Question

Find X if AX=B where
$A = \begin{bmatrix}
1 & 2 & 3\\
-1 & 1 & 2\\
1 & 2 & 4
\end{bmatrix}
{\rm{and}}\;
B = \begin{bmatrix}
1\\
2\\
3
\end{bmatrix}$

Answer 1

Hint: Here, we will be using an operation on the matrix by elementary transformation on the rows and columns of the matrix and assuming the value of X.

Complete step-by-step answer:
 Given: The matrix $A = \left[ {\begin{array}{*{20}{c}}
1&2&3\\
{ - 1}&1&2\\
1&2&4
\end{array}} \right]\;\;{\rm{and}}\;B = \begin{bmatrix}
1\\
2\\
3
\end{bmatrix} $
An \[m \times n\] matrix , the \[m\] rows are horizontal and the \[n\] columns are vertical. Each element of a matrix is denoted by a variable with two subscripts. For example, \[{a_{(2,1)}}\] represents the element at the second row and first column of the matrix.
We can assume $X$ as $X = \begin{bmatrix}
x\\
y\\
z\
\end{bmatrix}$
Now, $AX = B$
Here, the matrix \[A\] will be of order \[3 \times 3\] square matrix and the matrix will be of order \[3 \times 1\]. The matrix \[B\] will be of order \[3 \times 1\]. On putting the values of \[A,X,B\] we get,
$\begin{bmatrix}
1&2&3\\
{ - 1}&1&2\\
1&2&4
\end{bmatrix}$
$ \begin{bmatrix}
x\\
y\\
z
\end{bmatrix} $ =$ \begin{bmatrix}
1\\
2\\
3
\end{bmatrix} $
We can multiply the matrices when the column of the first matrix should be equal to the row of the second matrix. Multiply the matrices \[AX\]. The matrix formed will be of order \[3 \times 3\].
Now, formulate them into an equation,
$ \begin{bmatrix}
x + 2y + 3z\\
 - x + y + 2z\\
x + 2y + 4z
\end{bmatrix} = \begin{bmatrix}
1\\
2\\
3
\end{bmatrix} $
We can know equate them individually to the given equations,
$\begin{array}{c}
\Rightarrow x + 2y + 3z = 1\\
\Rightarrow - x + y + 2z = 2\\
\Rightarrow x + 2y + 4z = 3\
\end{array}$
On naming the equations
$\Rightarrow x + 2y + 3z = 1$ ----- (1)
$\Rightarrow - x + y + 2z = 2$ ----- (2)
$\Rightarrow x + 2y + 4z = 3$ ----- (3)
On subtracting the equation (1) and (3) we get,
$\begin{array}{l}
\Rightarrow x + 2y + 3z = 1\\
\Rightarrow x + 2y + 4z = 3\
\end{array}$
Hence, this results into $ z = 2$
Now, on adding the equation (1) and (2)
$\begin{array}{l}
\Rightarrow x + 2y + 3z = 1\\
\Rightarrow - x + y + 2z = 2\
\end{array}$
These equations gives $3y + 5z = 3$ ---- (4)
On putting the value of $z = 2$ in equation (4) we get,
$\Rightarrow 3y + 5 \times 2 = 3$
$\Rightarrow 3y = 3 - 10$
$\Rightarrow 3y = - 7$
$\Rightarrow y = \dfrac{{ - 7}}{3}$
By Putting the values of $y,z$ in (1),
$\Rightarrow x + 2 \times \dfrac{{ - 7}}{3} + 3 \times 2 = 1$
$\Rightarrow x - \dfrac{{14}}{3} + 6 = 1$
$\Rightarrow x = - 5 + \dfrac{{14}}{3}$
$\Rightarrow x = \dfrac{{ - 1}}{3}$
Hence, $X = \left[ {\begin{array}{*{20}{c}}
{\dfrac{{ - 1}}{3}}\\
{\dfrac{{ - 7}}{3}}\\
2
\end{array}} \right]$

Note: In this type of problems, we should be careful when multiplying the matrix. Check the columns of the first matrix and rows of the second matrix. Solve the equations carefully avoiding calculative problems.