Matrix and Equations::

Question

The cost of $3$ copies and $5$ pens is Rs. $21$. The cost of $2$ copies and $3$ pens is Rs. $13$. Find the cost of each copy and pen using matrix method.

My Attempt: Let the cost of each copy and each pen be $x$ and $y$ respectively.

Then $$3x+5y=21$$ $$2x+3y=13$$

I have made two equations which can be easily solved by either substitution, elimination or graphical method. But the question asks to solve by matrix method. Please some body help me in solving with matrix method..

Arnaldo · Accepted Answer · 2017-02-09T01:56:29.123

1

The system is equivalent to the expression:

$$\begin{pmatrix} 3&5\\ 2&3 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 21\\ 13 \end{pmatrix}$$

And applying the inverse on both sides we get:

$$\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 3&5\\ 2&3 \end{pmatrix}^{-1}\begin{pmatrix} 21\\ 13 \end{pmatrix}$$

edited Feb 09 '17 at 01:56

answered Feb 09 '17 at 01:42

Arnaldo

21,342

@ Arnaldo, Hello sir. Please help me with this ques.. math.stackexchange.com/questions/2134941/the-2nd-4th-and-9th-terms-of-an-ap – pi-π Feb 09 '17 at 01:47
@Euler, what about this one? Is it clear? – Arnaldo Feb 09 '17 at 01:48
You could at least explain something. – Alberto Miola Feb 09 '17 at 01:48
@ Arnaldo, Yeah sir. It is. Now, I can solve it by finding the inverse of the matrix and comparing the corresponding components... – pi-π Feb 09 '17 at 01:49
@Euler, exactly! – Arnaldo Feb 09 '17 at 01:50
@Euler: you already have many solutions for your question there. Why do you still need help? – Arnaldo Feb 09 '17 at 02:02
@ I did not get an elegant solution... – pi-π Feb 09 '17 at 02:05
Euler check out mine, maybe its something you know – K Split X Feb 09 '17 at 02:05
I don't think your fimilar with inverse matrices just yet – K Split X Feb 09 '17 at 02:05
@ K split X, And why is it so? – pi-π Feb 09 '17 at 02:07
@Arnaldo, Hello sir... – pi-π Feb 10 '17 at 15:49

score 0 · Answer 2 · answered Feb 09 '17 at 02:05

0

You must reduce it to row-echoleon form. These system of equations are equivalent to the augmented matrix:

\begin{bmatrix} 3& 5 & |21\\ 2&3 & |13 \end{bmatrix}

Perform a series of row operations now.

\begin{bmatrix} 1& \frac{5}{3} & |7\\ 2&3 & |13 \end{bmatrix} $$\text{Divided the first row by 3. Now let's mess with the second row. We will do $R_2 = -2R_1 + R2$}$$

\begin{bmatrix} 1& \frac{5}{3} & |7\\ 0&-\frac{1}{3} & |-1 \end{bmatrix}

Now we have an equation! We have: $$-\frac{1}{3}y = -1 \to y = 3$$ $$\text{Using substitution now, we can sub in $y=3$ in the first row}$$ $$1x + \frac{5}{3}y = 7 \to x + \frac{5}{3}*3 = 7 \to x = 2$$

So our solutions are: $$x=2, y=3$$

answered Feb 09 '17 at 02:05

K Split X

6,565

Is it a solution by matrix? – Arnaldo Feb 09 '17 at 02:08
This is a solution by matrix. I couldve went to RREF, but chose to not, since he also had option of substitution – K Split X Feb 09 '17 at 02:08
This is not a solution by matrix, That is just Gauss Elimination. – Arnaldo Feb 09 '17 at 02:10
What is solution by matrix? This is how I learned it .. – K Split X Feb 09 '17 at 02:12

Matrix and Equations::

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