Let A = begin{bmatrix} 2 & -1 & 3 1 & 1 & -1 | vedclass

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

Question

(B) Given the matrix equation $AX = D$,where $A = \begin{bmatrix} 2 & -1 & 3 \ 1 & 1 & -1 \ 0 & 0 & 1 \end{bmatrix}$ and $D = \begin{bmatrix} 0 \ 1

Vedclass · Accepted Answer

$A$ unique solution

Vedclass · Answer

(B) Given the matrix equation $AX = D$,where $A = \begin{bmatrix} 2 & -1 & 3 \ 1 & 1 & -1 \ 0 & 0 & 1 \end{bmatrix}$ and $D = \begin{bmatrix} 0 \ 1 \ 0 \end{bmatrix}$.To determine the nature of the solution,we calculate the determinant of matrix $A$,denoted by $\Delta = |A|$.$\Delta = \begin{vmatrix} 2 & -1 & 3 \ 1 & 1 & -1 \ 0 & 0 & 1 \end{vmatrix}$Expanding along the third row:$\Delta = 0 \cdot \begin{vmatrix} -1 & 3 \ 1 & -1 \end{vmatrix} - 0 \cdot \begin{vmatrix} 2 & 3 \ 1 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & -1 \ 1 & 1 \end{vmatrix}$$\Delta = 1 \cdot (2(1) - (-1)(1)) = 1 \cdot (2 + 1) = 3$.Since $\Delta = 3 
eq 0$,the matrix $A$ is non-singular and invertible.Therefore,the system $AX = D$ has a unique solution given by $X = A^{-1}D$.

Let $A = \begin{bmatrix} 2 & -1 & 3 \\ 1 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix}$ and $D = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$. The system $AX = D$ has

Similar Questions

For which of the following ordered pairs $(\mu, \delta)$ is the system of linear equations $x+2y+3z=1$,$3x+4y+5z=\mu$,and $4x+4y+4z=\delta$ inconsistent?

The following system of equations $x+y+z=9$, $2x+5y+7z=52$, $x+7y+11z=77$ has

If the system of equations
$2x + 7y + \lambda z = 3$
$3x + 2y + 5z = 4$
$x + \mu y + 32z = -1$
has infinitely many solutions,then $(\lambda - \mu)$ is equal to

If the system of linear equations $x_1 + 2x_2 + 3x_3 = 6$,$x_1 + 3x_2 + 5x_3 = 9$,and $2x_1 + 5x_2 + ax_3 = b$ is consistent and has an infinite number of solutions,then:

The system of equations $x + y + z = 2$,$3x - y + 2z = 6$ and $3x + y + z = -18$ has

Vedclass Test Series

Exam Paper Generator

Online Exam Module