The inverse of a square matrix - that is, a matrix with an equal number of rows and columns - if it exists, is equal in dimension to that matrix. Therefore, any inverse of a four-by-four matrix is itself a four-by-four matrix.

To prove a statement false, it suffices to find one case in which the statement does not hold. We show that

\(\displaystyle A = \begin{bmatrix} 1& 0\\ 0& 0 \end{bmatrix}\) and \(\displaystyle B= \begin{bmatrix} 0& 0\\ 0& 1 \end{bmatrix}\)

provide a counterexample.

A matrix is singular - that is, without an inverse - if and only if its determinant is equal to zero. The determinant of a two-by-two matrix is equal to the product of its upper left to lower right entries minus that of its upper right to lower left entries, so:

\(\displaystyle \det A = 1 \cdot 0 - 0 \cdot 0 = 0 - 0 = 0\)

\(\displaystyle \det B = 0 \cdot 1 - 0 \cdot 0 = 0 - 0 = 0\)

Both \(\displaystyle A\) and \(\displaystyle B\) are singular. 

Now add the matrices by adding them term by term. 

\(\displaystyle A+ B\)

\(\displaystyle = \begin{bmatrix} 1& 0\\ 0& 0 \end{bmatrix} + \begin{bmatrix} 0& 0\\ 0& 1 \end{bmatrix}\)

\(\displaystyle = \begin{bmatrix} 1+0& 0+0\\ 0+0 & 0+1 \end{bmatrix}\)

\(\displaystyle = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\)

This is simply the two-by-two identity, which has an inverse - namely, itself. 

The statement has been proved false by counterexample.

A matrix is nonsingular - that is, it has an inverse - if and only if its determinant is nonzero. Also, the determinant of the product of two matrices is equal to the product of their individual determinants. Combining these ideas:

\(\displaystyle \det AB = \det A \cdot \det B\)

If either \(\displaystyle \det A = 0\) or \(\displaystyle \det B = 0\), then it must hold that

\(\displaystyle \det AB = 0 \cdot 0 = 0\).

Equivalently, if either \(\displaystyle A\) or \(\displaystyle B\) has no inverse, then \(\displaystyle AB\) has no inverse. Contrapositively, if \(\displaystyle AB\) has an inverse, it must hold that each of \(\displaystyle A\) and \(\displaystyle B\) has an inverse.

We can prove that the sum of two nonsingular matrices need not be nonsingular by counterexample. 

Let \(\displaystyle A = \begin{bmatrix} 1& 0\\ 0& 1 \end{bmatrix}\) , \(\displaystyle B= \begin{bmatrix} -1& 0\\ 0& -1 \end{bmatrix}\).

A matrix is nonsingular - that is, with an inverse - if and only if its determinant is nonzero. The determinant of a two-by-two matrix is equal to the product of its upper left to lower right entries minus that of its upper right to lower left entries, so:

\(\displaystyle \det A = 1 \cdot 1 - 0 \cdot 0 = 1 -0 = 1\)

\(\displaystyle \det B = -1 \cdot (-1) - 0 \cdot 0 = 1 -0 = 1\)

Both \(\displaystyle A\) and \(\displaystyle B\) are nonsingular.

Now add the matrices by adding them term by term. 

\(\displaystyle A + B= \begin{bmatrix} 1& 0\\ 0& 1 \end{bmatrix}+ \begin{bmatrix} -1& 0\\ 0& -1 \end{bmatrix}\)

\(\displaystyle = \begin{bmatrix} 1+ (-1)& 0+0 \\ 0+0& 1+ (-1) \end{bmatrix}\)

\(\displaystyle = \begin{bmatrix} 0 & 0 \\0 & 0 \end{bmatrix}\),

the zero matrix, whose determinant is 0 and which is therefore not nonsingular.

A matrix is singular - that is, it has no inverse - if and only if its determinant is equal to 0. \(\displaystyle R\) is singular, so 

\(\displaystyle \det R = 0\).

