2.5E: Matrix Inverses Exercises

Exercises for 1

solutions

2

In each case, show that the matrices are inverses of each other.

1. \(\left[ \begin{array}{rr} 3 & 5 \\ 1 & 2 \end{array} \right]\), \(\left[ \begin{array}{rr} 2 & -5 \\ -1 & 3 \end{array} \right]\)
2. \(\left[ \begin{array}{rr} 3 & 0 \\ 1 & -4 \end{array} \right]\), \(\frac{1}{2}\left[ \begin{array}{rr} 4 & 0 \\ 1 & -3 \end{array} \right]\)
3. \(\left[ \begin{array}{rrr} 1 & 2 & 0 \\ 0 & 2 & 3 \\ 1 & 3 & 1 \end{array} \right]\), \(\left[ \begin{array}{rrr} 7 & 2 & -6 \\ -3 & -1 & 3 \\ 2 & 1 & -2 \end{array} \right]\)
4. \(\left[ \begin{array}{rr} 3 & 0 \\ 0 & 5 \end{array} \right]\), \(\left[ \begin{array}{rr} \frac{1}{3} & 0 \\ 0 & \frac{1}{5} \end{array} \right]\)

Find the inverse of each of the following matrices.

\(\left[ \begin{array}{rr} 1 & -1 \\ -1 & 3 \end{array} \right]\) \(\left[ \begin{array}{rr} 4 & 1 \\ 3 & 2 \end{array} \right]\) \(\left[ \begin{array}{rrr} 1 & 0 & -1 \\ 3 & 2 & 0 \\ -1 & -1 & 0 \end{array} \right]\) \(\left[ \begin{array}{rrr} 1 & -1 & 2 \\ -5 & 7 & -11 \\ -2 & 3 & -5 \end{array} \right]\) \(\left[ \begin{array}{rrr} 3 & 5 & 0 \\ 3 & 7 & 1 \\ 1 & 2 & 1 \end{array} \right]\) \(\left[ \begin{array}{rrr} 3 & 1 & -1 \\ 2 & 1 & 0 \\ 1 & 5 & -1 \end{array} \right]\) \(\left[ \begin{array}{rrr} 2 & 4 & 1 \\ 3 & 3 & 2 \\ 4 & 1 & 4 \end{array} \right]\) \(\left[ \begin{array}{rrr} 3 & 1 & -1 \\ 5 & 2 & 0 \\ 1 & 1 & -1 \end{array} \right]\) \(\left[ \begin{array}{rrr} 3 & 1 & 2 \\ 1 & -1 & 3 \\ 1 & 2 & 4 \end{array} \right]\) \(\left[ \begin{array}{rrrr} -1 & 4 & 5 & 2 \\ 0 & 0 & 0 & -1 \\ 1 & -2 & -2 & 0 \\ 0 & -1 & -1 & 0 \end{array} \right]\) \(\left[ \begin{array}{rrrr} 1 & 0 & 7 & 5 \\ 0 & 1 & 3 & 6 \\ 1 & -1 & 5 & 2 \\ 1 & -1 & 5 & 1 \end{array} \right]\) \(\left[ \begin{array}{rrrrr} 1 & 2 & 0 & 0 & 0 \\ 0 & 1 & 3 & 0 & 0 \\ 0 & 0 & 1 & 5 & 0 \\ 0 & 0 & 0 & 1 & 7 \\ 0 & 0 & 0 & 0 & 1 \end{array} \right]\)

