(Sec:MatrixInv)= # The inverse of a matrix ## Introduction In {numref}`Sec:MatrixOps` we defined the sum and product of matrices (of compatible sizes), and we saw that to a certain extent matrix algebra is guided by the same rules as the arithmetic of real numbers. We can also subtract two matrices via $$ A - B = A + (-1)B, $$ but we did not mention division of matrices. For two numbers $a$ and $b$, with $a \neq 0$, the equation $$ ax = b $$ has the unique solution $$ x = \frac{b}{a} = a^{-1}b = ba^{-1}, $$ where $$ a^{-1} = \frac1a $$ is the (unique) solution to the equation $$ ax = 1. $$ The bad news: $$ \frac{A}{B}~\text{ cannot be defined in any useful way!} $$ First of all the corresponding matrix equation $$ AX = B, \,\, A\neq O $$ does not always have a solution, or the solution is not unique, not even in the case of two $n \times n$-matrices $A$ and $B$. The following two examples to illustrate this. ::::::{prf:example} :label: Ex:MatrixInv:MatrixEqNoSolution The matrix equation $$ AX = B $$ where $$ A = \begin{pmatrix} 1 & 2 \\ 1 & 2 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} $$ does not have a solution. Why? Well, any column of $AX$ is a linear combination of the columns of $A$, and the columns of $B$ obviously cannot be written as such linear combinations: $$ \begin{pmatrix} 1 \\ 0 \end{pmatrix} \neq c_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2 \begin{pmatrix} 2 \\ 2 \end{pmatrix} \quad \text{for all } c_1,c_2 \quad\text{in } \mathbb{R}. $$ :::::: ::::::{prf:example} :label: Ex:MatrixInv:MatrixEqInfSolutions The matrix equation $$ AX = B $$ where $$ A = \begin{pmatrix} 1 & 2 \\ 1 & 2 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 4 \\ 1 & 4 \end{pmatrix} $$ has infinitely many solutions. Two of those are for instance $$ X_1 = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \quad \text{and} \quad X_2 = \begin{pmatrix} -1 & 2 \\ 1 & 1 \end{pmatrix}. $$ :::::: And lastly, if there is a matrix $C$ for which $$ CA = I $$ and we would adopt the notation $$ C = A^{-1} $$ then $$ \begin{array}{rclcrcl} AX &=& B &\Rightarrow& C(AX) &=& CB \\ &&&\Rightarrow& (CA)X &=& A^{-1}B \\ &&&\Rightarrow& IX &=& A^{-1}B \\ &&&\Rightarrow& X &=& A^{-1}B. \end{array} $$ So $X = A^{-1}B$. However, it is in no way clear why $A^{-1}B$ and $BA^{-1}$ should be equal, and in general indeed they are not. So the notation $$ \dfrac{B}{A} $$ will still be ambiguous. For non-square matrices things are even worse. In this section we will only consider square matrices. (Subsec:MatrixInv:DefInverse)= ## Definition and basic properties of the inverse ::::::{prf:definition} :label: Dfn:MatrixInv:DefInverse A square matrix $A$ is called **invertible** if there exists a matrix $B$ for which $$ AB = BA = I. $$ In this situation the matrix $B$ is called the **inverse** of $A$ and we write $$ B = A^{-1}. $$ A matrix that is invertible is also called a **regular matrix**, and a non-invertible matrix is also called a **singular matrix**. :::::: Note the use of the definite article **the** in the sentence ''$B$ is called **the** inverse of $A$''. The following proposition justifies this choice of word. ::::::{prf:proposition} :label: Prop:MatrixInv:UniqueInverse If an inverse of a matrix $A$ exists, then it is unique. :::::: The proof is very short, when we plug in the right idea at the right place. ::::::{admonition} Proof of {prf:ref}`Prop:MatrixInv:UniqueInverse` :class: tudproof Suppose $B$ and $C$ are two matrices that satisfy the properties of being an inverse of $A$, i.e. $$ AB = BA = I \quad \text{and} \quad AC = CA = I. $$ Then the following chain of identities proves that $B$ and $C$ must be equal: $$ B = B\,I \,= B\,(AC) = (BA)\,C= I\,C = C. $$ :::::: ::::::{prf:remark} :label: Rem:MatrixInv:RightInvLeftInv2 Actually, the proof shows slightly more, as the assumptions $$ CA= I, \quad AB = I $$ are not used. In fact it shows that for three $n \times n$-matrices $A$, $B$ and $C$ $$ \text{if} \quad BA = I \quad \text{ and }\quad AC = I \quad\text{ then } \quad B = C. $$ :::::: ::::::{prf:example} :label: Ex:FirstInverse For the matrices $$ A = \begin{pmatrix} 1 & 2 \\ 3 & 5 \end{pmatrix} \quad \text{and} \quad B = \begin{pmatrix} -5 & 2 \\ 3 & -1 \end{pmatrix} $$ we see $$ \begin{pmatrix} 1 & 2 \\ 3 & 5 \end{pmatrix} \begin{pmatrix} -5 & 2 \\ 3 & -1 \end{pmatrix} = \begin{pmatrix} -5 & 2 \\ 3 & -1 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 5 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}. $$ So $A$ and $B$ are each other's inverse. Another example: $$ \begin{pmatrix} 1 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} 0 & 1 & -1 \\ 1 & -1 & 1 \\ -1 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}. $$ You may check for yourself that the product in the other order also gives $I$, so $$ \begin{pmatrix} 1 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \end{pmatrix}^{-1} = \begin{pmatrix} 0 & 1 & -1 \\ 1 & -1 & 1 \\ -1 & 1 & 0 \end{pmatrix}. $$ It will appear ({prf:ref}`Rem:MatrixInv:RightInvLeftInv`) that for square matrices, a one-sided inverse is automatically a two-sided inverse, by which we mean $$ \text{if} \quad AB = I \quad \text{then also}\quad BA = I. $$ :::::: The first example can be generalised: ::::::{prf:proposition} :label: Prop:MatrixInv:Inverse2x2 If $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, then $A^{-1}$ exists if and only if $$ ad - bc \neq 0. $$ In that case $$ A^{-1} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \begin{pmatrix} \dfrac{d}{ad - bc} & \dfrac{-b}{ad - bc} \\ \dfrac{-c}{ad - bc} & \dfrac{a}{ad - bc} \end{pmatrix} = \frac{1}{ad-bc}\begin{pmatrix} d &- b \\ -c & a \end{pmatrix}. $$ :::::: We leave the verification as an exercise. ::::::{exercise} :label: Exc:MatrixInv:CheckBA=I Verify that the matrix $B=A^{-1}$ proposed in {prf:ref}`Prop:MatrixInv:Inverse2x2` indeed satisfies $$ AB = BA = I. $$ Also check that the first matrix in {prf:ref}`Ex:FirstInverse` illustrates the formula. :::::: ::::::{admonition} Solution to {numref}`Exc:MatrixInv:CheckBA=I` :class: solution, dropdown $$ \begin{array}{rcl} BA &=& \dfrac{1}{ad-bc}\begin{pmatrix} d &-b \\ -c & a \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix}\\ &=& \dfrac{1}{ad-bc}\begin{pmatrix} da-bc &db- bd \\ -ca+ac & -cb+ad \end{pmatrix} \\ &=& \begin{pmatrix} \dfrac{da-bc}{ad-bc} &0 \\ 0 & \dfrac{-cb+ad}{ad-bc} \end{pmatrix} = \begin{pmatrix} 1&0 \\ 0 & 1 \end{pmatrix}. \end{array} $$ Which is one of the two identities. Applying the formula of {prf:ref}`Prop:MatrixInv:Inverse2x2` to the matrix $A = \begin{pmatrix} 1 & 2 \\ 3 & 5 \end{pmatrix}$ of {prf:ref}`Ex:FirstInverse` gives $$ A^{-1} = \dfrac{1}{1\cdot5 - 2\cdot 3}\begin{pmatrix} 5 & -2 \\ -3&1 \end{pmatrix} = -\begin{pmatrix} 5 & -2 \\ -3&1 \end{pmatrix} = \begin{pmatrix} -5 & 2 \\ 3&-1 \end{pmatrix}, $$ which is indeed the matrix $B$ that was proposed there. :::::: ::::::{prf:remark} :label: Rem:MatrixInvDetZeroDependentColumns The condition $$ ad - bc \neq 0 $$ is equivalent to the statement $$ \text{the vectors } \begin{pmatrix} a \\ c \end{pmatrix} \text{ and } \begin{pmatrix} b \\ d \end{pmatrix} \text{ are linearly independent.} $$ First we show that $$ ad - bc = 0 \text{ implies that } \begin{pmatrix} a &b\\ c&d \end{pmatrix} \text{ has linearly dependent columns.} $$ It is best to split this in two cases: $$ ad = 0 \quad \text{and} \quad ad \neq 0. $$ If we assume $$ ad - bc = 0 \quad \text{and} \quad ad = 0, $$ then we have $$ b = 0 \quad \text{or} \quad c = 0 $$ which leads to a matrix $$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} $$ with either a zero row or a zero column, which will indeed have linearly dependent columns. Second, if we assume $$ ad - bc = 0 \quad\text{and} \quad ad \neq 0 $$ then both $a \neq 0$ and $d \neq 0$, in which case $$ d = \frac{bc}{a}, \quad \text{so } \begin{pmatrix} b \\ d \end{pmatrix} = \begin{pmatrix} b \\ \frac{bc}{a} \end{pmatrix} = \dfrac{b}{a}\begin{pmatrix} a \\ c \end{pmatrix}, $$ hence the columns are again linearly dependent. Thus we have shown: $$ ad-bc = 0 \quad \Longrightarrow \quad \begin{pmatrix} a & b \\ c & d \end{pmatrix} \text{ has linearly dependent columns.} $$ Next let us consider the converse, i.e. $$ \begin{pmatrix} a & b \\ c&d \end{pmatrix} \text{ has linearly dependent columns} \quad \text{implies: } \quad ad - bc = 0. $$ If a $2 \times 2$-matrix has two linearly dependent columns, then one of the columns will be a multiple of the other column, e.g. $$ \text{either } \quad \begin{pmatrix} a \\ c \end{pmatrix} = k \begin{pmatrix} b \\ d \end{pmatrix} \quad \text{or}\quad \begin{pmatrix} b \\ d \end{pmatrix} = k \begin{pmatrix} a \\ c \end{pmatrix} . $$ In both cases it is easily checked that $$ ad-bc = 0. $$ :::::: The following proposition shows that the above considerations can be generalised. ::::::{prf:proposition} :label: Prop:MatrixInv:InvertibleIndepCols If $A$ is a square matrix, then $$ AX = I $$ has a unique solution if and only if $$ A \text{ has linearly independent columns.} $$ :::::: ::::::{admonition} Proof of {prf:ref}`Prop:MatrixInv:InvertibleIndepCols` :class: tudproof As in the proof in {prf:ref}`Rem:MatrixInvDetZeroDependentColumns` we have to prove two implications: $$ AX = I \text{ has a unique solution } \quad\Longrightarrow \quad A \text{ has linearly independent columns} $$ and $$ A \text{ has linearly independent columns} \quad\Longrightarrow \quad AX = I \text{ has a unique solution.} $$ For the first part, assume that $$ AX = I \quad \text{has a (unique) solution.