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Maths Matrices Maths Class 12th Matrices

Consider the following system of equations:

x + 2y – 3z = a

2x + 6y – 11z = b

x – 2y + 7z = c,

where a, b and c are real constants. Then the system of equations:

Option 1 -

Has infinite number of solution when 5a = 2b +c

Option 2 -

Has no solution for all a, b, and c

Option 3 -

Has a unique solution when 5a = 2b + c

Option 4 -

Has a unique solution for all a, b and c

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1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

a month ago

Correct Option - 1

Detailed Solution:

Given x + 2y – 3z = a

2x + 6y – 11z = b

x – 2y + 7z = c

Here $\begin{matrix} \end{matrix} \begin{matrix} \begin{array}{l} \end{array} \end{matrix}$

$Δ = | \begin{matrix} 1 & 2 & - 3 \\ 2 & 6 & - 11 \\ 1 & - 2 & 7 \end{matrix} | \begin{matrix} \begin{array}{l} = (42 - 22) - 2 (14 + 11) - 3 (- 4 - 6) \end{array} & = 20 - 50 + 30 = 0 \end{matrix}$

For infinite solution $_{}_{}_{}$

20a – 8b – 4c = 0 5a = 2b + c

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If $A [\begin{matrix} 1 & 0 & 1 \end{matrix}] = 2 [\begin{matrix} 1 & 0 & 1 \end{matrix}]$ , $A [\begin{matrix} - 1 & 0 & 1 \end{matrix}] = 4 [\begin{matrix} - 1 & 0 & 1 \end{matrix}]$ and $A [\begin{matrix} 0 & 1 & 0 \end{matrix}] = 2 [\begin{matrix} 1 & 0 & 0 \end{matrix}]$ where, A is a 3 × 3 matrix and (A –3I) $[\begin{matrix} x & y & z \end{matrix}] = [\begin{matrix} - 1 & 2 & 3 \end{matrix}]$ then the value of (x, y, z) is

alok kumar singh

Let $A = [\begin{matrix} x_{1} & y_{1} & z_{1} \\ x_{2} & y_{2} & z_{2} \\ x_{3} & y_{3} & z_{3} \end{matrix}]$

Given $A = [\begin{matrix} 1 & 0 & 1 \end{matrix}] = [\begin{matrix} 2 & 0 & 2 \end{matrix}]$ ...(1)

$[\begin{matrix} x_{1} + z_{1} & x_{2} + z_{2} & x_{3} + z_{3} \end{matrix}] = [\begin{matrix} 2 & 0 & 2 \end{matrix}]$

∴ x₁ + z₁ = 2 … (2)

x₂ + z₂ = 0 … (3)

x₃ + z₃ = 0 … (4)

Given $A = [\begin{matrix} - 1 & 0 & 1 \end{matrix}] = [\begin{matrix} - 4 & 0 & 4 \end{matrix}]$

$[\begin{matrix} - x_{1} + z_{1} & - x_{2} + z_{2} & - x_{3} + z_{3} \end{matrix}] = [\begin{matrix} 4 & 0 & 4 \end{matrix}]$

⇒ – x₁ + z₁ = −4 … (5)

–x₂ + z₂ = 0 &nbs

...more

Let $A = [\begin{matrix} x_{1} & y_{1} & z_{1} \\ x_{2} & y_{2} & z_{2} \\ x_{3} & y_{3} & z_{3} \end{matrix}]$

Given $A = [\begin{matrix} 1 & 0 & 1 \end{matrix}] = [\begin{matrix} 2 & 0 & 2 \end{matrix}]$ ...(1)

$[\begin{matrix} x_{1} + z_{1} & x_{2} + z_{2} & x_{3} + z_{3} \end{matrix}] = [\begin{matrix} 2 & 0 & 2 \end{matrix}]$

∴ x₁ + z₁ = 2 … (2)

x₂ + z₂ = 0 … (3)

x₃ + z₃ = 0 … (4)

Given $A = [\begin{matrix} - 1 & 0 & 1 \end{matrix}] = [\begin{matrix} - 4 & 0 & 4 \end{matrix}]$

$[\begin{matrix} - x_{1} + z_{1} & - x_{2} + z_{2} & - x_{3} + z_{3} \end{matrix}] = [\begin{matrix} 4 & 0 & 4 \end{matrix}]$

⇒ – x₁ + z₁ = −4 … (5)

