DLA Exercises

Exercises 5.6 Exercises

1. Entry pattern in identity matrices.

As usual, let \(I\) represent the \(n \times n\) identity matrix. Describe the pattern of the entries of \(I\) in terms of the entry index pair \((i,j)\text{.}\) That is, what is the \(\nth{(i,j)}\) entry of \(I\text{,}\) and how does the value of that entry depend on the indices \(i\) and \(j\text{?}\)

Note: your response should be a description in words, not a formula in \(i\) and \(j\text{.}\)

Verifying inverses.

In each case, determine whether the two matrices are inverses of each other.

2.

\(\displaystyle A = \begin{abmatrix}{rr} 2 \amp 7 \\ -1 \amp -4 \end{abmatrix} \text{,}\) \(\displaystyle B = \begin{abmatrix}{rr} 4 \amp 7 \\ -1 \amp -2 \end{abmatrix} \text{.}\)

Solution.

Compute:

\begin{align*} A B \amp = \begin{abmatrix}{rr} 2 \amp 7 \\ -1 \amp -4 \end{abmatrix} \begin{abmatrix}{rr} 4 \amp 7 \\ -1 \amp -2 \end{abmatrix} = \bidentmattwo \text{,}\\ B A \amp = \begin{abmatrix}{rr} 4 \amp 7 \\ -1 \amp -2 \end{abmatrix} \begin{abmatrix}{rr} 2 \amp 7 \\ -1 \amp -4 \end{abmatrix} = \bidentmattwo \text{.} \end{align*}

Since both \(A B = I\) and \(B A = I\) are true, these matrices are inverses of each other.

3.

\(\displaystyle A = \begin{abmatrix}{rr} -2 \amp 3 \\ -1 \amp 2 \end{abmatrix} \text{,}\) \(\displaystyle C = \begin{abmatrix}{rr} 4 \amp 7 \\ -1 \amp 0 \end{abmatrix} \text{.}\)

Solution.

The matrix \(A\) is the same as in Exercise 2, and that exercise confirms that \(\inv{A} = B\) (where \(B\) is also as in that exercise). Therefore, the matrix \(C\) in the present exercise cannot be inverse with \(A\text{,}\) according to Theorem 5.5.2.

4.

\(\displaystyle A = \begin{abmatrix}{rr} 0 \amp 1 \\ 1 \amp -3 \end{abmatrix} \text{,}\) \(\displaystyle B = \begin{abmatrix}{rrr} 3 \amp 1 \amp 0 \\ 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \end{abmatrix} \text{.}\)

Solution.

These matrices cannot be inverses of each other because the are not the same size.

5.

\(\displaystyle A = \begin{abmatrix}{rrr} 1 \amp 3 \amp 1 \\ 0 \amp -2 \amp 3 \\ 2 \amp 7 \amp 1 \end{abmatrix} \text{,}\) \(\displaystyle B = \begin{abmatrix}{rrr} 23 \amp -4 \amp -11 \\ -6 \amp 1 \amp 3 \\ -4 \amp 1 \amp 2 \end{abmatrix} \text{.}\)

Solution.

Compute:

\begin{align*} A B \amp = \begin{abmatrix}{rrr} 1 \amp 3 \amp 1 \\ 0 \amp -2 \amp 3 \\ 2 \amp 7 \amp 1 \end{abmatrix} \begin{abmatrix}{rrr} 23 \amp -4 \amp -11 \\ -6 \amp 1 \amp 3 \\ -4 \amp 1 \amp 2 \end{abmatrix} = \bidentmatthree \text{,}\\ B A \amp = \begin{abmatrix}{rrr} 23 \amp -4 \amp -11 \\ -6 \amp 1 \amp 3 \\ -4 \amp 1 \amp 2 \end{abmatrix} \begin{abmatrix}{rrr} 1 \amp 3 \amp 1 \\ 0 \amp -2 \amp 3 \\ 2 \amp 7 \amp 1 \end{abmatrix} = \bidentmatthree \text{.} \end{align*}

Since both \(A B = I\) and \(B A = I\) are true, these matrices are inverses of each other.