} $$ That means that every column $\vect{e}_j$ of the identity matrix is a linear combination of columns $\vect{a}_1, \ldots, \vect{a}_n$ of $A$. So the span of the columns of $A$ contains the span of the columns of $\vect{e}_1, \ldots, \vect{e}_n$, which is the whole $\mathbb{R}^n$. Thus **every** linear system $$ A\vect{x} =\vect{b}, \quad \vect{b} \in \mathbb{R}^{n} $$ has a solution. Then the reduced echelon form of $A$ must have a pivot in every row, and, since it is a square matrix, it must be the identity matrix. Consequently, it has a pivot in every column, so the linear system $$ A\vect{x} =\vect{0} $$ only has the trivial solution, which proves that indeed the columns of $A$ are linearly independent. For the converse, suppose that $A$ has linearly independent columns. Then the reduced echelon form of $A$ must be the identity matrix. This implies that for each $\vect{b}$ in $\mathbb{R}^n$ $$ (\,A\,|\,\vect{b}\,) \sim(\,I\,|\,\vect{b'}\,), $$ and in particular, each linear system $$ A\vect{x} =\vect{e}_j $$ has a unique solution. If we denote this solution by $\vect{c}_j$ we have that $$ A(\,\vect{c}_1\,\,\vect{c}_2\,\, \cdots \,\, \vect{c}_n\,) = (\,A\vect{c}_1\,\,A\vect{c}_2\,\, \cdots \,\, A\vect{c}_n\,) = (\,\vect{e}_1\,\,\vect{e}_2\,\, \cdots \,\, \vect{e}_n\,) = I. $$ Since all solutions $\vect{c}_j$ are unique, the solution to the equation $$ AX = I $$ is unique as well. :::::: It makes sense that the solution $B$ of this matrix equation will be the inverse of $A$, and it is, but it takes some effort to show that the other requirement, $$ BA = I $$ is also fulfilled. In the next subsection we will see that the matrix equation $$ AX = I $$ will lead the way to an algorithm to compute the inverse of a matrix. Before we go there we will look at some general properties of invertible matrices. ::::::{prf:proposition} :label: Prop:SolutionViaInverse If the $n \times n$-matrix $A$ is invertible and $B$ is an $n \times p$-matrix, then the solution to the matrix equation $$ AX = B $$ is unique, and given by $$ X = A^{-1}B. $$ In particular, if the matrix $B$ has only one column, i.e., if it is a vector, then $$ A\mathbf{x} = \mathbf{b} \quad \text{has the unique solution} \quad \mathbf{x} = A^{-1}\mathbf{b}. $$ :::::: ::::::{admonition} Proof of {prf:ref}`Prop:SolutionViaInverse` :class: tudproof We multiply both sides of the equation $$ AX = B $$ by $A^{-1}$ and use the fact that the matrix product has the associative property: $$ \begin{array}{rclcrcl} AX &=& B &\Longrightarrow& A^{-1}(AX) &=& A^{-1}B \\ &&& \Longrightarrow& (A^{-1}A)X &=& A^{-1}B \\ &&& \Longrightarrow& IX &=& A^{-1}B \\ &&&\Longrightarrow& X &=& A^{-1}B. \end{array} $$ :::::: We illustrate the proposition by an example. ::::::{prf:example} :label: Ex:MatrixInv:SolutionViaInverse Suppose the matrix $A$ and the vectors $\mathbf{b}_1$ and $\mathbf{b}_2$ are given by $$ A=\begin{pmatrix}1 & 2 \\ 3 & 4 \end{pmatrix}, \quad \mathbf{b}_1= \begin{pmatrix}-1 \\ 1 \end{pmatrix} \quad \text{and} \quad \mathbf{b}_2=\begin{pmatrix}2 \\ 10 \end{pmatrix}. $$ Consider the two linear systems $$ A\mathbf{x} =\mathbf{b}_1\quad \text{and} \quad A\mathbf{x} = \mathbf{b}_2. $$ Using the inverse matrix $$ A^{-1} = \frac{1}{-2}\begin{pmatrix}4 & -2 \\ -3 & 1 \end{pmatrix} = \frac{1}{2}\begin{pmatrix}-4 & 2 \\ 3 & -1 \end{pmatrix}, $$ the two solutions are quickly written down: $$ \vect{x}_1= A^{-1}\vect{b}_1= \frac{1}{2}\begin{pmatrix}-4 & 2 \\ 3 & -1 \end{pmatrix} \begin{pmatrix}-1 \\ 1 \end{pmatrix} = \begin{pmatrix}3 \\ -2 \end{pmatrix} $$ and likewise $$ \vect{x}_2= A^{-1}\vect{b}_2= \frac{1}{2}\begin{pmatrix}-4 & 2 \\ 3 & -1 \end{pmatrix} \begin{pmatrix}2 \\ 10 \end{pmatrix} = \begin{pmatrix}6 \\ -2 \end{pmatrix}. $$ :::::: ::::::{warning} The proof of {prf:ref}`Prop:SolutionViaInverse` is based on the **existence** of the inverse of the matrix $A$. Beware of this: never start using the expression $A^{-1}$ unless you have made sure first that the matrix $A$ is indeed invertible. If not, you may lead yourself into inconsistencies like in the following example. :::::: ::::::{prf:example} :label: Ex:MatrixInv:FallaciousProof What goes wrong in the following 'proof' of the statement: $$ \text{ if } \quad A^2 = A \quad\text{ and } \quad A\neq O, \quad\text{ then } \quad A = I. $$ :::::{admonition} 'Fallacious' proof :class: tudproof Assume $A^2 = A$. Then $$ A^{-1}A^2 = A^{-1}A = I. $$ On the other hand $$ A^{-1}A^2 = A^{-1}(A\,A) = (A^{-1}A)A = IA = A. $$ So $$ I = A^{-1}A^2 = A, $$ which 'proves' that $A=I$. ::::: Somewhere something **must** have gone wrong, as the following counterexample shows. For the matrix $B = \begin{pmatrix} \frac12 & \frac12 \\ \frac12 & \frac12 \end{pmatrix}$ it can be checked that $$ B^2 = B, $$ whereas obviously $$ B \neq O, \quad B \neq I. $$ So, where exactly did it go wrong?! :::::: The next proposition contains a few rules to manipulate inverse matrices. ::::::{prf:proposition} :label: Prop:MatrixInv:ElemProperties If $A$ is invertible and $c \neq 0$, then the following is true
  1. The matrix $cA$ is invertible, and $$ (cA)^{-1} = \dfrac1c A^{-1}. $$
  2. The matrix $A^T$ is invertible, and $$ (A^T)^{-1} = (A^{-1})^T. $$
  3. The matrix $A^{-1}$ is invertible, and $$ (A^{-1})^{-1} = A. $$
:::::: ::::::{admonition} Proof of {prf:ref}`Prop:MatrixInv:ElemProperties` :class: tudproof All statements can be proved by verifying that the relevant products are equal to $I$.
