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

Hint:
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 $X$