2. \(\frac{1}{5} \left[ \begin{array}{rr} 2 & -1 \\ -3 & 4 \end{array} \right]\)
3. \(\left[ \begin{array}{rrr} 2 & -1 & 3 \\ 3 & 1 & -1 \\ 1 & 1 & -2 \end{array} \right]\)
4. \(\frac{1}{10} \left[ \begin{array}{rrr} 1 & 4 & -1 \\ -2 & 2 & 2 \\ -9 & 14 & -1 \end{array} \right]\)
5. \(\frac{1}{4} \left[ \begin{array}{rrr} 2 & 0 & -2 \\ -5 & 2 & 5 \\ -3 & 2 & -1 \end{array} \right]\)
6. \(\left[ \begin{array}{rrrr} 0 & 0 & 1 & -2 \\ -1 & -2 & -1 & -3 \\ 1 & 2 & 1 & 2 \\ 0 & -1 & 0 & 0 \end{array} \right]\)
7. \(\left[ \begin{array}{rrrrr} 1 & -2 & 6 & -30 & 210 \\ 0 & 1 & -3 & 15 & -105 \\ 0 & 0 & 1 & -5 & 35 \\ 0 & 0 & 0 & 1 & -7 \\ 0 & 0 & 0 & 0 & 1 \end{array} \right]\)

In each case, solve the systems of equations by finding the inverse of the coefficient matrix.

\( \begin{array}[t]{rrrrr} 3x & - & y & = & 5 \\ 2x & + & 2y & = & 1 \end{array}\) \( \begin{array}[t]{rrrrr} 2x & - & 3y & = & 0 \\ x & - & 4y & = & 1 \end{array}\) \( \begin{array}[t]{rrrrrrr} x & + & y & + & 2z & = & 5 \\ x & + & y & + & z & = & 0 \\ x & + & 2y & + & 4z & = & -2 \end{array}\) \( \begin{array}[t]{rrrrrrr} x & + & 4y & + & 2z & = & 1 \\ 2x & + & 3y & + & 3z & = & -1 \\ 4x & + & y & + & 4z & = & 0 \end{array}\)

2. \(\left[ \begin{array}{c} x \\ y \end{array} \right] = \frac{1}{5} \left[ \begin{array}{rr} 4 & -3 \\ 1 & -2 \end{array} \right] \left[ \begin{array}{r} 0 \\ 1 \end{array} \right] = \frac{1}{5} \left[ \begin{array}{r} -3 \\ -2 \end{array} \right]\)
3. \(\left[ \begin{array}{c} x \\ y \\ z \end{array} \right] = \frac{1}{5} \left[ \begin{array}{rrr} 9 & -14 & 6 \\ 4 & -4 & 1 \\ -10 & 15 & -5 \end{array} \right] \left[ \begin{array}{r} 1 \\ -1 \\ 0 \end{array} \right] = \frac{1}{5} \left[ \begin{array}{r} 23 \\ 8 \\ -25 \end{array} \right]\)

Given \(A^{-1} = \left[ \begin{array}{rrr} 1 & -1 & 3 \\ 2 & 0 & 5 \\ -1 & 1 & 0 \end{array} \right]\):

1. Solve the system of equations \(A\mathbf{x} = \left[ \begin{array}{r} 1 \\ -1 \\ 3 \end{array} \right]\).

2. Find a matrix \(B\) such that
   \(AB = \left[ \begin{array}{rrr} 1 & -1 & 2 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{array} \right]\).

3. Find a matrix \(C\) such that
   \(CA = \left[ \begin{array}{rrr} 1 & 2 & -1 \\ 3 & 1 & 1 \end{array} \right]\).

2. \(B = A^{-1}AB = \left[ \begin{array}{rrr} 4 & -2 & 1 \\ 7 & -2 & 4 \\ -1 & 2 & -1 \end{array} \right]\)

Find \(A\) when

\((3A)^{-1} = \left[ \begin{array}{rr} 1 & -1 \\ 0 & 1 \end{array} \right]\) \((2A)^{T} = \left[ \begin{array}{rr} 1 & -1 \\ 2 & 3 \end{array} \right]^{-1}\) \((I + 3A)^{-1} = \left[ \begin{array}{rr} 2 & 0 \\ 1 & -1 \end{array} \right]\) \((I - 2A^{T})^{-1} = \left[ \begin{array}{rr} 2 & 1 \\ 1 & 1 \end{array} \right]\) \(\left(A \left[ \begin{array}{rr} 1 & -1 \\ 0 & 1 \end{array} \right] \right)^{-1} = \left[ \begin{array}{rr} 2 & 3 \\ 1 & 1 \end{array} \right]\) \(\left(\left[ \begin{array}{rr} 1 & 0 \\ 2 & 1 \end{array} \right] A\right)^{-1} = \left[ \begin{array}{rr} 1 & 0 \\ 2 & 2 \end{array} \right]\) \(\left(A^{T} -2I \right)^{-1} = 2 \left[ \begin{array}{rr} 1 & 1 \\ 2 & 3 \end{array} \right]\) \(\left(A^{-1} -2I \right)^{T} = -2 \left[ \begin{array}{rr} 1 & 1 \\ 1 & 0 \end{array} \right]\)