6.

\(\displaystyle A = \begin{abmatrix}{rrr} 2 \amp -4 \amp 7 \\ 1 \amp -2 \amp 3 \\ 0 \amp 1 \amp -3 \end{abmatrix} \text{,}\) \(\displaystyle B = \begin{abmatrix}{rrr} 3 \amp -5 \amp 2 \\ 3 \amp -6 \amp 1 \\ 1 \amp -2 \amp 1 \end{abmatrix} \text{.}\)

Solution.

Compute:

\begin{equation*} A B = \begin{abmatrix}{rrr} 2 \amp -4 \amp 7 \\ 1 \amp -2 \amp 3 \\ 0 \amp 1 \amp -3 \end{abmatrix} \begin{abmatrix}{rrr} 3 \amp -5 \amp 2 \\ 3 \amp -6 \amp 1 \\ 1 \amp -2 \amp 1 \end{abmatrix} = \begin{abmatrix}{rrr} 1 \amp 0 \amp 7 \\ 0 \amp 1 \amp 3 \\ 0 \amp 0 \amp -2 \end{abmatrix}\text{.} \end{equation*}

Since \(A B\) does not evaluate to \(I\text{,}\) these matrices are not inverses of each other.

7.

\(\displaystyle A = \begin{abmatrix}{rrrr} 1 \amp 3 \amp -2 \amp 2 \\ 0 \amp 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 1 \amp 0 \\ 1 \amp 4 \amp 2 \amp -1 \end{abmatrix} \text{,}\) \(\displaystyle B = \begin{abmatrix}{rrrr} 4 \amp -11 \amp 14 \amp -3 \\ -1 \amp 3 \amp -4 \amp 1 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \end{abmatrix} \text{.}\)

Solution.

Compute:

\begin{align*} A B \amp = \begin{abmatrix}{rrrr} 1 \amp 3 \amp -2 \amp 2 \\ 0 \amp 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 1 \amp 0 \\ 1 \amp 4 \amp 2 \amp -1 \end{abmatrix} \begin{abmatrix}{rrrr} 4 \amp -11 \amp 14 \amp -3 \\ -1 \amp 3 \amp -4 \amp 1 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \end{abmatrix} = \bidentmatfour \text{,}\\ B A \amp = \begin{abmatrix}{rrrr} 4 \amp -11 \amp 14 \amp -3 \\ -1 \amp 3 \amp -4 \amp 1 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \end{abmatrix} \begin{abmatrix}{rrrr} 1 \amp 3 \amp -2 \amp 2 \\ 0 \amp 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 1 \amp 0 \\ 1 \amp 4 \amp 2 \amp -1 \end{abmatrix} = \bidentmatfour \text{.} \end{align*}

Since both \(A B = I\) and \(B A = I\) are true, these matrices are inverses of each other.

8.

\(\displaystyle A = \begin{bmatrix} 5 \end{bmatrix} \text{,}\) \(\displaystyle B = \begin{bmatrix} \frac{1}{5} \end{bmatrix} \text{.}\)

Solution.

Compute:

\begin{align*} A B \amp = \begin{bmatrix} 5 \end{bmatrix} \begin{bmatrix} \frac{1}{5} \end{bmatrix} = \begin{bmatrix} 1 \end{bmatrix} \text{,} \amp B A \amp = \begin{bmatrix} \frac{1}{5} \end{bmatrix} \begin{bmatrix} 5 \end{bmatrix} = \begin{bmatrix} 1 \end{bmatrix} \text{.} \end{align*}

Since both \(A B = I\) and \(B A = I\) are true, these matrices are inverses of each other.

The inversion problem as systems of equations.

In each case, carry out the following steps:

Compute the product matrix \(A B\text{.}\) (The result will be a matrix whose entries are formulas in the variables that make up the entries of \(B\text{.}\))
Use the definition of equal matrices to convert the matrix equality \(A B = I \) into equations in the variables that make up the entries of \(B\) (where, as usual, \(I\) represents the identity matrix of appropriate dimensions).
Solve the equations to compute the entries of \(B\text{,}\) which will then be equal to \(\inv{A}\text{,}\) or use the equations to decide that \(A\) has no inverse.

