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 $kx + y + z = 1$,$x + ky + z = k$,and $x + y + kz = k^2$ has no solution if $k$ is equal to

The bookshop of a particular school has $10$ dozen chemistry books,$8$ dozen physics books,and $10$ dozen economics books. Their selling prices are Rs. $80$,Rs. $60$,and Rs. $40$ each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

If $A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}$,find $A^{-1}$. Using $A^{-1}$,solve the system of equations: $2x - 3y + 5z = 11$,$3x + 2y - 4z = -5$,and $x + y - 2z = -3$.

If the system of linear equations $x-2y+z=-4$; $2x+\alpha y+3z=5$; $3x-y+\beta z=3$ has infinitely many solutions,then $12\alpha+13\beta$ is equal to

The number of solutions of the equations $x + y - z = 0$,$3x - y - z = 0$,and $x - 3y + z = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module