2. \(\frac{1}{10} \left[ \begin{array}{rr} 3 & -2 \\ 1 & 1 \end{array} \right]\)
3. \(\frac{1}{2} \left[ \begin{array}{rr} 0 & 1 \\ 1 & -1 \end{array} \right]\)
4. \(\frac{1}{2} \left[ \begin{array}{rr} 2 & 0 \\ -6 & 1 \end{array} \right]\)
5. \(-\frac{1}{2} \left[ \begin{array}{rr} 1 & 1 \\ 1 & 0 \end{array} \right]\)

Find \(A\) when:

\(A^{-1} = \left[ \begin{array}{rrr} 1 & -1 & 3 \\ 2 & 1 & 1 \\ 0 & 2 & -2 \end{array} \right]\) \(A^{-1} = \left[ \begin{array}{rrr} 0 & 1 & -1 \\ 1 & 2 & 1 \\ 1 & 0 & 1 \end{array} \right]\)

2. \(A = \frac{1}{2} \left[ \begin{array}{rrr} 2 & -1 & 3 \\ 0 & 1 & -1 \\ -2 & 1 & -1 \end{array} \right]\)

Given \(\left[ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array} \right] = \left[ \begin{array}{rrr} 3 & -1 & 2 \\ 1 & 0 & 4 \\ 2 & 1 & 0 \end{array} \right] \left[ \begin{array}{c} y_{1} \\ y_{2} \\ y_{3} \end{array} \right]\) and \(\left[ \begin{array}{c} z_{1} \\ z_{2} \\ z_{3} \end{array} \right] = \left[ \begin{array}{rrr} 1 & -1 & 1 \\ 2 & -3 & 0 \\ -1 & 1 & -2 \end{array} \right] \left[ \begin{array}{c} y_{1} \\ y_{2} \\ y_{3} \end{array} \right]\), express the variables \(x_{1}\), \(x_{2}\), and \(x_{3}\) in terms of \(z_{1}\), \(z_{2}\), and \(z_{3}\).

1. In the system \( \begin{array}{rrrrr} 3x & + & 4y & = & 7 \\ 4x & + & 5y & = & 1 \end{array}\), substitute the new variables \(x^\prime\) and \(y^\prime\) given by \( \begin{array}{rrrrr} x & = & -5x^\prime & + & 4y^\prime \\ y & = & 4x^\prime & - & 3y^\prime \end{array}\). Then find \(x\) and \(y\).
2. Explain part (a) by writing the equations as \(A \left[ \begin{array}{c} x \\ y \end{array} \right] = \left[ \begin{array}{r} 7 \\ 1 \end{array} \right]\) and \(\left[ \begin{array}{r} x \\ y \end{array} \right] = B \left[ \begin{array}{c} x^\prime \\ y^\prime \end{array} \right]\). What is the relationship between \(A\) and \(B\)?

2. \(A\) and \(B\) are inverses.

In each case either prove the assertion or give an example showing that it is false.

1. If \(A \neq 0\) is a square matrix, then \(A\) is invertible.
2. If \(A\) and \(B\) are both invertible, then \(A + B\) is invertible.
3. If \(A\) and \(B\) are both invertible, then \((A^{-1}B)^{T}\) is invertible.
4. If \(A^{4} = 3I\), then \(A\) is invertible.
5. If \(A^{2} = A\) and \(A \neq 0\), then \(A\) is invertible.
6. If \(AB = B\) for some \(B \neq 0\), then \(A\) is invertible.
7. If \(A\) is invertible and skew symmetric (\(A^{T} = -A\)), the same is true of \(A^{-1}\).
8. If \(A^{2}\) is invertible, then \(A\) is invertible.
9. If \(AB = I\), then \(A\) and \(B\) commute.