9.

\(\displaystyle A = \begin{abmatrix}{rr} 3 \amp -8 \\ 2 \amp -5 \end{abmatrix} \text{,}\) \(\displaystyle B = \begin{bmatrix} a \amp b \\ c \amp d \end{bmatrix} \text{,}\)

Solution.

\begin{equation*} A B = \begin{abmatrix}{rr} 3 \amp -8 \\ 2 \amp -5 \end{abmatrix} \begin{bmatrix} a \amp b \\ c \amp d \end{bmatrix} = \begin{bmatrix} 3 a - 8 c \amp 3 b - 8 d \\ 2 a - 5 c \amp 2 b - 5 d \end{bmatrix} \end{equation*}
Matrix equality \(A B = I\) is

\begin{equation*} \begin{bmatrix} 3 a - 8 c \amp 3 b - 8 d \\ 2 a - 5 c \amp 2 b - 5 d \end{bmatrix} = \bidentmattwo\text{.} \end{equation*}

Equating corresponding entries leads to a pair of systems of equations:

\begin{align*} \amp \begin{sysofeqns}{rcrcr} 3 a \amp - \amp 8 c \amp = \amp 1 \\ 2 a \amp - \amp 5 c \amp = \amp 0 \end{sysofeqns} \text{,} \amp \begin{sysofeqns}{rcrcr} 3 b \amp - \amp 8 d \amp = \amp 0 \\ 2 b \amp - \amp 5 d \amp = \amp 1 \end{sysofeqns}\text{.} \end{align*}
Put each of the systems of equations into an augmented matrices and reduce to solve. (Notice how \(A\) ends up being the coefficient matrix of both systems, and the column of constants is one of the columns of the identity matrix!) First,

\begin{equation*} \begin{abmatrix}{rr|r} 3 \amp -8 \amp 1 \\ 2 \amp -5 \amp 0 \end{abmatrix} \quad \rowredarrow \quad \begin{abmatrix}{rr|r} 1 \amp 0 \amp -5 \\ 0 \amp 1 \amp -2 \end{abmatrix}\text{,} \end{equation*}

from which we obtain \(a = -5\text{,}\) \(c = -2\text{.}\) Next,

\begin{equation*} \begin{abmatrix}{rr|r} 3 \amp -8 \amp 0 \\ 2 \amp -5 \amp 1 \end{abmatrix} \quad \rowredarrow \quad \begin{abmatrix}{rr|r} 1 \amp 0 \amp 8 \\ 0 \amp 1 \amp 3 \end{abmatrix}\text{,} \end{equation*}

from which we obtain \(b = 8\text{,}\) \(d = 3\text{.}\) Therefore, we have

\begin{equation*} \inv{A} = B = \begin{bmatrix} -5 \amp 8 \\ -2 \amp 3 \end{bmatrix}\text{.} \end{equation*}

10.

\(\displaystyle A = \begin{abmatrix}{rrr} 6 \amp -5 \amp -6 \\ 0 \amp -3 \amp -6 \\ 6 \amp -2 \amp 0 \end{abmatrix} \text{,}\) \(\displaystyle B = \begin{bmatrix} a \amp b \amp c \\ d \amp e \amp f \\ g \amp h \amp i \end{bmatrix} \text{,}\)

Solution.

\begin{align*} A B \amp = \begin{abmatrix}{rrr} 6 \amp -5 \amp -6 \\ 0 \amp -3 \amp -6 \\ 6 \amp -2 \amp 0 \end{abmatrix} \begin{bmatrix} a \amp b \amp c \\ d \amp e \amp f \\ g \amp h \amp i \end{bmatrix}\\ \amp = \begin{bmatrix} 6 a - 5 d - 6 g \amp 6 b - 5 e - 6 h \amp 6 c - 5 f - 6 i \\ -3 d - 6 g \amp -3 e - 6 h \amp -3 f - 6 i \\ 6 a - 2 d \amp 6 b - 2 e \amp 6 c - 2 f \end{bmatrix} \end{align*}
Matrix equality \(A B = I\) is

\begin{equation*} \begin{bmatrix} 6 a - 5 d - 6 g \amp 6 b - 5 e - 6 h \amp 6 c - 5 f - 6 i \\ -3 d - 6 g \amp -3 e - 6 h \amp -3 f - 6 i \\ 6 a - 2 d \amp 6 b - 2 e \amp 6 c - 2 f \end{bmatrix} = \bidentmatthree\text{.} \end{equation*}

