26 & 19\\9 & -16 & 5\\5 & -2 & -11\\\end{pmatrix}^T$

$or, Adj(A) = \begin{pmatrix}-3 & 9 & 5\\26 & -16 & -2\\19 & 5 & -11\\\end{pmatrix}$

Therefore,

$A^{-1} = \frac{ \begin{pmatrix}-3 & 9 & 5\\26 & -16 & -2\\19 & 5 & -11\\\end{pmatrix}}{62}$

$= \begin{pmatrix}-3/62 & 9/62 & 5/62\\13/31 & -8/31 & -1/31\\19/62 & 5/62 & -11/62\\\end{pmatrix}$

Therefore

X = $\begin{pmatrix}-3/62 & 9/62 & 5/62\\13/31 & -8/31 & -1/31\\19/62 & 5/62 & -11/62\\\end{pmatrix} * \begin{pmatrix}6\\5\\7\\\end{pmatrix}$

$= \begin{pmatrix}\frac{-18 + 45 + 35}{62} \\\frac{78-40-7}{31}\\ \frac{114 + 25 -77}{62} \\\end{pmatrix}$

$= \begin{pmatrix}\frac{62}{62} \\\frac{31}{31}\\ \frac{62}{62} \\\end{pmatrix}$

$= \begin{pmatrix}1\\1\\1\\\end{pmatrix}$

Therefore x=1, y=1, z=1

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