  1. The matrix $A^{-1}$ exists, and so does $\dfrac1c A^{-1}$. We find: $$ (cA)\,\dfrac1c A^{-1} = c\,\dfrac1c A\,A^{-1} = 1 \, I = I, $$ and likewise $\dfrac1c A^{-1}\,(cA) = I$, which proves that indeed $\dfrac1c A^{-1} = (cA)^{-1}$.
  2. Since it is given that $A^{-1}$ exists we can proceed as follows, where we make use of the characteristic property $ B^TA^T = (AB)^T$. $$ (A^{-1})^TA^T = ( AA^{-1})^T = I^T = I $$ and $$ A^T(A^{-1})^T =( A^{-1}A)^T = I^T = I, $$ which settles the second statement.
To prove iii., see {numref}`Exc:MatrixInv:Ainvinv`. :::::: ::::::{exercise} :label: Exc:MatrixInv:Ainvinv Prove the last statement of the previous proposition. :::::: ::::::{admonition} Solution to {numref}`Exc:MatrixInv:Ainvinv` :class: solution, dropdown For the inverse $C = (A^{-1})^{-1}$ of $A^{-1}$, it should hold that $$ CA^{-1} = A^{-1}C = I. $$ The matrix $C = A$ has these properties. :::::: The next example gives an illustration of [ii.](#Item:MatrixInv:TransposeInverse) in {prf:ref}`Prop:MatrixInv:ElemProperties`. ::::::{prf:example} :label: Ex:MatrixInv:TransposeInverse We consider the matrix $$ A = \begin{pmatrix} 2 & 6 & 5 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{pmatrix}. $$ It has the inverse matrix $$ B = \begin{pmatrix} 1/2 & -3/2 & 1/6 \\ 0 & 1/2 & -1/3 \\ 0 & 0 & 1/3 \end{pmatrix}, $$ which can be checked by showing that $AB$ and $BA$ are equal to $I$. So $B = A^{-1}$, and $B^T = (A^{-1})^T$. We also have $$ A^TB^T = \begin{pmatrix} 2 & 0 & 0 \\ 6 & 2 & 0 \\ 5 & 2 & 3 \end{pmatrix} \begin{pmatrix} 1/2 & 0 & 0 \\ -3/2 & 1/2 & 0 \\ 1/6 & -1/3 & 1/3 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}, $$ as well as $B^TA^T = I$, which proves that $B^T = (A^T)^{-1}$. As we already saw that $B^T = (A^{-1})^{T}$, the matter is settled: $$ (A^{-1})^T = (A^T)^{-1}. $$ :::::: The last property we mention and prove is the product rule for the matrix inverse. ::::::{prf:proposition} :label: Prop:MatrixInv:ProductRule If $A$ and $B$ are invertible $n \times n$-matrices then the matrix $AB$ is also invertible, and $$ (AB)^{-1} = B^{-1}A^{-1}. $$ :::::: ::::::{admonition} Proof of {prf:ref}`Prop:MatrixInv:ProductRule` :class: tudproof Again we just check that the properties of the definition hold. Suppose that $A$ and $B$ are invertible with inverses $A^{-1}$ and $B^{-1}$. Then using the associative property we find $$ (B^{-1}A^{-1})(AB) = B^{-1}A^{-1}AB = B^{-1}(A^{-1}A)B = B^{-1}IB = B^{-1}B = I, $$ and along the same lines $$ (AB) B^{-1}A^{-1} = I. $$ This shows that $B^{-1}A^{-1}$ is indeed the inverse of $AB$. :::::: ::::::{exercise} :label: Exc:MatrixInv:(AB)Tinv Is the identity $$ ((AB)^T)^{-1} = (A^T)^{-1}(B^T)^{-1} $$ true or false? In case it is true, give an argument, when false, give a counterexample. :::::: ::::::{admonition} Solution to {numref}`Exc:MatrixInv:(AB)Tinv` :class: solution, dropdown The statement is _true_. From the two properties $$ (AB)^T = B^TA^T, \quad (AB)^{-1} = B^{-1}A^{-1} $$ it follows that $$ ((AB)^T)^{-1} = (B^TA^T)^{-1} = (A^T)^{-1}(B^T)^{-1}. $$ :::::: (Subsec:MatrixInverse:ComputeInverse)= ## How to compute the inverse The construction of the inverse of a matrix was already present implicitly in {prf:ref}`Prop:MatrixInv:InvertibleIndepCols`. The inverse of the matrix $A$ must satisfy the equation $AX = I$. Written out column by column this means that $$ AX = I \quad \iff \quad A(\,\mathbf{x}_1\,\,\mathbf{x}_2\, \cdots\, \mathbf{x}_n\,) = (\,\mathbf{e}_1\,\mathbf{e}_2\, \cdots\, \mathbf{e}_n\,). $$ For the existence of a solution to this Equation {prf:ref}`Prop:MatrixInv:InvertibleIndepCols` tells us it is *necessary* that $A$ has linearly independent columns, and we can furthermore read off that the columns of the matrix $X$ will be the (unique) solutions of the linear systems $$ A\mathbf{x}_k = \mathbf{e}_k, $$ where $k = 1,2,\ldots, n$. Let us first focus on this equation by considering a fairly general $3\times 3$-matrix $A$. ::::::{prf:example} :label: Ex:MatrixInv:SolveAX=I For the matrix $$ A = \begin{pmatrix} 1 & 1 & 4 \\ 1 & -1 & -1 \\ 2 & -2 & -4 \end{pmatrix} $$ we find the solution $B$ of the matrix equation (which will appear to exist) $$ AX = I $$ and then check whether $$ BA = I $$ also holds. In which case we can truthfully assert that $B = A^{-1}$. Instead of finding the solution $X$ column by column, which gives three linear systems with the same coefficient matrix, $$ \left(\begin{array}{rrr|r}1 & 1 & 4 & 1\\1 & -1 & -1 & 0\\2 & -2 & -4 & 0\\\end{array}\right), \quad \left(\begin{array}{rrr|r}1 & 1 & 4 & 0\\1 & -1 & -1 & 1\\2 & -2 & -4 & 0\\\end{array}\right), \quad \left(\begin{array}{rrr|r}1 & 1 & 4 & 0\\1 & -1 & -1 & 0\\2 & -2 & -4 & 1\\\end{array}\right), $$ we can solve the three linear systems simultaneously using a combined augmented matrix which we may denote by either $$ \left(\begin{array}{rrr|r|r|r}1 & 1 & 4 & 1 & 0 & 0\\1 & -1 & -1 & 0 & 1 & 0\\2 & -2 & -4 & 0 & 0 & 1\\\end{array}\right) \quad \text{or} \quad \left(\begin{array}{rrr|rrr}1 & 1 & 4 & 1 & 0 & 0\\1 & -1 & -1 & 0 & 1 & 0\\2 & -2 & -4 & 0 & 0 & 1\\\end{array}\right)= \left(\, A \rule[-1.5ex]{0ex}{4ex}\,\,\,\,\middle|\,\,\rule[-1.5ex]{0ex}{4ex}\,\,I\,\right). $$ Let us first row reduce this matrix and then draw conclusions: $$ \begin{array}{rcl} \left(\, A \rule[-1.5ex]{0ex}{4ex}\,\,\,\,\middle|\,\,\rule[-1.5ex]{0ex}{4ex}\,\,I\,\right) &=& \left(\begin{array}{rrr|rrr}1 & 1 & 4 & 1 & 0 & 0\\1 & -1 & -1 & 0 & 1 & 0\\2 & -2 & -4 & 0 & 0 & 1 \end{array}\right)\begin{array}{l} [R_1] \\ {[R_2-1R_1]} \\ {[R_3-2R_1]} \end{array} \\ &\sim& \left(\begin{array}{rrr|rrr}1 & 1 & 4 & 1 & 0 & 0\\0 & -2 & -5 & -1 & 1 & 0\\0 & -4 & -12 & -2 & 0 & 1 \end{array}\right)\begin{array}{l} [R_1+\nicefrac12R_2] \\ {[R_2]} \\ {[R_3-2R_2]} \end{array} \\ &\sim& \left(\begin{array}{rrr|rrr}1 & 0 & 3/2 & 1/2 & 1/2 & 0\\0 & -2 & -5 & -1 & 1 & 0\\0 & 0 & -2 & 0 & -2 & 1 \end{array}\right)\begin{array}{l} [R_1+\nicefrac34R_3] \\ {[R_2-\nicefrac52R_3]} \\ {[R_3]} \end{array} \\ &\sim& \left(\begin{array}{rrr|rrr}1 & 0 & 0 & 1/2 & -1 & 3/4\\0 & -2 & 0 & -1 & 6 & -5/2\\0 & 0 & -2 & 0 & -2 & 1 \end{array}\right)\begin{array}{l} [R_1] \\ {[(-\nicefrac12)R_2]} \\ {[(-\nicefrac12)R_3]} \end{array} \\ &\sim& \left(\begin{array}{rrr|rrr}1 & 0 & 0 & 1/2 & -1 & 3/4\\0 & 1 & 0 & 1/2 & -3 & 5/4\\0 & 0 & 1 & 0 & 1 & -1/2 \end{array}\right) \\ &=& \left(\, I \rule[-1.5ex]{0ex}{4ex}\,\,\,\,\middle|\,\,\rule[-1.5ex]{0ex}{4ex}\,\,B\,\right). \end{array} $$ By construction we have that the matrix $$ B = \begin{pmatrix} 1/2 & -1 & 3/4 \\ 1/2 & -3 & 5/4 \\ 0 & 1 & -1/2 \end{pmatrix} = \frac14 \begin{pmatrix} 2 & -4 & 3 \\ 2 & -12 & 5 \\ 0 & 4 & -2 \end{pmatrix} $$ satisfies $$ AB = I. $$ Let us check the product in the other order $$ BA = \frac14 \begin{pmatrix} 2 & -4 & 3 \\ 2 & -12 & 5 \\ 0 & 4 & -2 \end{pmatrix} \begin{pmatrix} 1 & 1 & 4 \\ 1 & -1 & -1 \\ 2 & -2 & -4 \end{pmatrix} = \frac14 \begin{pmatrix} 4 & 0 & 0 \\0 & 4 & 0 \\ 0 & 0 & 4 \end{pmatrix} = I. $$ So indeed we can conclude $$ \begin{pmatrix} 1 & 1 & 4 \\ 1 & -1 & -1 \\ 2 & -2 & -4 \end{pmatrix}^{-1} \,=\, \frac14 \begin{pmatrix} 2 & -4 & 3 \\ 2 & -12 & 5 \\ 0 & 4 & -2 \end{pmatrix}\,. $$ :::::: Now was this just beginners' luck? It wasn't, as the next proposition shows. ::::::{prf:proposition} :label: Prop:MatrixInv:Algorithm A square matrix $A$ is invertible if and only it has linearly independent columns. In that case the inverse can be found by reducing the matrix $$ \left(\, A \rule[-1.5ex]{0ex}{4ex}\,\,\,\,\middle|\,\,\rule[-1.5ex]{0ex}{4ex}\,\,I\,\right) $$ to the reduced echelon form $$ \left(\, I \rule[-1.5ex]{0ex}{4ex}\,\,\,\,\middle|\,\,\rule[-1.5ex]{0ex}{4ex}\,\,B\,\right), $$ and then $$ B = A^{-1}. $$ :::::: ::::::{admonition} Proof of {prf:ref}`Prop:MatrixInv:Algorithm` :class: tudproof We have already seen ({prf:ref}`Prop:MatrixInv:InvertibleIndepCols`) that an invertible matrix has linearly independent columns, which implies that the reduced echelon form of $A$ is indeed the identity matrix. It is then clear that via row operations we get $$ \left(\, A \rule[-1.5ex]{0ex}{4ex}\,\,\,\,\middle|\,\,\rule[-1.5ex]{0ex}{4ex}\,\,I\,\right)\sim \quad .