Equating corresponding entries leads to a triple of systems of equations:

\begin{equation*} \begin{sysofeqns}{rcrcrcr} 6 a \amp - \amp 5 d \amp - \amp 6 g \amp = \amp 1 \\ \amp \amp -3 d \amp - \amp 6 g \amp = \amp 0 \\ 6 a \amp - \amp 2 d \amp \amp \amp = \amp 0 \end{sysofeqns}\text{,} \end{equation*}

\begin{equation*} \begin{sysofeqns}{rcrcrcr} 6 b \amp - \amp 5 e \amp - \amp 6 h \amp = \amp 0 \\ \amp \amp -3 e \amp - \amp 6 h \amp = \amp 1 \\ 6 b \amp - \amp 2 e \amp \amp \amp = \amp 0 \end{sysofeqns}\text{,} \end{equation*}

\begin{equation*} \begin{sysofeqns}{rcrcrcr} 6 c \amp - \amp 5 f \amp - \amp 6 i \amp = \amp 0 \\ \amp \amp -3 f \amp - \amp 6 i \amp = \amp 0 \\ 6 c \amp - \amp 2 f \amp \amp \amp = \amp 1 \end{sysofeqns} \end{equation*}
Let’s put the first system of equations into an augmented matrices and reduce to solve. (Notice how \(A\) ends up being the coefficient matrix, and the column of constants is the first column of the identity matrix!)

\begin{equation*} \begin{abmatrix}{rrr|r} 6 \amp -5 \amp -6 \amp 1 \\ 0 \amp -3 \amp -6 \amp 0 \\ 6 \amp -2 \amp 0 \amp 0 \end{abmatrix} \quad \rowredarrow \quad \begin{abmatrix}{ccc|c} 1 \amp 0 \amp \frac{2}{3} \amp 0 \\ 0 \amp 1 \amp 2 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{abmatrix} \end{equation*}

This RREF matrix has a leading one in the “equals” column, which indicates that the system is inconsistent. It doesn’t matter whether the other two systems are consistent or not, because the inconsistency of this first system means it is impossible to determine the values of entries \(a,d,g\) in \(B\) so that \(A B = I\) is true. Therefore, \(A\) is not invertible.

11.

\(\displaystyle A = \begin{abmatrix}{rrr} 1 \amp 1 \amp -3 \\ 0 \amp 1 \amp -3 \\ -1 \amp 1 \amp -2 \end{abmatrix} \text{,}\) \(\displaystyle B = \begin{bmatrix} a \amp b \amp c \\ d \amp e \amp f \\ g \amp h \amp i \end{bmatrix} \text{,}\)

Solution.

\begin{align*} A B \amp = \begin{abmatrix}{rrr} 1 \amp 1 \amp -3 \\ 0 \amp 1 \amp -3 \\ -1 \amp 1 \amp -2 \end{abmatrix} \begin{bmatrix} a \amp b \amp c \\ d \amp e \amp f \\ g \amp h \amp i \end{bmatrix}\\ \amp = \begin{abmatrix}{rrr} a + d - 3 g \amp b + e - 3 h \amp c + f - 3 i \\ d - 3 g \amp e - 3 h \amp f - 3 i \\ -a + d - 2 g \amp -b + e - 2 h \amp -c + f - 2 i \end{abmatrix} \end{align*}
Matrix equality \(A B = I\) is

\begin{equation*} \begin{bmatrix} a + d - 3 g \amp b + e - 3 h \amp c + f - 3 i \\ d - 3 g \amp e - 3 h \amp f - 3 i \\ -a + d - 2 g \amp -b + e - 2 h \amp -c + f - 2 i \end{bmatrix} = \bidentmatthree\text{.} \end{equation*}