–x₂ + z₂ = 0 … (6)

–x₃ + z₃ = 4 … (7)

Given $A = [\begin{matrix} 0 & 1 & 0 \end{matrix}] = [\begin{matrix} 0 & 2 & 0 \end{matrix}]$

$[\begin{matrix} y_{1} & y_{2} & y_{3} \end{matrix}] = [\begin{matrix} 0 & 2 & 0 \end{matrix}]$

∴ y₁ = 0, y₂ = 2, y₃ = 0

∴ from (2), (3), (4), (5), (6) and (7)

x₁ = 3, x₂ = 0, x₃ = –1

y₁ = 0, y₂ = 2, y₃ = 0

z₁ = –1, z₂ = 0, z₃ = 3

$A = [\begin{matrix} 3 & 0 & - 1 \\ 0 & 2 & 0 \\ - 1 & 0 & 3 \end{matrix}]$

Now (A – 3I) $[\begin{matrix} x & y & z \end{matrix}]$ = $[\begin{matrix} - 1 & 2 & 3 \end{matrix}]$

$[\begin{matrix} 0 & 0 & - 1 \\ 0 & - 1 & 0 \\ - 1 & 0 & 0 \end{matrix}] [\begin{matrix} x & y & z \end{matrix}] = [\begin{matrix} - 1 & 2 & 3 \end{matrix}]$

$[\begin{matrix} - z & - y & - x \end{matrix}]$ = $[\begin{matrix} - 1 & 2 & 3 \end{matrix}]$

[z = 1], [y = –2], [x = –3]

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Let $M = [\begin{matrix} 0 & - α \\ α & 0 \end{matrix}],$ where a is a non-zero real number an $N = \sum_{k = 1}^{49} M^{2 k} .$ If (I – M²)N = -2I, then the positive integral value of α is…………….

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$f (x) = a x^{2} + b x + c$

g (x) = px + q

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$\Rightarrow 8 x^{2} - 2 x = a {(p x + q)}^{2} + b (p x + q) + c$

Compare 8 = ap² …………… (i)

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0 = aq² + bq + c

$9 (f (x)) = p (a x^{2} + b x + c) + q$

=>4x² + 6x + 1 = apx² + bpx + cp + q

=> Andhra Pradesh = 4 ……………. (ii)

6 = bp

1 = cp + q

From (i) & (ii), p = 2, q = -1

=> b = 3, c = 1, a = 2

f (x) = 2x² + 3x + 1

f (2) = 8 + 6 + 1 = 15

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Let $A = (\begin{matrix} 2 & - 1 \\ 0 & 2 \end{matrix}) .$ If $B = I -^{5} ? C_{1} (a d j A) +^{5} ? C_{2} {(a d j A)}^{2} - . . .^{5} ? C_{5} {(a d j A)}^{5},$ then the sum of all elements of the matrix B is

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Kindly consider the following figure

B = (I – adjA)⁵

Vishal Baghel

Kindly consider the following figure

B = (I – adjA)⁵

If the system of linear equations

2x + 3y – z = -2

x + y + z = 4

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where $λ \in R,$ has no solution, then

Raj Pandey

System of equation is

$(\begin{matrix} 2 & 3 & - 1 \\ 1 & 1 & 1 \\ 1 & - 1 & | λ | \end{matrix}) (\begin{array}{l} x \\ y \\ z \end{array}) = [\begin{array}{l} - 2 \\ 4 \\ 4 λ - 4 \end{array}]$

R₁ – 2 R₂, R₃ – R₂

$(\begin{matrix} 0 & 1 & - 3 \\ 1 & 1 & 1 \\ 0 & - 2 & | λ | - 1 \end{matrix}) (\begin{array}{l} x \\ y \\ z \end{array}) = (\begin{array}{l} - 10 \\ 4 \\ 4 λ - 8 \end{array})$

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Consider the following system of equations:

x + 2y – 3z = a

2x + 6y – 11z = b

x – 2y + 7z = c,

where a, b and c are real constants. Then the system of equations:

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Consider the following system of equations: x + 2y – 3z = a 2x + 6y – 11z = b x – 2y + 7z = c, where a, b and c are real constants. Then the system of equations:

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Learn more about...

Most viewed information

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Your Question

Consider the following system of equations:

x + 2y – 3z = a

2x + 6y – 11z = b

x – 2y + 7z = c,

where a, b and c are real constants. Then the system of equations: