Solutions of Linear Algebraic Equations MCQ Quiz in हिन्दी - Objective Question with Answer for Solutions of Linear Algebraic Equations - मुफ्त [PDF] डाउनलोड करें

दिया गया है:

l + m + n = 0         ----- (1)

समीकरणों की प्रणाली

-2x + y + z = l    

x - 2y + z = m        

x + y - 2z = n     

गणना​:

⇒ \(A= \begin{bmatrix}\ -2&\ 1& \ 1 \\\ \ 1&\ -2&\ 1\\\ 1&\ 1&\ -2\end{bmatrix}x= \begin{bmatrix}\ x\\\ \ y\\\ z\end{bmatrix}b= \begin{bmatrix}\ l\\\ \ m\\\ n\end{bmatrix}\)

⇒ |A| = -6 + 3 + 3 = 0

यदि |A| 0 है, तो प्रणाली असंगत है।

अब, हम adj (A) प्राप्त करेंगे

⇒ Adj (A) = \(\begin{bmatrix}\ 3&\ 3& \ 3 \\\ \ 3&\ 3&\ 3\\\ 3&\ 3&\ 3\end{bmatrix}\)

⇒ Adj(A).b = \(\begin{bmatrix}\ 3&\ 3& \ 3 \\\ \ 3&\ 3&\ 3\\\ 3&\ 3&\ 3\end{bmatrix} \begin{bmatrix}\ l\\\ \ m\\\ n\end{bmatrix}\)

⇒ \(\begin{bmatrix}\ 3l&\ 3m& \ 3n \\\ \ 3l&\ 3m&\ 3n\\\ 3l&\ 3m&\ 3n\end{bmatrix}\)

⇒ 3× 3× 3 \(\begin{bmatrix}\ l&\ m& \ n \\\ \ l&\ m&\ n\\\ l&\ m&\ n\end{bmatrix}\)

समीकरण (1) का प्रयोग करते हुए, हम प्राप्त करते हैं

⇒ Adj(A).b = 0

∴ प्रणाली समीकरण के अनंत कई हल हैं।

Concept:

Gauss-Seidel Method:

In the Gauss-Seidel method, the value of x calculated is used in the next calculation putting other variable as 0.

2x - 5y + z = 8

Putting y = 0, z = 0 ⇒ x = 4

x + 3y - 2z = 6

Putting x = 4, z = 0 ⇒ y = 0.6

3x + 2y + z = 9

Putting x = 4, y = 0.6 ⇒ z = 9 – 3 x 4 – 2(0.6)

z = - 4.2

Mistake Point:

Don’t arrange them diagonally because It is given in question solve as per given order.

Explanation:

Equation are

20x + y – 2z = 17

3x + 20y – z = -18

2x – 3y + 20z = 25

Re-writing the above equations.

\(x = \frac{1}{{20}}\left( {17 - y + 2z} \right)\)         ---(1)

\(y = \frac{1}{{20}}\left( { - 18 - 3x + z} \right)\)        ---(2)

\(z = \frac{1}{{20}}\left( {25 - 2x + 3y} \right)\)          ---(3)

Putting y0 = z0 = 0 in equation (1)

\({x_1} = \frac{1}{{20}}\left( {17 - 0 + 0} \right)\)

∴ x₁ = 0.85

Putting x = 0.85, z = 0 in equation (2)

\({y_1} = \frac{1}{{20}}\left( { - 18 - \left( {3 \times 0.85} \right) + 0} \right)\)

∴ y₁ = - 1.0275

Putting x = 0.85, y = -1.0275 in eq. (3)

\({z_1} = \frac{1}{{20}}\left[ {25 - \left( {2 \times 0.85} \right) + 3\left( { - 1.0275} \right)} \right]\)

∴ z₁ = 1.010875

व्याख्या:

समीकरण हैं:

20x + y - 2z = 17

3x + 20y - z = -18

2x - 3y + 20z = 25

ऊपर दिए गए समीकरणों को पुनः लिखने पर:

\(x = \frac{1}{{20}}\left( {17 - y + 2z} \right)\) ---(1)

\(y = \frac{1}{{20}}\left( { - 18 - 3x + z} \right)\) ---(2)

\(z = \frac{1}{{20}}\left( {25 - 2x + 3y} \right)\) ---(3)

समीकरण (1) में y₀ = z₀ = 0 रखने पर:

\({x_1} = \frac{1}{{20}}\left( {17 - 0 + 0} \right)\)

∴ x1 = 0.85

समीकरण (2) में x = 0.85, z = 0 रखने पर:

\({y_1} = \frac{1}{{20}}\left( { - 18 - \left( {3 \times 0.85} \right) + 0} \right)\)

∴ y1 = - 1.0275

\({z_1} = \frac{1}{{20}}\left[ {25 - \left( {2 \times 0.85} \right) + 3\left( { - 1.0275} \right)} \right]\)