\,.\,.\,.\,. \quad \sim \left(\, I \rule[-1.5ex]{0ex}{4ex}\,\,\,\,\middle|\,\,\rule[-1.5ex]{0ex}{4ex}\,\,B\,\right), $$ where the matrix $B$ satisfies $AB = I$. What we have to show is that $$ BA = I $$ as well. To understand that this is indeed true, we recall ({prf:ref}`Dfn:MatrixOps:ElementaryMatrix`) that row operations can be effectuated via multiplications with elementary matrices. Furthermore, since the matrix product is defined column by column, i.e. $$ MX = M\left(\begin{array}{cccc}\mathbf{x_1} &\mathbf{x_2} &\cdots &\mathbf{x_p} \end{array}\right)= \left(\begin{array}{cccc}M\mathbf{x_1} &M\mathbf{x_2} &\cdots &M\mathbf{x_p} \end{array}\right), $$ we also have $$ E\left(\, A_1 \,\, \middle|\,\,\,A_2\,\right)= \left(\, EA_1 \,\, \middle|\,\,\,\,EA_2\,\right). $$ A series of $k$ row operations can be mimicked by $k$ multiplications with elementary matrices: $$ \begin{array}{ccl} \left(\, A \,\, \middle|\,\, I\,\right)&\sim& \left(\, E_1A \,\, \middle|\,\, E_1I\,\right) \sim \left(\,E_2 E_1A \,\, \middle|\,\, \,E_2E_1I\,\right)\sim \cdots \sim \\ &\sim& \left(\,E_k\cdots E_2 E_1A \,\,\, \middle|\,\,\, E_k\cdots E_2E_1I\,\right) = \left(\, I \,\, \middle|\,\,B\,\right). \end{array} $$ So the matrix $B$ that was found as the solution to the matrix equation $$ AX = I $$ is the product of all the elementary matrices by which $A$ is reduced to the identity matrix. Thus we have shown that indeed $$ BA = (E_k\cdots E_2 E_1)A = I. $$ :::::: ::::::{prf:remark} :label: Rem:MatrixInv:RightInvLeftInv In the proof we in fact showed that for a **square** matrix $A$: $$ \text{if} \quad AB = I \quad \text{then} \quad BA = I. $$ For non-square matrices this statement is not correct. The interested reader is invited to take a look at the last exercises in the {numref}`Subsection %s ` containing the Grasple exercises. %{prf:ref}`grasple_exercise_3_4_23` and {prf:ref}`grasple_exercise_3_4_24`. :::::: ::::::{prf:remark} :label: Rem:MatrixInv:RowReductionInverse If $A$ is not invertible, then the outcome of the row reduction of $$ \left(\, A \,\,\middle|\,\,\,I\,\right) $$ will also lead to the correct answer: as soon as it is clear that $A$ cannot be row reduced to $I$ we can conclude that $A$ is not invertible. :::::: To help understand the above exposition let us run through the whole procedure for a specific matrix . ::::::{prf:example} :label: Ex:MatrixInverse:InverseInTwoWays We want to compute the inverse of the matrix $$ A = \begin{pmatrix} 1 & 4 \\ 2 & 6 \end{pmatrix}. $$ The short way: $$ \begin{array}{rcl} \left(\begin{array}{rr|rr}1 & 4 & 1 & 0\\2 & 6 & 0 & 1 \end{array}\right) \begin{array}{l} [R_1] \\ {[R_2-2R_1]} \\ \end{array} \!\!\! &\sim& \left(\begin{array}{rr|rr} 1 & 4 & 1 & 0 \\ 0 & -2 & -2 & 1 \end{array}\right) \begin{array}{l} [R_1+2R_2] \\ {[R_2]} \\ \end{array} \\ &\sim& \left(\begin{array}{rr|rr}1 & 0 & -3 & 2\\0 & -2 & -2 & 1 \end{array}\right) \begin{array}{l} [R_1] \\ {[(-\frac12)R_2]} \\ \end{array} \\ &\sim& \left(\begin{array}{rr|rr}1 & 0 & -3 & 2\\0 & 1 & 1 & -\nicefrac12 \end{array}\right) \end{array} $$ So: $$ A^{-1} = \begin{pmatrix} -3 & 2 \\ 1 & -\frac12 \end{pmatrix}. $$ End of story. To see how the proof of {prf:ref}`Prop:MatrixInv:Algorithm` works for this specific matrix, we will give a derivation using elementary matrices. First step: row replacement with the entry on position (1,1) as a first pivot: $$ \begin{pmatrix} 1 & 0 \\ -2 & 1 \end{pmatrix}\, \left(\begin{array}{rr|rr}1 & 4 & 1 & 0\\2 & 6 & 0 & 1 \end{array}\right) \,\,=\,\, \left(\begin{array}{rr|rr}1 & 4 & 1 & 0\\0 & -2 & -2 & 1 \end{array}\right), \quad E_1 = \begin{pmatrix} 1 & 0 \\ -2 & 1 \end{pmatrix}. $$ Second step: another