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# Matrices Calculator

## Get detailed solutions to your math problems with our Matrices step-by-step calculator. Practice your math skills and learn step by step with our math solver. Check out all of our online calculators here.

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###  Difficult Problems

1

Here, we show you a step-by-step solved example of matrices. This solution was automatically generated by our smart calculator:

$\int\frac{dy}{y}=\int\frac{dx}{x\left(x-1\right)}$
2

Rewrite the fraction $\frac{1}{x\left(x-1\right)}$ in $2$ simpler fractions using partial fraction decomposition

$\frac{1}{x\left(x-1\right)}=\frac{A}{x}+\frac{B}{x-1}$
3

Find the values for the unknown coefficients: $A, B$. The first step is to multiply both sides of the equation from the previous step by $x\left(x-1\right)$

$1=x\left(x-1\right)\left(\frac{A}{x}+\frac{B}{x-1}\right)$
4

Multiplying polynomials

$1=\frac{x\left(x-1\right)A}{x}+\frac{x\left(x-1\right)B}{x-1}$
5

Simplifying

$1=\left(x-1\right)A+xB$
6

Assigning values to $x$ we obtain the following system of equations

$\begin{matrix}1=-A&\:\:\:\:\:\:\:(x=0) \\ 1=B&\:\:\:\:\:\:\:(x=1)\end{matrix}$
7

Proceed to solve the system of linear equations

$\begin{matrix} -1A & + & 0B & =1 \\ 0A & + & 1B & =1\end{matrix}$
8

Rewrite as a coefficient matrix

$\left(\begin{matrix}-1 & 0 & 1 \\ 0 & 1 & 1\end{matrix}\right)$
9

Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & -1 \\ 0 & 1 & 1\end{matrix}\right)$
10

The integral of $\frac{1}{x\left(x-1\right)}$ in decomposed fractions equals

$\frac{-1}{x}+\frac{1}{x-1}$
11

Expand the integral $\int\left(\frac{-1}{x}+\frac{1}{x-1}\right)dx$ into $2$ integrals using the sum rule for integrals, to then solve each integral separately

$\int\frac{1}{y}dy=\int\frac{-1}{x}dx+\int\frac{1}{x-1}dx$

The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$

$\ln\left|y\right|$
12

Solve the integral $\int\frac{1}{y}dy$ and replace the result in the differential equation

$\ln\left|y\right|=\int\frac{-1}{x}dx+\int\frac{1}{x-1}dx$

The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$

$-\ln\left|x\right|+\int\frac{1}{x-1}dx$

Apply the formula: $\int\frac{n}{x+b}dx$$=nsign\left(x\right)\ln\left(x+b\right)+C$, where $b=-1$ and $n=1$

$-\ln\left|x\right|+\ln\left|x-1\right|$

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

$-\ln\left|x\right|+\ln\left|x-1\right|+C_0$
13

Solve the integral $\int\frac{-1}{x}dx+\int\frac{1}{x-1}dx$ and replace the result in the differential equation

$\ln\left|y\right|=-\ln\left|x\right|+\ln\left|x-1\right|+C_0$

The difference of two logarithms of equal base $b$ is equal to the logarithm of the quotient: $\log_b(x)-\log_b(y)=\log_b\left(\frac{x}{y}\right)$

$\ln\left(y\right)=\ln\left(\frac{x-1}{x}\right)+C_0$

Take the variable outside of the logarithm

$e^{\ln\left(y\right)}=e^{\left(\ln\left(\frac{x-1}{x}\right)+C_0\right)}$

Simplifying the logarithm

$y=e^{\left(\ln\left(\frac{x-1}{x}\right)+C_0\right)}$

Simplify $e^{\left(\ln\left(\frac{x-1}{x}\right)+C_0\right)}$ by applying the properties of exponents and logarithms

$y=e^{C_0}\frac{x-1}{x}$

Multiplying the fraction by $e^{C_0}$

$y=\frac{e^{C_0}\left(x-1\right)}{x}$

We can rename $e^{C_0}$ as other constant

$y=\frac{C_1\left(x-1\right)}{x}$
14

Find the explicit solution to the differential equation. We need to isolate the variable $y$

$y=\frac{C_1\left(x-1\right)}{x}$

##  Final answer to the problem

$y=\frac{C_1\left(x-1\right)}{x}$

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