Solution. Compute \(A^2\) first, because it makes everything else shorter: \[A^2=\begin {pmatrix}-1&2\\0&1\end {pmatrix} \begin {pmatrix}-1&2\\0&1\end {pmatrix} =\begin {pmatrix}1&0\\0&1\end {pmatrix}=I.\] So \(A\) is its own inverse, and therefore \[A^3=A^2A=IA=A=\begin {pmatrix}-1&2\\0&1\end {pmatrix},\qquad A^{-1}=A.\] For the second expression, the pieces are \[AB=\begin {pmatrix}-3&-1\\-1&0\end {pmatrix},\qquad B^2=\begin {pmatrix}0&1\\-1&-1\end {pmatrix},\qquad B^{-1}=\begin {pmatrix}0&-1\\1&1\end {pmatrix},\] the last because \(\det B=1\). Then \(A^{-1}B^{-1}=AB^{-1}=\begin {pmatrix}2&3\\1&1\end {pmatrix}\), and adding everything up, \begin {align*} A^2+2AB+B^2-A^{-1}B^{-1}+3I &=\begin {pmatrix}1&0\\0&1\end {pmatrix} +\begin {pmatrix}-6&-2\\-2&0\end {pmatrix} +\begin {pmatrix}0&1\\-1&-1\end {pmatrix} -\begin {pmatrix}2&3\\1&1\end {pmatrix} +\begin {pmatrix}3&0\\0&3\end {pmatrix}\\ &=\begin {pmatrix}-4&-4\\-4&2\end {pmatrix}. \end {align*}

Note that \(2AB\) is genuinely needed here: \(A\) and \(B\) do not commute, so \((A+B)^2\) is not \(A^2+2AB+B^2\) and the expression cannot be shortened that way.

Solution. The computations give \[AA^t=\begin {pmatrix}2&2&-1\\2&5&0\\-1&0&1\end {pmatrix},\qquad A^tA=\begin {pmatrix}5&-2&-1\\-2&1&0\\-1&0&2\end {pmatrix},\] and both are visibly equal to their own transposes. Also \[\tfrac {1}{2}(A+A^t)=\begin {pmatrix}1&1&-1/2\\1&-1&0\\-1/2&0&1\end {pmatrix}, \qquad \tfrac {1}{2}(A-A^t)=\begin {pmatrix}0&-1&-1/2\\1&0&0\\1/2&0&0\end {pmatrix},\] the first symmetric, the second skew-symmetric (note its diagonal is zero, as it must be: \(s_{ii}=-s_{ii}\) forces \(s_{ii}=0\)).

None of this depends on the particular \(A\). Using \((XY)^t=Y^tX^t\) and \((X^t)^t=X\), \[(AA^t)^t=(A^t)^tA^t=AA^t,\] so \(AA^t\) is symmetric for every \(A\), and the same argument applies to \(A^tA\). Likewise \[\left (\tfrac {1}{2}(A+A^t)\right )^t=\tfrac {1}{2}(A^t+A)=\tfrac {1}{2}(A+A^t), \qquad \left (\tfrac {1}{2}(A-A^t)\right )^t=\tfrac {1}{2}(A^t-A)=-\tfrac {1}{2}(A-A^t).\] Adding the last two gives \(A\) itself, so every square matrix splits as a symmetric part plus a skew-symmetric part, and that splitting is unique.

Solution. Make the leading entry \(1\), clear the column below it, then clear upwards. \begin {align*} \begin {pmatrix}3&-1&2\\-5&7&3\end {pmatrix} &\xrightarrow {\ r_1\rightarrow \frac {1}{3}r_1\ } \begin {pmatrix}1&-1/3&2/3\\-5&7&3\end {pmatrix} \xrightarrow {\ r_2\rightarrow r_2+5r_1\ } \begin {pmatrix}1&-1/3&2/3\\0&16/3&19/3\end {pmatrix}\\[2mm] &\xrightarrow {\ r_2\rightarrow \frac {3}{16}r_2\ } \begin {pmatrix}1&-1/3&2/3\\0&1&19/16\end {pmatrix} \xrightarrow {\ r_1\rightarrow r_1+\frac {1}{3}r_2\ } \begin {pmatrix}1&0&17/16\\0&1&19/16\end {pmatrix}. \end {align*}

So the reduced echelon matrix is \(\begin {pmatrix}1&0&17/16\\0&1&19/16\end {pmatrix}\), and \(A\) has rank \(2\).

Solution. Adjoin \(I\) and reduce the left half to \(I\); whatever the right half becomes is \(A^{-1}\). Carrying this out gives \[A^{-1}=\begin {pmatrix}1/4&9/4&-1/2\\1/4&-3/4&1/2\\1/4&5/4&-1/2\end {pmatrix}.\] It is worth checking the answer rather than trusting the arithmetic: \[AA^{-1}=\begin {pmatrix}-1&2&3\\1&0&-1\\2&1&-3\end {pmatrix} \begin {pmatrix}1/4&9/4&-1/2\\1/4&-3/4&1/2\\1/4&5/4&-1/2\end {pmatrix} =\begin {pmatrix}1&0&0\\0&1&0\\0&0&1\end {pmatrix}.\] For instance the \((1,1)\) entry is \((-1)\frac 14+2\cdot \frac 14+3\cdot \frac 14=\frac {-1+2+3}{4}=1\), and the \((1,2)\) entry is \((-1)\frac 94+2\left (-\frac 34\right )+3\cdot \frac 54 =\frac {-9-6+15}{4}=0\).