Equating corresponding entries leads to a triple of systems of equations:

\begin{equation*} \begin{sysofeqns}{rcrcrcr} a \amp + \amp d \amp - \amp 3 g \amp = \amp 1 \\ \amp \amp d \amp - \amp 3 g \amp = \amp 0 \\ -a \amp + \amp d \amp - \amp 2 g \amp = \amp 0 \end{sysofeqns}\text{,} \end{equation*}

\begin{equation*} \begin{sysofeqns}{rcrcrcr} b \amp + \amp e \amp - \amp 3 h \amp = \amp 0 \\ \amp \amp e \amp - \amp 3 h \amp = \amp 1 \\ -b \amp + \amp e \amp - \amp 2 h \amp = \amp 0 \end{sysofeqns}\text{,} \end{equation*}

\begin{equation*} \begin{sysofeqns}{rcrcrcr} c \amp + \amp f \amp - \amp 3 i \amp = \amp 0 \\ \amp \amp f \amp - \amp 3 i \amp = \amp 0 \\ -c \amp + \amp f \amp - \amp 2 i \amp = \amp 1 \end{sysofeqns}\text{.} \end{equation*}
Put each of the systems of equations into an augmented matrices and reduce to solve. (Notice how \(A\) ends up being the coefficient matrix of both systems, and the column of constants is one of the columns of the identity matrix!) First,

\begin{equation*} \begin{abmatrix}{rrr|r} 1 \amp 1 \amp -3 \amp 1 \\ 0 \amp 1 \amp -3 \amp 0 \\ -1 \amp 1 \amp -2 \amp 0 \end{abmatrix} \quad \rowredarrow \quad \begin{abmatrix}{ccc|c} 1 \amp 0 \amp 0 \amp 1 \\ 0 \amp 1 \amp 0 \amp 3 \\ 0 \amp 0 \amp 1 \amp 1 \end{abmatrix}\text{,} \end{equation*}

from which we obtain \(a = 1\text{,}\) \(d = 3\text{,}\) \(g = 1\text{.}\) Next,

\begin{equation*} \begin{abmatrix}{rrr|r} 1 \amp 1 \amp -3 \amp 0 \\ 0 \amp 1 \amp -3 \amp 1 \\ -1 \amp 1 \amp -2 \amp 0 \end{abmatrix} \quad \rowredarrow \quad \begin{abmatrix}{ccc|r} 1 \amp 0 \amp 0 \amp -1 \\ 0 \amp 1 \amp 0 \amp -5 \\ 0 \amp 0 \amp 1 \amp -2 \end{abmatrix}\text{,} \end{equation*}

from which we obtain \(b = -1\text{,}\) \(e = -5\text{,}\) \(h = -2\text{.}\)

\begin{equation*} \begin{abmatrix}{rrr|r} 1 \amp 1 \amp -3 \amp 0 \\ 0 \amp 1 \amp -3 \amp 0 \\ -1 \amp 1 \amp -2 \amp 1 \end{abmatrix} \quad \rowredarrow \quad \begin{abmatrix}{ccc|r} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 3 \\ 0 \amp 0 \amp 1 \amp 1 \end{abmatrix}\text{,} \end{equation*}

from which we obtain \(c = 0\text{,}\) \(f = 3\text{,}\) \(i = 1\text{.}\) Therefore, we have

\begin{equation*} \inv{A} = B = \begin{abmatrix}{rrr} 1 \amp -1 \amp 0 \\ 3 \amp -5 \amp 3 \\ 1 \amp -2 \amp 1 \end{abmatrix}\text{.} \end{equation*}

Computing \(2 \times 2\) inverses.

Use the \(2 \times 2\) inversion formula, if possible, to compute the inverse of each matrix. In the case that the matrix is invertible, multiply it against your computed inverse to verify that your inverse is correct.

12.

\(\displaystyle \begin{abmatrix}{rr} 1 \amp -1 \\ 3 \amp -2 \end{abmatrix}\)

Answer.

