Define Upper and Lower Triangular Matrices.

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6 months ago

An upper triangular matrix is a square matrix where all the elements above the diagonal are non-zero, and below it is zero. A lower triangular matrix is a square matrix where all the elements above the diagonal are zero.

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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 &nb

Vishal Baghel

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

g (x) = px + q

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

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

Compare 8 = ap² …………… (i)

-2 = a (2pq) + bp

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

g (x) = 2x – 1

g (2) = 3

Vishal Baghel

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

$x - y + | λ | z = 4 λ - 4$

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

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