Question

$AX = B$ where A = \left[ {\begin{array}{*{20}{l}} 1&2&3 \\\ { - 1}&1&2 \\\ 1&2&4 \end{array}} \right] and B = \left[ {\begin{array}{*{20}{l}} 1 \\\ 2 \\\ 3 \end{array}} \right] where $X$ is
A. \left[ {\begin{array}{*{20}{l}} {\dfrac{1}{3}} \\\ { - \dfrac{7}{3}} \\\ 2 \end{array}} \right]
B. \left[ {\begin{array}{*{20}{l}} { - \dfrac{1}{3}} \\\ {\dfrac{7}{3}} \\\ 2 \end{array}} \right]
C. \left[ {\begin{array}{*{20}{l}} { - \dfrac{1}{3}} \\\ { - \dfrac{7}{3}} \\\ 2 \end{array}} \right]
D. \left[ {\begin{array}{*{20}{l}} {\dfrac{1}{3}} \\\ {\dfrac{7}{3}} \\\ 2 \end{array}} \right]

Answer 1

let us assume $X$ as \left[ {\begin{array}{*{20}{l}} a \\\ b \\\ c \end{array}} \right] elements
Now when we multiply $AX$ we get another matrix which is $B$ and now we can compare and find $a,b,c$

Complete step by step solution:
Here we are given the certain equation $AX = B$ where $A,X,B$ all represent the matrix. Now we know that $A$ is $3 \times 3$ matrix and $B$ is $3 \times 1$ matrix. So $X$ must be $3 \times 1$ matrix. So that we get the product as $AX = B$
Now we are given that A = \left[ {\begin{array}{*{20}{l}} 1&2&3 \\\ { - 1}&1&2 \\\ 1&2&4 \end{array}} \right]
We need to use the multiplication of the matrix theorem.
So as we assumed X = \left[ {\begin{array}{*{20}{l}} a \\\ b \\\ c \end{array}} \right]
Now when we multiply $AX$ we get
$AX =$ \left[ {\begin{array}{*{20}{l}} 1&2&3 \\\ { - 1}&1&2 \\\ 1&2&4 \end{array}} \right] \left[ {\begin{array}{*{20}{l}} a \\\ b \\\ c \end{array}} \right]
= \left[ {\begin{array}{*{20}{l}} {a + 2b + 3c} \\\ { - a + b + 2c} \\\ {a + 2b + 4c} \end{array}} \right]
So we get $AX =$ \left[ {\begin{array}{*{20}{l}} {a + 2b + 3c} \\\ { - a + b + 2c} \\\ {a + 2b + 4c} \end{array}} \right]
Now we are given that $AX = B$
So we can write that
\left[ {\begin{array}{*{20}{l}} {a + 2b + 3c} \\\ { - a + b + 2c} \\\ {a + 2b + 4c} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 1 \\\ 2 \\\ 3 \end{array}} \right]
Upon comparing we will get the three equations
$a + 2b + 3c = 1 - - - - - (1)$
$- a + b + 2c = 2 - - - - - (2)$
$a + 2b + 4c = 3 - - - - - (3)$
On adding 1 and 2 we get that
$3b + 5c = 3 - - - - (4)$
On adding (2) and (3) we get that
$3b + 6c = 5 - - - - - (5)$
So $3b = 5 - 6c$
Putting the value of 3b in equation (4) we get that
$5 - 6c + 5c = 3 \\\ \- c = 3 - 5 \\\ c = 2 \\\$
Now for $b{\text{ }}b = \dfrac{{5 - 6c}}{3} = \dfrac{{5 - 12}}{3} = - \dfrac{7}{3}$
Now we know that $a + 2b + 3c = 1$
So $a - \dfrac{{14}}{3} + 6 = 1$
$a = - 6 + 1 + \dfrac{{14}}{3} \\\ a = - \dfrac{1}{3} \\\$
Hence we get that X = \left[ {\begin{array}{*{20}{l}} a \\\ b \\\ c \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} { - \dfrac{1}{3}} \\\ { - \dfrac{7}{3}} \\\ 2 \end{array}} \right]

Note:
We can also solve it like this
$AX = B$
\left[ {\begin{array}{*{20}{l}} 1&2&3 \\\ { - 1}&1&2 \\\ 1&2&4 \end{array}} \right] $X$ = \left[ {\begin{array}{*{20}{l}} 1 \\\ 2 \\\ 3 \end{array}} \right]
${R_1} \to {R_1} + {R_2},{R_3} \to {R_3} - {R_1}$
\left[ {\begin{array}{*{20}{l}} 1&2&3 \\\ 0&3&5 \\\ 0&0&1 \end{array}} \right] $X$ = \left[ {\begin{array}{*{20}{l}} 1 \\\ 3 \\\ 2 \end{array}} \right]
${R_2} \to \dfrac{{{R_2}}}{3}$
We get \left[ {\begin{array}{*{20}{l}} 1&2&3 \\\ 0&1&{\dfrac{5}{3}} \\\ 0&0&1 \end{array}} \right] $X$ = \left[ {\begin{array}{*{20}{l}} 1 \\\ 1 \\\ 2 \end{array}} \right]
${R_1} \to {R_1} - 2{R_2}$
\left[ {\begin{array}{*{20}{l}} 1&0&{\dfrac{{ - 1}}{3}} \\\ 0&1&{\dfrac{5}{3}} \\\ 0&0&1 \end{array}} \right] $X$ = \left[ {\begin{array}{*{20}{l}} { - 1} \\\ 1 \\\ 2 \end{array}} \right]
${R_1} \to {R_1} + \dfrac{{{R_3}}}{3},{R_2} \to {R_2} - \dfrac{{5{R_1}}}{3}$
\left[ {\begin{array}{*{20}{l}} 1&0&0 \\\ 0&1&0 \\\ 0&0&1 \end{array}} \right] $X$ = \left[ {\begin{array}{*{20}{l}} { - \dfrac{1}{3}} \\\ { - \dfrac{7}{3}} \\\ 2 \end{array}} \right]
Hence we got $X$