2. False. \(\left[ \begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array} \right] + \left[ \begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array} \right]\)
3. True. \(A^{-1} = \frac{1}{3}A^{3}\)
4. False. \(A = B = \left[ \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array} \right]\)
5. True. If \((A^{2})B = I\), then \(A(AB) = I\); use Theorem [thm:004553].

[ex:ex2_4_10]

1. If \(A\), \(B\), and \(C\) are square matrices and \(AB = I\), \(I = CA\), show that \(A\) is invertible and \(B = C = A^{-1}\).
2. If \(C^{-1} = A\), find the inverse of \(C^{T}\) in terms of \(A\).

2. \((C^{T})^{-1} = (C^{-1})^{T} = A^{T}\) because \(C^{-1} = (A^{-1})^{-1} = A\).

Suppose \(CA = I_{m}\), where \(C\) is \(m \times n\) and \(A\) is \(n \times m\). Consider the system \(A\mathbf{x} = \mathbf{b}\) of \(n\) equations in \(m\) variables.

1. Show that this system has a unique solution \(CB\) if it is consistent.
2. If \(C = \left[ \begin{array}{rrr} 0 & -5 & 1 \\ 3 & 0 & -1 \end{array} \right]\) and \(A = \left[ \begin{array}{rr} 2 & -3 \\ 1 & -2 \\ 6 & -10 \end{array} \right]\), find \(\mathbf{x}\) (if it exists) when (i) \(\mathbf{b} = \left[ \begin{array}{r} 1 \\ 0 \\ 3 \end{array} \right]\); and (ii) \(\mathbf{b} = \left[ \begin{array}{r} 7 \\ 4 \\ 22 \end{array} \right]\).

2. (ii) \(\left[ \begin{array}{c} x_{1} \\ x_{2} \end{array} \right] = \left[ \begin{array}{r} 2 \\ -1 \end{array} \right]\)

Verify that \(A = \left[ \begin{array}{rr} 1 & -1 \\ 0 & 2 \end{array} \right]\) satisfies \(A^{2} - 3A + 2I = 0\), and use this fact to show that
\(A^{-1} = \frac{1}{2}(3I - A)\).

Let \(Q = \left[ \begin{array}{rrrr} a & -b & -c & -d \\ b & a & -d & c \\ c & d & a & -b \\ d & -c & b & a \end{array} \right]\). Compute \(QQ^{T}\) and so find \(Q^{-1}\) if \(Q \neq 0\).

Let \(U = \left[ \begin{array}{rr} 0 & 1 \\ 1 & 0 \end{array} \right]\). Show that each of \(U\), \(-U\), and \(-I_{2}\) is its own inverse and that the product of any two of these is the third.

Consider \(A = \left[ \begin{array}{rr} 1 & 1 \\ -1 & 0 \end{array} \right]\),
\(B = \left[ \begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array} \right]\), \(C = \left[ \begin{array}{rrr} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 5 & 0 & 0 \end{array} \right]\). Find the inverses by computing (a) \(A^{6}\); (b) \(B^{4}\); and (c) \(C^{3}\).

2. \(B^{4} = I\), so \(B^{-1} = B^{3} = \left[ \begin{array}{rr} 0 & 1 \\ -1 & 0 \end{array} \right]\)

Find the inverse of \(\left[ \begin{array}{rrr} 1 & 0 & 1 \\ c & 1 & c \\ 3 & c & 2 \end{array} \right]\) in terms of \(c\).

\(\left[ \begin{array}{crr} c^{2} - 2 & -c & 1 \\ -c & 1 & 0 \\ 3 - c^{2} & c & -1 \end{array} \right]\)

If \(c \neq 0\), find the inverse of \(\left[ \begin{array}{rrr} 1 & -1 & 1 \\ 2 & -1 & 2 \\ 0 & 2 & c \end{array} \right]\) in terms of \(c\).

