For a matrix A=begin{bmatrix} 1 & 0 & 0 2 & | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $A U_{1} = \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix}$, $A U_{2} = \begin{bmatrix} 2 \ 3 \ 0 \end{bmatrix}$, and $A U_{3} = \begin{bmatrix}

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) Given $A U_{1} = \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix}$, $A U_{2} = \begin{bmatrix} 2 \ 3 \ 0 \end{bmatrix}$, and $A U_{3} = \begin{bmatrix} 2 \ 3 \ 1 \end{bmatrix}$.This can be written as $A U = \begin{bmatrix} 1 & 2 & 2 \ 0 & 3 & 3 \ 0 & 0 & 1 \end{bmatrix}$, where $U = [U_{1} U_{2} U_{3}]$.We know that $A U = B$, where $B = \begin{bmatrix} 1 & 2 & 2 \ 0 & 3 & 3 \ 0 & 0 & 1 \end{bmatrix}$.Thus, $U = A^{-1} B$.Taking the inverse of both sides, $U^{-1} = (A^{-1} B)^{-1} = B^{-1} A$.First, find $A^{-1}$. Since $A = \begin{bmatrix} 1 & 0 & 0 \ 2 & 1 & 0 \ 3 & 2 & 1 \end{bmatrix}$, it is a lower triangular matrix.$A^{-1} = \begin{bmatrix} 1 & 0 & 0 \ -2 & 1 & 0 \ 1 & -2 & 1 \end{bmatrix}$.Next, find $B^{-1}$. Since $B = \begin{bmatrix} 1 & 2 & 2 \ 0 & 3 & 3 \ 0 & 0 & 1 \end{bmatrix}$, $|B| = 1(3-0) = 3$.$B^{-1} = \frac{1}{3} \begin{bmatrix} 3 & -2 & 0 \ 0 & 1 & -3 \ 0 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 1 & -2/3 & 0 \ 0 & 1/3 & -1 \ 0 & 0 & 1 \end{bmatrix}$.Now, $U^{-1} = B^{-1} A = \begin{bmatrix} 1 & -2/3 & 0 \ 0 & 1/3 & -1 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \ 2 & 1 & 0 \ 3 & 2 & 1 \end{bmatrix} = \begin{bmatrix} -1/3 & -2/3 & 0 \ -7/3 & -5/3 & -1 \ 3 & 2 & 1 \end{bmatrix}$.Sum of elements = $(-1/3 - 2/3 + 0) + (-7/3 - 5/3 - 1) + (3 + 2 + 1) = -1 - 5 + 6 = 0$.

Similar Questions

If the system of linear equations $8x + y + 4z = -2$,$x + y + z = 0$,and $\lambda x - 3y = \mu$ has infinitely many solutions,then the distance of the point $\left(\lambda, \mu, -\frac{1}{2}\right)$ from the plane $8x + y + 4z + 2 = 0$ is:

If the system of linear equations $x - 4y + 7z = g$,$3y - 5z = h$,and $-2x + 5y - 9z = k$ is consistent,then:

Let $A = \{X = (x, y, z)^{T} : PX = 0 \text{ and } x^{2} + y^{2} + z^{2} = 1\}$ where $P = \begin{bmatrix} 1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & -1 \end{bmatrix}$,then the set $A$:

If the solution of the system of simultaneous linear equations $x+y-z=6$,$3x+2y-z=5$ and $2x-y-2z+3=0$ is $x=\alpha, y=\beta, z=\gamma$,then $\alpha+\beta=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module