row replacement, using the entry on position (2,2) as pivot: $$ \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\, \left(\begin{array}{rr|rr}1 & 4 & 1 & 0\\0 & -2 & -2 & 1 \end{array}\right) \,\,=\,\, \left(\begin{array}{rr|rr}1 & 0 & -3 & 2\\0 & -2 & -2 & 1 \end{array}\right), \quad E_2 = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}. $$ Third step: the scaling of the second row: $$ \begin{pmatrix} 1 & 0 \\ 0 & -\nicefrac12 \end{pmatrix}\, \left(\begin{array}{rr|rr}1 & 0 & -3 & 2\\0 & -2 & -2 & 1 \end{array}\right) \,\,=\,\, \left(\begin{array}{rr|rr}1 & 0 & -3 & 2\\0 & 1 & 1 & -\nicefrac12 \end{array}\right),\quad E_3 = \begin{pmatrix} 1 & 0 \\ 0 & -\nicefrac12 \end{pmatrix}. $$ All in all $$ (E_3E_2E_1)A = \left(\begin{pmatrix} 1 & 0 \\ 0 & -\nicefrac12 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ -2 & 1 \end{pmatrix}\right)\,A = \begin{pmatrix} -3 & 2 \\ 1 & -\nicefrac12 \end{pmatrix}A = I, $$ which reconfirms $$ E_3E_2E_1 = A^{-1} = \begin{pmatrix} -3 & 2 \\ 1 & -\tfrac12 \end{pmatrix}. $$ :::::: ::::::{exercise} :label: Exc:MatrixInv:ConverseProdRule Prove the following converse of {prf:ref}`Prop:MatrixInv:ProductRule`. If $A$ and $B$ are $n\times n$-matrices for which the product $AB$ is invertible, then $A$ and $B$ are both invertible. Make sure that you do not use $A^{-1}$ or $B^{-1}$ prematurely, i.e., before you have established that they exist. :::::: ::::::{admonition} Solution to {numref}`Exc:MatrixInv:ConverseProdRule` :class: solution, dropdown Suppose $A$ and $B$ are two $n \times n$-matrices for which $AB$ is invertible. Let $C=(AB)^{-1}$ be the inverse of $AB$. We claim that $BC$ is the inverse of $A$. Now since $$ A(BC) = (AB)C = AB(AB)^{-1} = I, $$ it follows from {prf:ref}`Rem:MatrixInv:RightInvLeftInv` that $(BC)A=I$ also holds. So we have $$ A(BC) = (BC)A = I, $$ which means that $A$ is invertible and has as inverse the matrix $BC$. In the same vein it is shown that $CA$ is the inverse of $B$. :::::: (Subsec:MatrixInverse:Summary)= ## Characterisations of invertibility In the previous subsections quite a few properties of invertible matrices came along, either explicitly or implicitly. For future reference we list them in a theorem. Recall that by definition a (square) matrix $A$ is invertible (or regular) if and only if there exists a matrix $B$ for which $$ AB = BA= I. $$ ::::::{prf:theorem} :label: Thm:MatrixInv:InvertibilityCharacterizations For an $n\times n$-matrix $A$, the following statements are equivalent.
That is, each of the following properties is a characterisation of invertibility of a square matrix $A$.
  1. $A$ is invertible.
  2. there exists a matrix $B$ for which $AB = I$.
  3. for each $\mathbf{b}\in\mathbb{R}^n$ the linear system $A\mathbf{x} = \mathbf{b}$ has a unique solution.
  4. $A$ is row equivalent to the identity matrix $I_n$.
  5. $A$ has linearly independent columns.
  6. the equation $A\vect{x} = \vect{0}$ has only the trivial solution $\vect{x} = \vect{0}$.
  7. $A$ can be written as a product of elementary matrices: $A = E_1E_2\cdots E_k$.
:::::: ::::::{admonition} Proof of {prf:ref}`Thm:MatrixInv:InvertibilityCharacterizations` :class: tudproof Using {prf:ref}`Dfn:MatrixInv:DefInverse`, we directly have that 1. and 2. are equivalent. The equivalence of 1., 3. and 6. follows from {prf:ref}`Prop:SolutionViaInverse`. The equivalence of 1., 4., 5. and 7. follows from {prf:ref}`Prop:MatrixInv:Algorithm`. :::::: There are many variations on {prf:ref}`Thm:MatrixInv:InvertibilityCharacterizations`. The following exercise contains a few. ::::::{exercise} :label: Exc:MatrixInv:LastEquivalencesInvertibility Show that the following statements are equivalent for an $n\times n$-matrix $A$:
  1. $A$ is invertible.
  2. there exists a matrix $B$ such that $BA = I$.
  3. $A$ has linearly independent rows.
  4. each column of the matrix $A$ is a pivot column.
  5. the columns of $A$ span the whole $\mathbb{R}^n$.