Solution. Row reducing \(A\) gives \(R=I_3\): the matrix has rank \(3\). This is forced, not a coincidence — a square reduced echelon matrix of full rank has a leading \(1\) in every row and zeros elsewhere in those columns, and the only such matrix is the identity.

The reduction can be done with the six operations \[r_2\rightarrow r_2-2r_1,\quad r_3\rightarrow r_3+3r_1,\quad r_3\rightarrow r_3-2r_2,\quad r_3\rightarrow -\tfrac {1}{14}r_3,\quad r_1\rightarrow r_1+r_3,\quad r_2\rightarrow r_2-4r_3.\] The fourth is easy to forget: after the third operation the last row is \((0,0,-14)\), and it must be scaled to a leading \(1\) before it can be used to clear the column above it.

With \(E'_t\) the elementary matrix of the \(t\)-th operation, \(E'_6E'_5\cdots E'_1A=I\). By Theorem 2.6.3(2) each \(E'_t\) is invertible with elementary inverse, so multiplying back through, \[A=(E'_1)^{-1}(E'_2)^{-1}\cdots (E'_6)^{-1}R,\] which is the required form with \(E_t=(E'_t)^{-1}\). Concretely \[(E'_1)^{-1}=\begin {pmatrix}1&0&0\\2&1&0\\0&0&1\end {pmatrix},\qquad (E'_2)^{-1}=\begin {pmatrix}1&0&0\\0&1&0\\-3&0&1\end {pmatrix},\qquad (E'_4)^{-1}=\begin {pmatrix}1&0&0\\0&1&0\\0&0&-14\end {pmatrix}.\] An inverse of an “add a multiple” operation is the identity with the sign of the multiplier flipped; the inverse of a scaling is the reciprocal scaling.

Since \(R=I\) here, the factorisation also says \(A\) is a product of elementary matrices, which is true of every invertible matrix.

Solution. \(A\) is \(3\) by \(2\), so \(P\) is \(3\) by \(3\) and \(Q\) is \(2\) by \(2\). Row reduce, recording the operations on a copy of \(I_3\): \[r_2\rightarrow r_2-2r_1,\qquad r_3\rightarrow r_3+3r_1,\qquad r_2\rightarrow \tfrac {1}{2}r_2,\qquad r_3\rightarrow r_3+r_2,\qquad r_1\rightarrow r_1+r_2,\] which carries \(A\) to \(\begin {pmatrix}1&0\\0&1\\0&0\end {pmatrix}\). That is already the normal form \(N=\begin {pmatrix}I_2\\0\end {pmatrix}\), so no column operations are needed and \(Q=I_2\). Applying the same operations to \(I_3\) gives \[P=\begin {pmatrix}0&1/2&0\\-1&1/2&0\\2&1/2&1\end {pmatrix},\qquad Q=I_2, \qquad PAQ=\begin {pmatrix}1&0\\0&1\\0&0\end {pmatrix}.\] The rank of \(A\) is \(2\), which is what the number of leading ones in \(N\) records — and by Theorem 2.7.5 that number is the same whatever sequence of operations is used.

Solution.

Uniqueness

Suppose \(B\) and \(C\) both satisfy \(AB=BA=I\) and \(AC=CA=I\). Then \[B=BI=B(AC)=(BA)C=IC=C,\] so there is only one such matrix. The argument uses only associativity and the two-sided identity property; note that it needs \(B\) to be a left inverse and \(C\) a right inverse, which is why both equations are assumed.

The inverse of a product

It suffices to check that \(B^{-1}A^{-1}\) works, since by the first part nothing else can: \[(AB)(B^{-1}A^{-1})=A(BB^{-1})A^{-1}=AIA^{-1}=AA^{-1}=I,\] \[(B^{-1}A^{-1})(AB)=B^{-1}(A^{-1}A)B=B^{-1}IB=B^{-1}B=I.\] Hence \(AB\) is invertible with \((AB)^{-1}=B^{-1}A^{-1}\). The order reverses, for the same reason it reverses for the inverse of a composition of functions.

Solution. Let \(R\) be the reduced echelon form of \(A\). By Theorem 2.6.3(3) there are elementary matrices with \(R=E_k\cdots E_1A\), and by part (2) of the same theorem each is invertible, so \(R=PA\) with \(P=E_k\cdots E_1\) invertible.

If \(R=I_n\) then \(PA=I\), and since \(P\) is invertible, \(A=P^{-1}\), which is invertible.

Conversely suppose \(A\) is invertible. Then \(R=PA\) is a product of invertible matrices and so is invertible. Now \(R\) is a square reduced echelon matrix; if it had a row of zeros it would send some non-zero vector to \(0\) and could not be invertible, so every row of \(R\) contains a leading \(1\). With \(n\) leading ones in \(n\) columns, and zeros above and below each, \(R=I_n\).