in Linear Algebra edited by
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17 votes
17 votes

What values of x, y and z satisfy the following system of linear equations?

$$\begin{bmatrix} 1 &2 &3 \\ 1& 3 &4 \\ 2& 2 &3 \end{bmatrix} \begin{bmatrix} x\\y \\ z \end{bmatrix} = \begin{bmatrix} 6\\8 \\ 12 \end{bmatrix}$$

  1. $x = 6$, $y = 3$, $z = 2$
  2. $x = 12$, $y = 3$, $z = - 4$
  3. $x = 6$, $y = 6$, $z = - 4$
  4. $x = 12$, $y = - 3$, $z = 0$
in Linear Algebra edited by
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1 comment

Just apply row transformation on (A | B) and directly check option.
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4 Answers

18 votes
18 votes
Best answer

$\begin{bmatrix} 1 &2 &3 \\ 1& 3 &4 \\ 2& 2 &3 \end{bmatrix} \begin{bmatrix} x\\y \\ z \end{bmatrix} = \begin{bmatrix} 6\\8 \\ 12 \end{bmatrix}$

  • $x+2y+3z=6\quad \to (1)$
  • $x+3y+4z=8\quad \to(2)$
  • $2x+2y+3z=12\quad \to (3)$

Apply $(3)-(1)$
$\implies x=6$

Put value of $x$ in any of the $2$ equations. Let's take (1) and (2)

  • $2y+3z=0$
  • $3y +4z=2$
  • $\implies y=6 , z=-4$

Hence, Option(C) $x=6,y=6,z=-4$.

edited by

2 Comments

How x = 6 satisfied if we put x=6 in equation no 2 then we get 3y + 4z = 0 but according to u 3y + 4z = 2 ????

@LeenSharma
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Btw u have taken equation no 2 wrong it is x + 3y + 4z = 8 not x + 3y + 4z =6
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19 votes
19 votes
Correct answer is (C). It can be easily verified by keeping the value of variables in the equations.
by
19 votes
19 votes

I will tell bit faster method to solve such questions

System is

$\begin{bmatrix} 1 & 2&3 \\ 1 & 3 &4 \\ 2& 2 & 3 \end{bmatrix}$  $\begin{bmatrix} x\\ y\\ z \end{bmatrix}$ = $\begin{bmatrix} 6\\ 8\\ 12 \end{bmatrix}$

So, what question asks is what combination of (X*column1) + (Y*column2) +(Z*column3) produces the vector on the right-hand side.

$X\begin{bmatrix} 1\\ 1\\ 2 \end{bmatrix}$ + $Y\begin{bmatrix} 2\\ 3\\ 2 \end{bmatrix}$ + $Z\begin{bmatrix} 3\\ 4\\ 3 \end{bmatrix}$ =$\begin{bmatrix} 6\\ 8\\ 12 \end{bmatrix}$

(A)X=6, Y=3, Z=2

LHS comes to be while evaluating

$\begin{bmatrix} {\color{Red} 6} &{\color{Red} 6} &{\color{Red} 6} \\ & & \\ & & \end{bmatrix}$

The first column will not evaluate to be 6 on RHS(6+6+6 is not equal to 6), so we stop here only and reject option (A).

Option (B) X=12, Y=3, Z=-4

$\begin{bmatrix} {\color{Green} 12} & {\color{Green} 6} & {\color{Green} -12}\\ {\color{Red} 12} &{\color{Red} 9} & {\color{Red} -16} \\ & & \end{bmatrix}$

The second entry of column2 does not add up to match 8 on the RHS so we stop our calculation here and reject this option too.

Option (c) X=6, Y=6, Z=-4

$\begin{bmatrix} 6 & 12 &-12 \\ 6 & 18 &-16 \\ 12 &12 & -12 \end{bmatrix}$ = $\begin{bmatrix} 6\\ 8\\ 12 \end{bmatrix}$

Yes this matches our answer. Ans-(C)

At first this method may look complicated but this is fastest method to solve such question.Not more than 50Secs. :P

2 Comments

(y) The magic of Prof. Gilbert Strang. :P
Even Option A is eliminated straightforward.
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This the solution I was looking for. Thank you @Ayush sir. At the end of the day, this is how matrices actually work.

For more intuition refer:- 

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1 vote
1 vote
putting values ,it is seen Option C is correct.
Answer:

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