Question

Solved the
following system of equations by matrix
method
X-y+2z=7
3 x +4y-5z=-5
2x -y +3z = 1​

Answer 1

Answer:

x = 2

y = 1

z = 3

Step-by-step explanation:

Given:

x - y + 2z = 7

3x + 4y - 5z = -5

2x - y + 3z = 12

To Find:

The solutions of the equations

Solution:

$\sf Let\:A=\left[\begin{array}{ccc}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{array}\right]$

$\sf Let\:X=\left[\begin{array}{c}x\\y\\z\end{array}\right]$

$\sf Let\:B=\left[\begin{array}{c}d_1\\d_2\\d_3\end{array}\right]$

where a₁ = 1, a₂ = 3, a₃ = 2, b₁ = -1, b₂ = 4, b₃ = -1, c₁ = 2, c₂ = -5, c₃ = 3,

d₁ = 7, d₂ = -5, d₃ = 1

Substitute the data,

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

$\sf B=\left[\begin{array}{c}7\\-5\\1\end{array}\right]$

$\sf X=\left[\begin{array}{c}x\\y\\z\end{array}\right]$

Now we know that,

X = A⁻¹ B

First finding A⁻¹,

We know that,

$\sf A^{-1}=\dfrac{adj\:A}{|A|}$

Here we have to first find the co factor matrix of A and then the transpose of it.

$\sf adj\:A=\left[\begin{array}{ccc}A_{11}&A_{21}&A_{31}\\A_{12}&A_{22}&A_{32}\\A_{13}&A_{23}&A_{33}\end{array}\right]$

Hence,

$\sf adj\:A=\left[\begin{array}{ccc}12-5&-(-3+2)&5-8\\-(9+10)&3-4&-(-5-6)\\-3-8&-(-1+2)&4+3\end{array}\right]$

$\sf adj\:A=\left[\begin{array}{ccc}7&1&-3\\-19&-1&11\\-11&-1&7\end{array}\right]$

Now finding |A|

|A| = 1 (12 - 5) - 3( -3 + 2) + 2 (5 - 8)

= 7 + 3 - 6

= 4

Therefore,

$\sf A^{-1}=\dfrac{1}{4} \left[\begin{array}{ccc}7&1&-3\\-19&-1&11\\-11&-1&7\end{array}\right]$

Now,

$\sf X=\dfrac{1}{4} \left[\begin{array}{ccc}7&1&-3\\-19&-1&11\\-11&-1&7\end{array}\right]\times \left[\begin{array}{c}7\\-5\\12\end{array}\right]$

$\sf X=\dfrac{1}{4} \left[\begin{array}{c}49-5-36\\-133+5+132\\-77+5+84\end{array}\right]$

$\sf X=\dfrac{1}{4} \left[\begin{array}{c}8\\4\\12\end{array}\right]$

$\sf X=\left[\begin{array}{c}2\\1\\3\end{array}\right]$

$\sf \left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}2\\1\\3\end{array}\right]$

Equating it we get,

x = 2

y = 1

z = 3

Hence the values of x, y, z are 2, 1, 3 respectively.

Answer 2

$\begin{gathered}\begin{gathered}\bf Given: - \begin{cases} &\sf{x - y + 2z = 7 } \\ &\sf{3x + 4y - 5z = -5 }\\ &\sf{2x - y + 3z = 12} \end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\begin{gathered}\bf To \: find \: the \: values \: of \: - \begin{cases} &\sf{x} \\ &\sf{y}\\ &\sf{z} \end{cases}\end{gathered}\end{gathered}$

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$\large\underline\purple{\bold{Solution :- }}$

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$\large\underline{\bold{❥︎ \: Step :- 1 }}$

☆ x - y + 2z = 7

☆ 3x + 4y - 5z = -5

☆ 2x - y + 3z = 12

☆ The matrix form of the above equation is

$\begin{gathered}\sf \left[\begin{array}{ccc}1&-1&2\\3&4&-5\\2&-1&3\end{array}\right]\end{gathered}\begin{gathered}\sf \left[\begin{array}{c}x\\y\\z\end{array}\right]\end{gathered} = \begin{gathered}\sf \left[\begin{array}{c}7\\-5\\1\end{array}\right]\end{gathered}$

where,

$\begin{gathered}\bf A=\left[\begin{array}{ccc}1&-1&2\\3&4&-5\\2&-1&3\end{array}\right]\end{gathered}$

$\begin{gathered}\bf B=\left[\begin{array}{c}7\\-5\\1\end{array}\right]\end{gathered}$

$\begin{gathered}\bf X=\left[\begin{array}{c}x\\y\\z\end{array}\right]\end{gathered}$

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$\large\underline{\bold{❥︎Step :- 2}}$

☆ Calculate :- |A|

$\begin{gathered}\sf A=\left[\begin{array}{ccc}1&-1&2\\3&4&-5\\2&-1&3\end{array}\right]\end{gathered}$

$\sf\implies \:|A| = 1 \times (12 - 5) - 3 \times ( -3 + 2) + 2 \times (5 - 8)$

$\sf \:  ⟼|A| \: = 7 + 3 - 6$

$\sf \:  ⟼|A| = 4$

☆ Since, |A| ≠ 0, so system of equations is consistent having unique solution.

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$\large\underline{\bold{❥︎ \: Step :- 3}}$

☆Calculation to find Adjoint of matrix A

$\begin{gathered}\bf adj\:A=\left[\begin{array}{ccc}12-5&-(-3+2)&5-8\\-(9+10)&3-4&-(-5-6)\\-3-8&-(-1+2)&4+3\end{array}\right]\end{gathered}$

$\begin{gathered}\sf adj\:A=\left[\begin{array}{ccc}7&1&-3\\-19&-1&11\\-11&-1&7\end{array}\right]\end{gathered}$

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$\large\underline{\bold{❥︎Step :- 4 }}$

☆ Calculation of A⁻¹

$\bf \:We \: know \: that, \: A^{-1}=\dfrac{adj\:A}{|A|}$

$\begin{gathered} \therefore \: \sf A^{-1}=\dfrac{1}{4} \left[\begin{array}{ccc}7&1&-3\\-19&-1&11\\-11&-1&7\end{array}\right]\end{gathered}$

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$\large\underline{\bold{❥︎Step :- 5 }}$

☆ Solution of equations is given by

$\bf \:  ⟼ \: X = {A}^{ - 1} \: B$

$\begin{gathered} \therefore \: \sf X=\dfrac{1}{4} \left[\begin{array}{ccc}7&1&-3\\-19&-1&11\\-11&-1&7\end{array}\right] \left[\begin{array}{c}7\\-5\\12\end{array}\right]\end{gathered}$

$\sf\implies \:\begin{gathered}\sf X=\dfrac{1}{4} \left[\begin{array}{c}49-5-36\\-133+5+132\\-77+5+84\end{array}\right]\end{gathered}</p><p>$

$\sf \:  ⟼\begin{gathered}\sf X=\dfrac{1}{4} \left[\begin{array}{c}8\\4\\12\end{array}\right]\end{gathered}</p><p>$

$\therefore \: \begin{gathered}\sf \left[\begin{array}{c}x\\y\\z\end{array}\right]\end{gathered} = \begin{gathered}\sf \left[\begin{array}{c}2\\1\\3\end{array}\right]\end{gathered}$

☆ So, on comparing we get

$\begin{gathered}\begin{gathered}\bf \begin{cases} &\sf{x \: = \: 2} \\ &\sf{y \: = \: 1}\\ &\sf{z \: = \: 3} \end{cases}\end{gathered}\end{gathered}$

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