MATH 240

Mon. September 23rd, 2019

Matrix Inverse

For a matrix AA, the its inverse A−1A^{-1} should have the following property: A−1A=AA−1=In.A^{-1}A=AA^{-1}=I_n.

Ex. Determine if
A=(25−3−7),B=(−7532)\begin{aligned} A&=\begin{pmatrix} 2&5\\-3&-7 \end{pmatrix},\\ B&=\begin{pmatrix} -7&5\\3&2 \end{pmatrix} \end{aligned} are inverses of each other.

AB=(25−3−7)(−7532)=(1001)=I2.\begin{aligned} AB&=\begin{pmatrix} 2&5\\-3&-7 \end{pmatrix}\begin{pmatrix} -7&5\\3&2 \end{pmatrix}\\ &=\begin{pmatrix} 1&0\\0&1 \end{pmatrix}\\ &=I_2. \end{aligned}

Therefore, B=A−1B=A^{-1} and A=B−1A=B^{-1}.

Application

If we had the equation 5x=15x=1, we would solve it by multiplying both sides by 15\frac{1}{5}, the multiplicative inverse of 5. We can use an analogous operation to solve systems of linear equations.

Suppose that AA is an invertible matrix. Then the equation Ax=bAx=b has a unique solution given by x=A−1bx=A^{-1}b. We can prove this using properties of matrix multiplication:

Ax=bA−1(Ax)=A−1b(A−1A)x=A−1bInx=A−1bx=A−1b.\begin{aligned} Ax&=b\\ A^{-1}(Ax)&=A^{-1}b\\ (A^{-1}A)x&=A^{-1}b\\ I_nx&=A^{-1}b\\ x&=A^{-1}b. \end{aligned}

Finding the Inverse of 2x2 Matrices

In the case of a 2x2 matrix A=(abcd)A=\begin{pmatrix} a&b\\c&d \end{pmatrix}, if ad−bc≠0ad-bc\neq 0, then AA has an inverse, which is given by
A−1=1ad−bc(d−b−ca).A^{-1}=\frac{1}{ad-bc}\begin{pmatrix} d&-b\\-c&a \end{pmatrix}.

Ex. Is (5348)\begin{pmatrix} 5&3\\4&8 \end{pmatrix} invertible? (5)(8)−(3)(4)=28≠0,(5)(8)-(3)(4)=28\neq 0, so it is invertible.

Finding the Inverse of Larger Matrices

You can also find the inverse of a matrix by putting an identity matrix to the right of it and row-reducing the left matrix into the identity matrix. Afterwards, the right matrix will be the inverse of the original left matrix.

(Any invertible matrix will be row-equivalent to the identity matrix. If you can’t row-reduce a matrix into the identity matrix, then it’s non-invertible.)

Ex. Find the inverse of AA.
A=(0121034−38)→(0121001030104−38001)→(1030100121004−38001)→(1030100121000−3−40−41)→(1030100121000023−41)→(10301001210000132−212)→(103010010−24−100132−212)→(100−927−32010−24−100132−212)\begin{aligned} A=&\begin{pmatrix} 0&1&2\\1&0&3\\4&-3&8 \end{pmatrix}\\ \rightarrow&\begin{pmatrix} 0&1&2&1&0&0\\1&0&3&0&1&0\\4&-3&8&0&0&1 \end{pmatrix}\\ \rightarrow&\begin{pmatrix} 1&0&3&0&1&0\\0&1&2&1&0&0\\4&-3&8&0&0&1 \end{pmatrix}\\ \rightarrow&\begin{pmatrix} 1&0&3&0&1&0\\0&1&2&1&0&0\\0&-3&-4&0&-4&1 \end{pmatrix}\\ \rightarrow&\begin{pmatrix} 1&0&3&0&1&0\\0&1&2&1&0&0\\0&0&2&3&-4&1 \end{pmatrix}\\ \rightarrow&\begin{pmatrix} 1&0&3&0&1&0\\0&1&2&1&0&0\\0&0&1&\frac{3}{2}&-2&\frac{1}{2} \end{pmatrix}\\ \rightarrow&\begin{pmatrix} 1&0&3&0&1&0\\0&1&0&-2&4&-1\\0&0&1&\frac{3}{2}&-2&\frac{1}{2} \end{pmatrix}\\ \rightarrow&\begin{pmatrix} 1&0&0&-\frac{9}{2}&7&-\frac{3}{2}\\0&1&0&-2&4&-1\\0&0&1&\frac{3}{2}&-2&\frac{1}{2} \end{pmatrix} \end{aligned}

Now that the left side is I3I_3, the right side must be A−1A^{-1}, so
A−1=(−927−32−24−132−212).A^{-1}=\begin{pmatrix} -\frac{9}{2}&7&-\frac{3}{2}\\-2&4&-1\\\frac{3}{2}&-2&\frac{1}{2} \end{pmatrix}.