For a system of simultaneous linear equations | vedclass

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

Question

(A) Given $A X=\left[\begin{array}{l}1 \ 1 \ 2\end{array}ight]$ and $\operatorname{Adj} A=\left[\begin{array}{ccc}1 & -1 & -1 \ 1 & 1 & -1 \ 1 &

Vedclass · Accepted Answer

$\left[\begin{array}{c}-1 \ 0 \ 2\end{array}ight]$

Vedclass · Answer

(A) Given $A X=\left[\begin{array}{l}1 \ 1 \ 2\end{array}ight]$ and $\operatorname{Adj} A=\left[\begin{array}{ccc}1 & -1 & -1 \ 1 & 1 & -1 \ 1 & 1 & 1\end{array}ight]$.We know that $A^{-1} = \frac{1}{|A|} \operatorname{Adj} A$.Multiplying $A X = B$ by $A^{-1}$ on both sides, we get $X = A^{-1} B = \frac{1}{|A|} \operatorname{Adj} A \cdot B$.$X = \frac{1}{|A|} \left[\begin{array}{ccc}1 & -1 & -1 \ 1 & 1 & -1 \ 1 & 1 & 1\end{array}ight] \left[\begin{array}{l}1 \ 1 \ 2\end{array}ight]$.Performing matrix multiplication:$X = \frac{1}{|A|} \left[\begin{array}{c}1(1) + (-1)(1) + (-1)(2) \ 1(1) + 1(1) + (-1)(2) \ 1(1) + 1(1) + 1(2)\end{array}ight] = \frac{1}{|A|} \left[\begin{array}{c}1 - 1 - 2 \ 1 + 1 - 2 \ 1 + 1 + 2\end{array}ight] = \frac{1}{|A|} \left[\begin{array}{c}-2 \ 0 \ 4\end{array}ight]$.Since $|A| > 0$, we check the determinant of the given $\operatorname{Adj} A$. We know $|\operatorname{Adj} A| = |A|^{n-1}$, where $n=3$.$|\operatorname{Adj} A| = 1(1+1) - (-1)(1+1) + (-1)(1-1) = 2 + 2 + 0 = 4$.So, $|A|^2 = 4 \implies |A| = 2$ (since $|A| > 0$).Substituting $|A| = 2$ into the expression for $X$:$X = \frac{1}{2} \left[\begin{array}{c}-2 \ 0 \ 4\end{array}ight] = \left[\begin{array}{c}-1 \ 0 \ 2\end{array}ight]$.Thus, option $A$ is correct.

For a system of simultaneous linear equations, if $A X=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right]$, $\operatorname{Adj} A=\left[\begin{array}{ccc}1 & -1 & -1 \\ 1 & 1 & -1 \\ 1 & 1 & 1\end{array}\right]$ and $\operatorname{det} A>0$, then $X=$

Similar Questions

The system of linear equations $3x - 2y - kz = 10$,$2x - 4y - 2z = 6$,and $x + 2y - z = 5m$ is inconsistent if

The number of real values of $\lambda$ for which the system of linear equations $2x + 4y - \lambda z = 0$,$4x + \lambda y + 2z = 0$,and $\lambda x + 2y + 2z = 0$ has infinitely many solutions is:

The system of equations $x+2y+3z=6$,$x+3y+5z=9$,and $2x+5y+az=12$ has no solution when $a=$

If the system of linear equations given by $x+y+z=3$,$2x+2y-z=3$,and $x+y-z=1$ is consistent and if $(x_0, y_0, z_0)$ is a solution,then $2x_0+2y_0+z_0=$

The system of equations $x_1 - x_2 + x_3 = 2$,$3x_1 - x_2 + 2x_3 = -6$ and $3x_1 + x_2 + x_3 = -18$ has

Vedclass Test Series

Exam Paper Generator

Online Exam Module