Show that \(A\) has no inverse when:

1. \(A\) has a row of zeros.
2. \(A\) has a column of zeros.
3. each row of \(A\) sums to \(0\).
4. [Hint: Corollary [cor:004537], Theorem [thm:004442].]

2. If column \(j\) of \(A\) is zero, \(A\mathbf{y} = \mathbf{0}\) where \(\mathbf{y}\) is column \(j\) of the identity matrix. Use Theorem [thm:004553].
3. If each column of \(A\) sums to \(0\), \(XA = 0\) where \(X\) is the row of \(1\)s. Hence \(A^{T}X^{T} = 0\) so \(A\) has no inverse by Theorem [thm:004553] (\(X^{T} \neq 0\)).

Let \(A\) denote a square matrix.

1. Let \(YA = 0\) for some matrix \(Y \neq 0\). Show that \(A\) has no inverse. [Hint: Corollary [cor:004537], Theorem [thm:004442].]
2. Use part (a) to show that (i) \(\left[ \begin{array}{rrr} 1 & -1 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 2 \end{array} \right]\); and (ii) \(\left[ \begin{array}{rrr} 2 & 1 & -1 \\ 1 & 1 & 0 \\ 1 & 0 & -1 \end{array} \right]\) have no inverse.

   [Hint: For part (ii) compare row 3 with the difference between row 1 and row 2.]

2. (ii) \((-1, 1, 1)A = 0\)

If \(A\) is invertible, show that

\(A^{2} \neq 0\). \(A^{k} \neq 0\) for all
\(k = 1, 2, \dots\).

2. Each power \(A^{k}\) is invertible by Theorem [thm:004442] (because \(A\) is invertible). Hence \(A^{k}\) cannot be \(0\).

Suppose \(AB = 0\), where \(A\) and \(B\) are square matrices. Show that:

1. If one of \(A\) and \(B\) has an inverse, the other is zero.
2. It is impossible for both \(A\) and \(B\) to have inverses.
3. \((BA)^{2} = 0\).

2. By (a), if one has an inverse the other is zero and so has no inverse.

Find the inverse of the \(x\)-expansion in Example [exa:003128] and describe it geometrically.

If \(A = \left[ \begin{array}{rr} a & 0 \\ 0 & 1 \end{array} \right]\), \(a > 1\), then \(A^{-1} = \left[ \begin{array}{rr} \frac{1}{a} & 0 \\ 0 & 1 \end{array} \right]\) is an x-compression because \(\frac{1}{a} < 1\).

Find the inverse of the shear transformation in Example [exa:003136] and describe it geometrically.

In each case assume that \(A\) is a square matrix that satisfies the given condition. Show that \(A\) is invertible and find a formula for \(A^{-1}\) in terms of \(A\).

1. \(A^{3} - 3A + 2I = 0\).
2. \(A^{4} + 2A^{3} - A - 4I = 0\).

2. \(A^{-1} = \frac{1}{4} (A^3 +2A^2-1)\)

Let \(A\) and \(B\) denote \(n \times n\) matrices.

1. If \(A\) and \(AB\) are invertible, show that \(B\) is invertible using only (2) and (3) of Theorem [thm:004442].
2. If \(AB\) is invertible, show that both \(A\) and \(B\) are invertible using Theorem [thm:004553].

2. If \(B\mathbf{x} = \mathbf{0}\), then \((AB)\mathbf{x} = (A)B\mathbf{x} = \mathbf{0}\), so \(\mathbf{x} = \mathbf{0}\) because \(AB\) is invertible. Hence \(B\) is invertible by Theorem [thm:004553]. But then \(A = (AB)B^{-1}\) is invertible by Theorem [thm:004442].

In each case find the inverse of the matrix \(A\) using Example [exa:004627].

