# Matrix Calculator

This solver will add, subtract, multiply, divide, and raise to power the two matrices, with steps shown. It will also find the determinant, inverse, rref (reduced row echelon form), null space, rank, eigenvalues and eigenvectors.

• In general, you can skip the multiplication sign, so 5x is equivalent to 5*x.
• In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x).
• Also, be careful when you write fractions: 1/x^2 ln(x) is 1/x^2 ln(x), and 1/(x^2 ln(x)) is 1/(x^2 ln(x)).
• If you skip parentheses or a multiplication sign, type at least a whitespace, i.e. write sin x (or even better sin(x)) instead of sinx.
• Sometimes I see expressions like tan^2xsec^3x: this will be parsed as tan^(2*3)(x sec(x)). To get tan^2(x)sec^3(x), use parentheses: tan^2(x)sec^3(x).
• Similarly, tanxsec^3x will be parsed as tan(xsec^3(x)). To get tan(x)sec^3(x), use parentheses: tan(x)sec^3(x).
• From the table below, you can notice that sech is not supported, but you can still enter it using the identity sech(x)=1/cosh(x).
• If you get an error, double check your expression, add parentheses and multiplication signs where needed, and consult the table below.
• All suggestions and improvements are welcome. Please leave them in comments.
The following table contains the supported operations and functions:
 Type Get Constants e e pi pi i i (imaginary unit) Operations a+b a+b a-b a-b a*b a*b a^b, a**b a^b sqrt(x), x^(1/2) sqrt(x) cbrt(x), x^(1/3) root(3)(x) root(x,n), x^(1/n) root(n)(x) x^(a/b) x^(a/b) abs(x) |x| Functions e^x e^x ln(x), log(x) ln(x) ln(x)/ln(a) log_a(x) Trigonometric Functions sin(x) sin(x) cos(x) cos(x) tan(x) tan(x), tg(x) cot(x) cot(x), ctg(x) sec(x) sec(x) csc(x) csc(x), cosec(x) Inverse Trigonometric Functions asin(x), arcsin(x), sin^-1(x) asin(x) acos(x), arccos(x), cos^-1(x) acos(x) atan(x), arctan(x), tan^-1(x) atan(x) acot(x), arccot(x), cot^-1(x) acot(x) asec(x), arcsec(x), sec^-1(x) asec(x) acsc(x), arccsc(x), csc^-1(x) acsc(x) Hyperbolic Functions sinh(x) sinh(x) cosh(x) cosh(x) tanh(x) tanh(x) coth(x) coth(x) 1/cosh(x) sech(x) 1/sinh(x) csch(x) Inverse Hyperbolic Functions asinh(x), arcsinh(x), sinh^-1(x) asinh(x) acosh(x), arccosh(x), cosh^-1(x) acosh(x) atanh(x), arctanh(x), tanh^-1(x) atanh(x) acoth(x), arccoth(x), cot^-1(x) acoth(x) acosh(1/x) asech(x) asinh(1/x) acsch(x)

Select an operation:

Choose the dimension of the matrix: $$\times$$$Enter the elements of the matrix or Choose the dimension of the second matrix: $$\times$$$

Enter the elements of the matrix or

:

Write all suggestions in comments below.

## Solution

Your input: calculate $$A + B=\left[ \begin{array}{ccc} 1 & 0 & 0 \\\\ 0 & 0 & 4 \\\\ 0 & 1 & 0 \end{array} \right] + \left[ \begin{array}{ccc} 2 & 1 & 4 \\\\ 5 & 7 & 1 \\\\ 1 & 2 & 5 \end{array} \right]$$$$$\left[ \begin{array}{ccc} \color{Crimson}{1} & \color{Fuchsia}{0} & \color{DeepPink}{0} \\\\ \color{Purple}{0} & \color{OrangeRed}{0} & \color{Blue}{4} \\\\ \color{Chartreuse}{0} & \color{Peru}{1} & \color{Magenta}{0} \end{array} \right] + \left[ \begin{array}{ccc} \color{Crimson}{2} & \color{Fuchsia}{1} & \color{DeepPink}{4} \\\\ \color{Purple}{5} & \color{OrangeRed}{7} & \color{Blue}{1} \\\\ \color{Chartreuse}{1} & \color{Peru}{2} & \color{Magenta}{5} \end{array} \right]=$$$

$$=\left[ \begin{array}{ccc}\left(\color{Crimson}{1}\right) + \left(\color{Crimson}{2}\right) & \left(\color{Fuchsia}{0}\right) + \left(\color{Fuchsia}{1}\right) & \left(\color{DeepPink}{0}\right) + \left(\color{DeepPink}{4}\right) \\\\ \left(\color{Purple}{0}\right) + \left(\color{Purple}{5}\right) & \left(\color{OrangeRed}{0}\right) + \left(\color{OrangeRed}{7}\right) & \left(\color{Blue}{4}\right) + \left(\color{Blue}{1}\right) \\\\ \left(\color{Chartreuse}{0}\right) + \left(\color{Chartreuse}{1}\right) & \left(\color{Peru}{1}\right) + \left(\color{Peru}{2}\right) & \left(\color{Magenta}{0}\right) + \left(\color{Magenta}{5}\right) \end{array} \right]$$$$$=\left[ \begin{array}{ccc} 3 & 1 & 4 \\\\ 5 & 7 & 5 \\\\ 1 & 3 & 5 \end{array} \right]$$$

Answer: $$A + B=\left[ \begin{array}{ccc} 3 & 1 & 4 \\\\ 5 & 7 & 5 \\\\ 1 & 3 & 5 \end{array} \right]$$\$