Inverse Matrix Calculator

Invert any square matrix step by step - exact fractions, the row operations spelled out, and the answer verified against the identity.

Size of the matrix:

Matrix A

Method:

Your input

Find the inverse of $[\begin{matrix} 2 & 1 \\ 1 & 3 \end{matrix}]$ .

Answer

The inverse matrix is

[\begin{matrix} \frac{3}{5} & - \frac{1}{5} \\ - \frac{1}{5} & \frac{2}{5} \end{matrix}]

Solution

So, augment the matrix with the identity matrix:

[\begin{matrix} 2 & 1 & 1 & 0 \\ 1 & 3 & 0 & 1 \end{matrix}]

Divide row 1 by 2: $R_{1} = \frac{R_{1}}{2}$ .

[\begin{matrix} 1 & \frac{1}{2} & \frac{1}{2} & 0 \\ 1 & 3 & 0 & 1 \end{matrix}]

Subtract row 1 from row 2: $R_{2} = R_{2} - R_{1}$ .

[\begin{matrix} 1 & \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & \frac{5}{2} & - \frac{1}{2} & 1 \end{matrix}]

Multiply row 2 by 2/5: $R_{2} = \frac{2}{5} * R_{2}$ .

[\begin{matrix} 1 & \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & 1 & - \frac{1}{5} & \frac{2}{5} \end{matrix}]

Subtract row 2 multiplied by 1/2 from row 1: $R_{1} = R_{1} - \frac{1}{2} R_{2}$ .

[\begin{matrix} 1 & 0 & \frac{3}{5} & - \frac{1}{5} \\ 0 & 1 & - \frac{1}{5} & \frac{2}{5} \end{matrix}]

We are done. On the left is the identity matrix. On the right is the inverse matrix.

Answer

The inverse matrix is $[\begin{matrix} \frac{3}{5} & - \frac{1}{5} \\ - \frac{1}{5} & \frac{2}{5} \end{matrix}]$ .

In decimal form: (0.6, −0.2), (−0.2, 0.4)

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How to use the inverse matrix calculator

Pick the size n - the grid is square, from 1×1 to 8×8.
Type a number (decimals and fractions like 1/5 both work) into every cell.
Choose Gauss–Jordan elimination to watch [A | I] reduce to [I | A⁻¹] one row operation at a time, or the adjugate method for the cofactor route.
Read the answer in exact fractions plus its decimal form, or copy it.

Singular matrices are detected automatically - if the determinant is 0, no inverse exists and the write-up says so.

Where inverses get used

Matrix inverses solve linear systems in one step, decode Hill ciphers, and reverse 3D graphics transforms. Computing them by hand is cubic drudgery, which is why Gauss-Jordan elimination is the taught method: augment with the identity, reduce, and the inverse appears on the right. The adjugate route suits small matrices and theory questions about determinants.