\(A = \left[ \begin{array}{rrr} -1 & 1 & 2 \\ 0 & 2 & -1 \\ 0 & 1 & -1 \end{array} \right]\) \(A = \left[ \begin{array}{rrr} 3 & 1 & 0 \\ 5 & 2 & 0 \\ 1 & 3 & -1 \end{array} \right]\) \(A = \left[ \begin{array}{rrrr} 3 & 4 & 0 & 0 \\ 2 & 3 & 0 & 0 \\ 1 & -1 & 1 & 3 \\ 3 & 1 & 1 & 4 \end{array} \right]\) \(A = \left[ \begin{array}{rrrr} 2 & 1 & 5 & 2 \\ 1 & 1 & -1 & 0 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 1 & -2 \end{array} \right]\)

2. \(\left[ \begin{array}{rr|r} 2 & -1 & 0 \\ -5 & 3 & 0 \\ \hline -13 & 8 & -1 \end{array} \right]\)
3. \(\left[ \begin{array}{rr|rr} 1 & -1 & -14 & 8 \\ -1 & 2 & 16 & -9 \\ \hline 0 & 0 & 2 & -1 \\ 0 & 0 & 1 & -1 \end{array} \right]\)

If \(A\) and \(B\) are invertible symmetric matrices such that \(AB = BA\), show that \(A^{-1}\), \(AB\), \(AB^{-1}\), and \(A^{-1}B^{-1}\) are also invertible and symmetric.

Let \(A\) be an \(n \times n\) matrix and let \(I\) be the \(n \times n\) identity matrix.

1. If \(A^{2} = 0\), verify that \((I - A)^{-1} = I + A\).
2. If \(A^{3} = 0\), verify that \((I - A)^{-1} = I + A + A^{2}\).
3. Find the inverse of \(\left[ \begin{array}{rrr} 1 & 2 & -1 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{array} \right]\).
4. If \(A^{n} = 0\), find the formula for \((I - A)^{-1}\).

4. If \(A^{n} = 0\), \((I - A)^{-1} = I + A + \cdots + A^{n-1}\).

[ex:ex2_4_29] Prove property 6 of Theorem [thm:004442]: If \(A\) is invertible and \(a \neq 0\), then \(aA\) is invertible and \((aA)^{-1} = \frac{1}{a}A^{-1}\)

Let \(A\), \(B\), and \(C\) denote \(n \times n\) matrices. Using only Theorem [thm:004442], show that:

1. If \(A\), \(C\), and \(ABC\) are all invertible, \(B\) is invertible.
2. If \(AB\) and \(BA\) are both invertible, \(A\) and \(B\) are both invertible.

2. \(A[B(AB)^{-1}] = I = [(BA)^{-1}B]A\), so \(A\) is invertible by Exercise [ex:ex2_4_10].

Let \(A\) and \(B\) denote invertible \(n \times n\) matrices.

1. If \(A^{-1} = B^{-1}\), does it mean that \(A = B\)? Explain.
2. Show that \(A = B\) if and only if \(A^{-1}B = I\).

Let \(A\), \(B\), and \(C\) be \(n \times n\) matrices, with \(A\) and \(B\) invertible. Show that

1. If \(A\) commutes with \(C\), then \(A^{-1}\) commutes with \(C\).
2. If \(A\) commutes with \(B\), then \(A^{-1}\) commutes with \(B^{-1}\).

1. Have \(AC = CA\). Left-multiply by \(A^{-1}\) to get \(C = A^{-1}CA\). Then right-multiply by \(A^{-1}\) to get \(CA^{-1} = A^{-1}C\).

Let \(A\) and \(B\) be square matrices of the same size.

1. Show that \((AB)^{2} = A^{2}B^{2}\) if \(AB = BA\).
2. If \(A\) and \(B\) are invertible and \((AB)^{2} = A^{2}B^{2}\), show that \(AB = BA\).
3. If \(A = \left[ \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array} \right]\) and \(B = \left[ \begin{array}{rr} 1 & 1 \\ 0 & 0 \end{array} \right]\), show that \((AB)^{2} = A^{2}B^{2}\) but \(AB \neq BA\).

