If AX=B,where A=left[begin{array}{ccc}1 & -1 | vedclass

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

Question

(B) Given the matrix equation $AX=B$,where $A=\left[\begin{array}{ccc}1 & -1 & 1 \ 2 & -1 & 0 \ 3 & 3 & -4\end{array}ight]$,$B=\left[\begin{array}

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Given the matrix equation $AX=B$,where $A=\left[\begin{array}{ccc}1 & -1 & 1 \ 2 & -1 & 0 \ 3 & 3 & -4\end{array}ight]$,$B=\left[\begin{array}{l}1 \ 1 \ 2\end{array}ight]$,and $X=\left[\begin{array}{l}x \ y \ z\end{array}ight]$.First,we find the determinant of $A$:$|A| = 1((-1)(-4) - (0)(3)) - (-1)((2)(-4) - (0)(3)) + 1((2)(3) - (-1)(3))$$|A| = 1(4) + 1(-8) + 1(6+3) = 4 - 8 + 9 = 5$.Since $|A| 
eq 0$,$A^{-1}$ exists.The adjoint of $A$,$adj(A)$,is the transpose of the cofactor matrix:$C_{11} = 4, C_{12} = 8, C_{13} = 9$$C_{21} = -1, C_{22} = -7, C_{23} = -6$$C_{31} = 1, C_{32} = 2, C_{33} = 1$$adj(A) = \left[\begin{array}{ccc}4 & -1 & 1 \ 8 & -7 & 2 \ 9 & -6 & 1\end{array}ight]$.Thus,$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{5} \left[\begin{array}{ccc}4 & -1 & 1 \ 8 & -7 & 2 \ 9 & -6 & 1\end{array}ight]$.Now,$X = A^{-1}B = \frac{1}{5} \left[\begin{array}{ccc}4 & -1 & 1 \ 8 & -7 & 2 \ 9 & -6 & 1\end{array}ight] \left[\begin{array}{l}1 \ 1 \ 2\end{array}ight] = \frac{1}{5} \left[\begin{array}{l}4(1) - 1(1) + 1(2) \ 8(1) - 7(1) + 2(2) \ 9(1) - 6(1) + 1(2)\end{array}ight] = \frac{1}{5} \left[\begin{array}{l}5 \ 5 \ 5\end{array}ight] = \left[\begin{array}{l}1 \ 1 \ 1\end{array}ight]$.Therefore,$x=1, y=1, z=1$.Finally,$x+y+z = 1+1+1 = 3$.

If $AX=B$,where $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & -1 & 0 \\ 3 & 3 & -4\end{array}\right]$,$B=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right]$ and $X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$,then $x+y+z=$

Similar Questions

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

The existence of a unique solution for the system of equations $x+y+z=\beta$,$5x-y+\alpha z=10$,and $2x+3y-z=6$ depends on:

For the system of equations $x+y+z=6$,$x+2y+\alpha z=10$,and $x+3y+5z=\beta$,which one of the following is $NOT$ true?

For what values of $a$ will the system of equations $x+y+z=1$,$2x+3y+2z=2$,and $ax+ay+2az=4$ have a unique solution?

The system of equations $x + 3y + 7 = 0$,$3x + 10y - 3z + 18 = 0$ and $3y - 9z + 2 = 0$ has

Vedclass Test Series

Exam Paper Generator

Online Exam Module