∴ z1 = 1.010875

व्याख्या:

समीकरण हैं:

20x + y - 2z = 17

3x + 20y - z = -18

2x - 3y + 20z = 25

ऊपर दिए गए समीकरणों को पुनः लिखने पर:

\(x = \frac{1}{{20}}\left( {17 - y + 2z} \right)\) ---(1)

\(y = \frac{1}{{20}}\left( { - 18 - 3x + z} \right)\) ---(2)

\(z = \frac{1}{{20}}\left( {25 - 2x + 3y} \right)\) ---(3)

समीकरण (1) में y₀ = z₀ = 0 रखने पर:

\({x_1} = \frac{1}{{20}}\left( {17 - 0 + 0} \right)\)

∴ x1 = 0.85

समीकरण (2) में x = 0.85, z = 0 रखने पर:

\({y_1} = \frac{1}{{20}}\left( { - 18 - \left( {3 \times 0.85} \right) + 0} \right)\)

∴ y1 = - 1.0275

\({z_1} = \frac{1}{{20}}\left[ {25 - \left( {2 \times 0.85} \right) + 3\left( { - 1.0275} \right)} \right]\)

∴ z1 = 1.010875

Explanation:

Equation are

20x + y – 2z = 17

3x + 20y – z = -18

2x – 3y + 20z = 25

Re-writing the above equations.

\(x = \frac{1}{{20}}\left( {17 - y + 2z} \right)\)         ---(1)

\(y = \frac{1}{{20}}\left( { - 18 - 3x + z} \right)\)        ---(2)

\(z = \frac{1}{{20}}\left( {25 - 2x + 3y} \right)\)          ---(3)

Putting y0 = z0 = 0 in equation (1)

\({x_1} = \frac{1}{{20}}\left( {17 - 0 + 0} \right)\)

∴ x₁ = 0.85

Putting x = 0.85, z = 0 in equation (2)

\({y_1} = \frac{1}{{20}}\left( { - 18 - \left( {3 \times 0.85} \right) + 0} \right)\)

∴ y₁ = - 1.0275

Putting x = 0.85, y = -1.0275 in eq. (3)

\({z_1} = \frac{1}{{20}}\left[ {25 - \left( {2 \times 0.85} \right) + 3\left( { - 1.0275} \right)} \right]\)

∴ z₁ = 1.010875

दिया गया है:

l + m + n = 0         ----- (1)

समीकरणों की प्रणाली

-2x + y + z = l    

x - 2y + z = m        

x + y - 2z = n     

गणना​:

⇒ \(A= \begin{bmatrix}\ -2&\ 1& \ 1 \\\ \ 1&\ -2&\ 1\\\ 1&\ 1&\ -2\end{bmatrix}x= \begin{bmatrix}\ x\\\ \ y\\\ z\end{bmatrix}b= \begin{bmatrix}\ l\\\ \ m\\\ n\end{bmatrix}\)

⇒ |A| = -6 + 3 + 3 = 0

यदि |A| 0 है, तो प्रणाली असंगत है।

अब, हम adj (A) प्राप्त करेंगे

⇒ Adj (A) = \(\begin{bmatrix}\ 3&\ 3& \ 3 \\\ \ 3&\ 3&\ 3\\\ 3&\ 3&\ 3\end{bmatrix}\)

⇒ Adj(A).b = \(\begin{bmatrix}\ 3&\ 3& \ 3 \\\ \ 3&\ 3&\ 3\\\ 3&\ 3&\ 3\end{bmatrix} \begin{bmatrix}\ l\\\ \ m\\\ n\end{bmatrix}\)

⇒ \(\begin{bmatrix}\ 3l&\ 3m& \ 3n \\\ \ 3l&\ 3m&\ 3n\\\ 3l&\ 3m&\ 3n\end{bmatrix}\)

⇒ 3× 3× 3 \(\begin{bmatrix}\ l&\ m& \ n \\\ \ l&\ m&\ n\\\ l&\ m&\ n\end{bmatrix}\)

समीकरण (1) का प्रयोग करते हुए, हम प्राप्त करते हैं

⇒ Adj(A).b = 0

∴ प्रणाली समीकरण के अनंत कई हल हैं।

Concept:

Gauss-Seidel Method:

In the Gauss-Seidel method, the value of x calculated is used in the next calculation putting other variable as 0.

2x - 5y + z = 8

Putting y = 0, z = 0 ⇒ x = 4

x + 3y - 2z = 6

Putting x = 4, z = 0 ⇒ y = 0.6

3x + 2y + z = 9

Putting x = 4, y = 0.6 ⇒ z = 9 – 3 x 4 – 2(0.6)

z = - 4.2

Mistake Point:

Don’t arrange them diagonally because It is given in question solve as per given order.