Question

Suppose the vectors ${\mathrm{x}}_1$, ${\mathrm{x}}_2$ and $x_3$ are the solutions of system of linear equation, $\mathrm{Ax}=\mathrm{b}$ when the vector is on the right side is equal to ${\mathrm{b}}_1$,${\mathrm{b}}_2$ and ${\mathrm{b}}_3$ respectively. If ${\mathrm{x}}_1=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$, ${\mathrm{x}}_2=\left[\begin{array}{c}0\\ 2\\ 1\end{array}\right]$,${\mathrm{x}}_3=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$,${\mathrm{b}}_1=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$,${\mathrm{b}}_2=\left[\begin{array}{c}0\\ 2\\ 0\end{array}\right]$,${\mathrm{b}}_3=\left[\begin{array}{c}0\\ 2\\ 2\end{array}\right]$ ,then the determinant of $\mathrm{A}$ is equal to

• A. $2$
• B. $\frac{1}{2}$
• C. $\frac{3}{2}$
• D. $4$

Answer 1

The correct option is A

$2$

Explanation for the correct option:

Find the determinant of $\mathrm{A}$:

Given, $\mathrm{Ax}=\mathrm{b}$,

By given, we know that $x$ is $3\times 1$ and $b$ is $3\times 1$ matrices

If the matrices are possible when the number of columns in the first matrix is the same as the number of rows in the second matrix and the resultant matrix is in the form of a number of rows in the first matrix and the number of columns in the second matrix.

Therefore, matrix $A$ is in the form of $3$ rows and $3$ columns.

Let,

$\mathrm{A}=\left[\begin{array}{ccc}{\mathrm{a}}_1& {\mathrm{a}}_2& {\mathrm{a}}_3\\ {\mathrm{a}}_4& {\mathrm{a}}_5& {\mathrm{a}}_6\\ {\mathrm{a}}_7& {\mathrm{a}}_8& {\mathrm{a}}_9\end{array}\right]$

By using this equation,

$\mathrm{Ax}=\mathrm{b}$

We solve,

${\mathrm{Ax}}_1={\mathrm{b}}_1$

$$\begin{gathered} \left[\begin{array}{ccc}{\mathrm{a}}_1& {\mathrm{a}}_2& {\mathrm{a}}_3\\ {\mathrm{a}}_4& {\mathrm{a}}_5& {\mathrm{a}}_6\\ {\mathrm{a}}_7& {\mathrm{a}}_8& {\mathrm{a}}_9\end{array}\right]\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]\left[\text{Given}\right] \\ \left[\begin{array}{c}{\mathrm{a}}_1+{\mathrm{a}}_2+{\mathrm{a}}_3\\ {\mathrm{a}}_4+{\mathrm{a}}_5+{\mathrm{a}}_6\\ {\mathrm{a}}_7+{\mathrm{a}}_8+{\mathrm{a}}_9\end{array}\right]=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right] \end{gathered}$$

After solving this we get,

$$\begin{gathered} {\mathrm{a}}_1+{\mathrm{a}}_2+{\mathrm{a}}_3=1\to \left(1\right) \\ {\mathrm{a}}_4+{\mathrm{a}}_5+{\mathrm{a}}_6=0\to \left(2\right) \\ {\mathrm{a}}_7+{\mathrm{a}}_8+{\mathrm{a}}_9=0\to \left(3\right) \end{gathered}$$

Similarly,

${\mathrm{Ax}}_2={\mathrm{b}}_2$

$$\begin{gathered} \left[\begin{array}{ccc}{\mathrm{a}}_1& {\mathrm{a}}_2& {\mathrm{a}}_3\\ {\mathrm{a}}_4& {\mathrm{a}}_5& {\mathrm{a}}_6\\ {\mathrm{a}}_7& {\mathrm{a}}_8& {\mathrm{a}}_9\end{array}\right]\left[\begin{array}{c}0\\ 2\\ 1\end{array}\right]=\left[\begin{array}{c}0\\ 2\\ 0\end{array}\right] \\ \left[\begin{array}{c}0.{\mathrm{a}}_1+2.{\mathrm{a}}_2+1.{\mathrm{a}}_3\\ 0.{\mathrm{a}}_4+2.{\mathrm{a}}_5+1.{\mathrm{a}}_6\\ 0.{\mathrm{a}}_7+2.{\mathrm{a}}_8+1.{\mathrm{a}}_9\end{array}\right]=\left[\begin{array}{c}0\\ 2\\ 0\end{array}\right] \\ \left[\begin{array}{c}2{\mathrm{a}}_2+{\mathrm{a}}_3\\ 2{\mathrm{a}}_5+{\mathrm{a}}_6\\ 2{\mathrm{a}}_8+{\mathrm{a}}_9\end{array}\right]=\left[\begin{array}{c}0\\ 2\\ 0\end{array}\right] \end{gathered}$$

After solving this we get,

$$\begin{gathered} 2{\mathrm{a}}_2+{\mathrm{a}}_3=0\to \left(4\right) \\ 2{\mathrm{a}}_5+{\mathrm{a}}_6=2\to \left(5\right) \\ 2{\mathrm{a}}_8+{\mathrm{a}}_9=0\to \left(6\right) \end{gathered}$$

Similarly,

${\mathrm{Ax}}_3={\mathrm{b}}_3$

$$\begin{gathered} \left[\begin{array}{ccc}{\mathrm{a}}_1& {\mathrm{a}}_2& {\mathrm{a}}_3\\ {\mathrm{a}}_4& {\mathrm{a}}_5& {\mathrm{a}}_6\\ {\mathrm{a}}_7& {\mathrm{a}}_8& {\mathrm{a}}_9\end{array}\right]\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]=\left[\begin{array}{c}0\\ 2\\ 2\end{array}\right] \\ \left[\begin{array}{c}0.{\mathrm{a}}_1+0.{\mathrm{a}}_2+1.{\mathrm{a}}_3\\ 0.{\mathrm{a}}_4+0.{\mathrm{a}}_5+1.{\mathrm{a}}_6\\ 0.{\mathrm{a}}_7+0.{\mathrm{a}}_8+1.{\mathrm{a}}_9\end{array}\right]=\left[\begin{array}{c}0\\ 2\\ 2\end{array}\right] \\ \left[\begin{array}{c}{\mathrm{a}}_3\\ {\mathrm{a}}_6\\ {\mathrm{a}}_9\end{array}\right]=\left[\begin{array}{c}0\\ 2\\ 2\end{array}\right] \end{gathered}$$

After solving this we get,

${\mathrm{a}}_3=0,{\mathrm{a}}_6=2,{\mathrm{a}}_9=2$

Substitute these values in equations $\left(4\right),\left(5\right)\text{and}\left(6\right)$

We get,

${\mathrm{a}}_2=0,{\mathrm{a}}_5=0,{\mathrm{a}}_8=-1$

Substitute all these values in the equation $\left(1\right),\left(2\right)\text{and}\left(3\right)$

${\mathrm{a}}_1=1,{\mathrm{a}}_4=-2,{\mathrm{a}}_7=-1$

Therefore,

$\mathrm{A}=\left[\begin{array}{ccc}1& 0& 0\\ -2& 0& 2\\ -1& -1& 2\end{array}\right]$

Find $\left|A\right|$

$$\begin{gathered} \mathrm{A}=\left|\begin{array}{ccc}1& 0& 0\\ -2& 0& 2\\ -1& -1& 2\end{array}\right| \\ =1(0+2) \\ =2 \end{gathered}$$

Hence, option (A) is the correct answer.