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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..

2 Answers 2

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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:

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

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    @ Arnaldo, Hello sir. Please help me with this ques.. math.stackexchange.com/questions/2134941/the-2nd-4th-and-9th-terms-of-an-ap – 2017-02-09
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    @Euler, what about this one? Is it clear? – 2017-02-09
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    You could at least explain something. – 2017-02-09
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    @ Arnaldo, Yeah sir. It is. Now, I can solve it by finding the inverse of the matrix and comparing the corresponding components... – 2017-02-09
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    @Euler, exactly! – 2017-02-09
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    @Euler: you already have many solutions for your question there. Why do you still need help? – 2017-02-09
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    @ I did not get an elegant solution... – 2017-02-09
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    Euler check out mine, maybe its something you know – 2017-02-09
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    I don't think your fimilar with inverse matrices just yet – 2017-02-09
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    @ K split X, And why is it so? – 2017-02-09
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    @Arnaldo, Hello sir... – 2017-02-10
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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: $$x=2, y=3$$

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    Is it a solution by matrix? – 2017-02-09
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    This is a solution by matrix. I couldve went to RREF, but chose to not, since he also had option of substitution – 2017-02-09
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    This is not a solution by matrix, That is just Gauss Elimination. – 2017-02-09
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    What is solution by matrix? This is how I learned it .. – 2017-02-09