2. Given \(ABAB = AABB\). Left multiply by \(A^{-1}\), then right multiply by \(B^{-1}\).

Let \(A\) and \(B\) be \(n \times n\) matrices for which \(AB\) is invertible. Show that \(A\) and \(B\) are both invertible.

If \(B\mathbf{x} = \mathbf{0}\) where \(\mathbf{x}\) is \(n \times 1\), then \(AB\mathbf{x} = \mathbf{0}\) so \(\mathbf{x} = \mathbf{0}\) as \(AB\) is invertible. Hence \(B\) is invertible by Theorem [thm:004553], so \(A = (AB)B^{-1}\) is invertible.

Consider \(A = \left[ \begin{array}{rrr} 1 & 3 & -1 \\ 2 & 1 & 5 \\ 1 & -7 & 13 \end{array} \right]\), \(B = \left[ \begin{array}{rrr} 1 & 1 & 2 \\ 3 & 0 & -3 \\ -2 & 5 & 17 \end{array} \right]\).

1. [Hint: Row 3 of \(A\) equals \(2\mbox{(row 2) }- 3\mbox{(row 1)}\).]
2. [Hint: Column 3 \(= 3\mbox{(column 2) } - \mbox{ column 1}\).]

2. \(B \left[ \begin{array}{r} -1 \\ 3 \\ -1 \end{array} \right] = 0\) so \(B\) is not invertible by Theorem [thm:004553].

Show that a square matrix \(A\) is invertible if and only if it can be left-cancelled: \(AB = AC\) implies \(B = C\).

If \(U^{2} = I\), show that \(I + U\) is not invertible unless \(U = I\).

1. If \(J\) is the \(4 \times 4\) matrix with every entry \(1\), show that \(I - \frac{1}{2}J\) is self-inverse and symmetric.
2. If \(X\) is \(n \times m\) and satisfies \(X^{T}X = I_{m}\), show that \(I_{n} - 2XX^{T}\) is self-inverse and symmetric.

2. Write \(U = I_{n} - 2XX^{T}\). Then \(U^{T} = I_{n}^{T} - 2X^{TT}X^{T} = U\), and \(U^{2} = I_{n}^{2} - (2XX^{T})I_{n} - I_{n}(2XX^{T}) + 4(XX^{T})(XX^{T}) = I_{n} - 4XX^{T} + 4XX^{T} = I_{n}\).

An \(n \times n\) matrix \(P\) is called an idempotent if \(P^{2} = P\). Show that:

1. \(I\) is the only invertible idempotent.
2. \(P\) is an idempotent if and only if \(I - 2P\) is self-inverse.
3. \(U\) is self-inverse if and only if \(U = I - 2P\) for some idempotent \(P\).
4. \(I - aP\) is invertible for any \(a \neq 1\), and that \((I - aP)^{-1} = I + \left(\frac{a}{1 - a}\right)^{P}\).

2. \((I - 2P)^{2} = I - 4P + 4P^{2}\), and this equals \(I\) if and only if \(P^{2} = P\).

If \(A^{2} = kA\), where \(k \neq 0\), show that \(A\) is invertible if and only if \(A = kI\).

Let \(A\) and \(B\) denote \(n \times n\) invertible matrices.

1. Show that \(A^{-1} + B^{-1} = A^{-1}(A + B)B^{-1}\).
2. If \(A + B\) is also invertible, show that \(A^{-1} + B^{-1}\) is invertible and find a formula for \((A^{-1} + B^{-1})^{-1}\).

2. \((A^{-1} + B^{-1})^{-1} = B(A + B)^{-1}A\)

Let \(A\) and \(B\) be \(n \times n\) matrices, and let \(I\) be the \(n \times n\) identity matrix.

1. Verify that \(A(I + BA) = (I + AB)A\) and that \((I + BA)B = B(I + AB)\).
2. If \(I + AB\) is invertible, verify that \(I + BA\) is also invertible and that \((I + BA)^{-1} = I - B(I + AB)^{-1}A\).