:::::: :::{admonition} Solution to {numref}`Exc:MatrixInv:LastEquivalencesInvertibility` :class: solution, dropdown Using {prf:ref}`Dfn:MatrixInv:DefInverse`, we directly have that 1. and 2. are equivalent. Using {prf:ref}`Prop:MatrixInv:ElemProperties` and 5. from {prf:ref}`Thm:MatrixInv:InvertibilityCharacterizations` shows that 1. and 3. are equivalent. Using 4. from {prf:ref}`Thm:MatrixInv:InvertibilityCharacterizations` shows that 1. and 4. are equivalent. Using 3. from {prf:ref}`Thm:MatrixInv:InvertibilityCharacterizations` shows that 1. and 5. are equivalent. ::: (Subsec:MatrixInverse:Grasple)= ## Grasple exercises The first exercises are quite straightfordwardly computational. The remaining exercises tend to be more theoretic. ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/6683a2f9-7b6b-4dd1-bec1-1e8b894fa3bb?id=71086 :label: grasple_exercise_3_4_1 :dropdown: :description: To compute the inverse of a $2 \times 2$-matrix. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/1bbca38b-a734-4049-b8a2-f79d4bf1b098?id=71087 :label: grasple_exercise_3_4_2 :dropdown: :description: To compute the inverse of a $2 \times 2$-matrix. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/045cd183-ab19-4675-908c-26c41298bade?id=83051 :label: grasple_exercise_3_4_3 :dropdown: :description: To compute the inverse of a $2 \times 2$-matrix. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/82c06a56-8ee8-4f36-8173-e5d56da1e8e3?id=71073 :label: grasple_exercise_3_4_4 :dropdown: :description: To compute the inverse of a $3 \times 3$-matrix step by step. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/551172d9-861c-4958-9b17-dfa828acdabe?id=71088 :label: grasple_exercise_3_4_5 :dropdown: :description: To compute the inverse of a $3 \times 3$-matrix. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/9174c68c-e2d5-4c23-af96-e3fe3dd36f42?id=71089 :label: grasple_exercise_3_4_6 :dropdown: :description: To compute the inverse of a $3 \times 3$-matrix. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/800dc2f9-227e-401b-818b-093fc9647dd9?id=83083 :label: grasple_exercise_3_4_7 :dropdown: :description: To compute the inverse of a $4 \times 4$-matrix. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/9146f49d-74a5-4fda-a641-181c4536fe01?id=83086 :label: grasple_exercise_3_4_8 :dropdown: :description: To find $p$ for which a matrix $A$ is singular. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/4b844244-2e9d-480f-b65a-88d8a2fef781?id=70276 :label: grasple_exercise_3_4_9new :dropdown: :description: To solve an equation $AX = B$ ($A$ and $B$ are $3\times3$-matrices). :::::: The remaining exercises have a more theoretic flavour. ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/677aa3ee-4594-4d77-ace6-583a1efcba59?id=71090 :label: grasple_exercise_3_4_9 :dropdown: :description: True/False question about invertibility versus consistent linear systems. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/f789ebd5-171b-4556-83a9-eefc5ef830ef?id=71092 :label: grasple_exercise_3_4_10 :dropdown: :description: To show if $A$ is invertible, then so is $A^T$. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/29dc7c2f-6636-493e-9c97-da1847a336b7?id=68908 :label: grasple_exercise_3_4_11 :dropdown: :description: To show: if $AB$ is invertible, then so are $A$ and $B$. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/8f3feb75-b41b-42e0-b574-f6442da253ce?id=70272 :label: grasple_exercise_3_4_12 :dropdown: :description: What about $(AB)^{-1} = A^{-1}B^{-1}$? :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/5185c5c0-4d92-4e0e-92a7-6dc5eed8f7cf?id=68896 :label: grasple_exercise_3_4_13 :dropdown: :description: What about $((AB)^T)^{-1} = (A^T)^{-1}(B^T)^{-1}$? :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/ee4bb61e-6939-4074-a556-b82f3d0e8c28?id=71091 :label: grasple_exercise_3_4_14 :dropdown: :description: 'True/False? Every elementary matrix is invertible.' :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/1732d75b-2027-4a92-b8bb-c98bda62475d?id=71093 :label: grasple_exercise_3_4_15 :dropdown: :description: 'True/False? If $A$ and $B$ are invertible, then so is $A+B$.' :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/f8602d4f-57b7-4752-9edc-69c83069fe36?id=71095 :label: grasple_exercise_3_4_16 :dropdown: :description: 'True/False? If $A$ and $B$ are singular, then so is $A+B$.' :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/a8ea864d-1164-4afc-9a24-c0a126ee8e54?id=71097 :label: grasple_exercise_3_4_17 :dropdown: :description: 'True/False? If $A$ is row equivalent to $I$, then so is $A^2$. ' :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/73a16f62-28d7-4a4c-baf5-7ce3be9272ce?id=71104 :label: grasple_exercise_3_4_18 :dropdown: :description: To find 'by inspection' inverses of elemenatry matrices. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/a8c2b8ed-9961-4779-8841-491a9529b71c?id=71466 :label: grasple_exercise_3_4_19 :dropdown: :description: To find the inverses of $AE$ and $EA$, when $A^{-1}$ is given. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/dfe429bd-1ab9-47f7-8f6c-06150c468645?id=71468 :label: grasple_exercise_3_4_20 :dropdown: :description: Finding the inverses of (almost) elementary matrices. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/9af5928a-7ecb-478e-a896-7c66d16d9d09?id=71463 :label: grasple_exercise_3_4_21 :dropdown: :description: Distilling $A^{-1}$ from a relation $c_2A^2 + c_1A + c_0I = 0$. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/2d363522-682f-4ef1-bd9f-7e0d1de30136?id=116691 :label: grasple_exercise_3_4_23new :dropdown: :description: Extricate $X$ from an equation containing $A$, $B$, $X$ and an inverse. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/864d2af1-e751-46b1-88de-f205d95e493c?id=116721 :label: grasple_exercise_3_4_23 :dropdown: :description: Extricate $X$ from an equation containing $A$, $B$, $X$ and a transpose. :::::: In the last two exercises (non-)invertibility of non-square matrices is considered. ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/4e9b4ec1-f775-430f-b81f-c76c42fcbc76?id=60136 :label: grasple_exercise_3_4_24 :dropdown: :description: To explore invertibility for a $3\times 2$-matrix. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/ca504661-cc62-454f-8035-04a9bef85f91?id=61170 :label: grasple_exercise_3_4_25 :dropdown: :description: To explore invertibility for a $2\times 3$-matrix. ::::::