The determinant of the scalar product of \(\displaystyle c\) and an \(\displaystyle n \times n\) matrix \(\displaystyle R\) is

\(\displaystyle \det (cR) = c^{n}(\det R)\); 

setting \(\displaystyle c = 5\), \(\displaystyle n = 4\), \(\displaystyle \det R = 0\):

\(\displaystyle \det (5R) = 5^{4 } \cdot 0\)

\(\displaystyle \det (5R) = 0\)

Therefore, \(\displaystyle 5R\), having determinant 0, is also singular.

By definition, 

\(\displaystyle R R^{-1} = I\) and \(\displaystyle R^{-1}R = I\).

Multiply:

\(\displaystyle 3R \cdot \frac{1}{3}R^{-1} = 3 \cdot \frac{1}{3} \cdot RR^{-1} = 1 \cdot I = I\)

Similarly,

\(\displaystyle \frac{1}{3}R^{-1}\cdot 3R = \frac{1}{3} \cdot 3 \cdot R^{-1} R= 1 \cdot I = I\)

Therefore, \(\displaystyle \frac{1}{3}R^{-1}\) is the inverse of \(\displaystyle 3R\).

To prove \(\displaystyle AB\) is invertible, we need to find another square matrix \(\displaystyle D\) such that \(\displaystyle D(AB)=(AB)D = I\).

Since \(\displaystyle A^{-1},B^{-1}\) exist, take \(\displaystyle D = B^{-1}A^{-1}\), then we have

\(\displaystyle D(AB)=(B^{-1}A^{-1})AB = B^{-1}(A^{-1}A)B = B^{-1}B = I\),

and

\(\displaystyle (AB)D = (AB)(B^{-1}A^{-1})=A(BB^{-1})A^{-1}=AA^{-1} = I\).

Hence \(\displaystyle AB\) is invertible.

To simplify

\(\displaystyle AAA^{-1}AAA^{-1}A^{-1}A\)

we used the identities:

\(\displaystyle AA^{-1}=A^{-1}A=I\)

\(\displaystyle AI=IA=A\)

so we get

\(\displaystyle AAA^{-1}AAA^{-1}A^{-1}A=A(AA^{-1})A(AA^{-1})(A^{-1}A)=AIAII=AA=A^2\)

The inverse of \(\displaystyle ABC\) is \(\displaystyle C^{-1}B^{-1}A^{-1}\) since we can multiply it by \(\displaystyle ABC\) to get:

\(\displaystyle C^{-1}B^{-1}A^{-1}ABC=C^{-1}B^{-1}(A^{-1}A)BC=C^{-1}B^{-1}IBC\)

\(\displaystyle =C^{-1}B^{-1}BC=C^{-1}(B^{-1}B)C=C^{-1}C=I\)

Therefore \(\displaystyle C^{-1}B^{-1}A^{-1}\) is the inverse of \(\displaystyle ABC\)

The inverse of a two-by-two matrix

\(\displaystyle A= \begin{bmatrix} a & b \\ c & d \end{bmatrix}\) 

is

\(\displaystyle A{-1}=\frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\)

Substituting the entries in the matrix for the variables:

\(\displaystyle A ^{-1}=\frac{1}{(1+i)(-2) - (1-i)(2)} \begin{bmatrix} -2 & -1 + i \\- 2 & 1 + i \end{bmatrix}\)

\(\displaystyle A ^{-1}=\frac{1}{-2-2i - (2-2i)} \begin{bmatrix} -2 & -1 + i \\- 2 & 1 + i \end{bmatrix}\)

\(\displaystyle A ^{-1}=\frac{1}{-4} \begin{bmatrix} -2 & -1 + i \\- 2 & 1 + i \end{bmatrix}\)

\(\displaystyle A ^{-1}= \begin{bmatrix} \frac{1}{2} & \frac{1}{4} - \frac{1}{4} i \\ \\ \frac{1}{2} & -\frac{1}{4} - \frac{1}{4} i\end{bmatrix}\)