\(\displaystyle \begin{abmatrix}{rr} -2 \amp 1 \\ -3 \amp 1 \end{abmatrix}\)

13.

\(\displaystyle \begin{abmatrix}{rr} -5 \amp -9 \\ 1 \amp 2 \end{abmatrix}\)

Answer.

\(\displaystyle \begin{abmatrix}{rr} -2 \amp -9 \\ 1 \amp 5 \end{abmatrix}\)

14.

\(\displaystyle \begin{bmatrix} 1 \amp 1 \\ 1 \amp 1 \end{bmatrix}\)

Answer.

This matrix is singular.

15.

\(\displaystyle \begin{bmatrix} 7 \amp 2 \\ 21 \amp 6 \end{bmatrix}\)

Answer.

This matrix is singular.

16.

\(\displaystyle \begin{abmatrix}{rr} 2 \amp -3 \\ -2 \amp 4 \end{abmatrix}\)

Answer.

\(\displaystyle \begin{bmatrix} 2 \amp \frac{3}{2} \\ 1 \amp 1 \end{bmatrix}\)

17.

\(\displaystyle \begin{bmatrix} \frac{1}{3} \amp \frac{2}{3} \\ 1 \amp 3 \end{bmatrix}\)

Answer.

\(\displaystyle \begin{abmatrix}{rr} 9 \amp -2 \\ -3 \amp 1 \end{abmatrix}\)

Solving systems using inverses.

In each case, solve the system using the inverse of the corresponding coefficient matrix.

18.

\(\displaystyle \begin{sysofeqns}{rcrcr} x \amp - \amp y \amp = \amp 0 \\ 3 x \amp - \amp 2 y \amp = \amp -8 \end{sysofeqns}\)

Solution.

This system is represented in matrix form \(A \uvec{x} = \uvec{b} \) where

\begin{align*} A \amp = \begin{abmatrix}{rr} 1 \amp -1 \\ 3 \amp -2 \end{abmatrix} \text{,} \amp \uvec{x} \amp = \begin{bmatrix} x \\ y \end{bmatrix} \text{,} \amp \uvec{b} \amp = \begin{abmatrix}{r} 0 \\ -8 \end{abmatrix} \text{.} \end{align*}

The coefficient matrix is identical to the matrix of Exercise 12 and has inverse

\begin{equation*} \begin{abmatrix}{rr} -2 \amp 1 \\ -3 \amp 1 \end{abmatrix}\text{.} \end{equation*}

The system therefore has unique solution

\begin{equation*} \uvec{x} = \inv{A} \uvec{b} = \begin{abmatrix}{rr} -2 \amp 1 \\ -3 \amp 1 \end{abmatrix} \begin{abmatrix}{r} 0 \\ -8 \end{abmatrix} = \begin{bmatrix} -8 \\ -8 \end{bmatrix}\text{.} \end{equation*}

In this unique solution, \(x = -8\) and \(y = -8\text{.}\)

19.

\(\displaystyle \begin{sysofeqns}{rcrcr} x \amp - \amp y \amp = \amp 6 \\ 3 x \amp - \amp 2 y \amp = \amp 5 \end{sysofeqns}\)

Solution.

This system has the same coefficient matrix and variable column as Exercise 18, but has column of constants

\begin{equation*} \uvec{b} = \begin{bmatrix} 6 \\ 5 \end{bmatrix} \text{.} \end{equation*}

We can use the same inverse matrix computed in Exercise 12 to compute the unique solution

\begin{equation*} \uvec{x} = \inv{A} \uvec{b} = \begin{abmatrix}{rr} -2 \amp 1 \\ -3 \amp 1 \end{abmatrix} \begin{abmatrix}{r} 6 \\ 5 \end{abmatrix} = \begin{bmatrix} 1 \\ 8 \end{bmatrix}\text{.} \end{equation*}

In this unique solution, \(x = 1\) and \(y = 8\text{.}\)

20.

\(\displaystyle \begin{sysofeqns}{rcrcr} 2 s \amp - \amp 3 t \amp = \amp -7 \\ -2 s \amp + \amp 4 t \amp = \amp 3 \end{sysofeqns}\)

Solution.

This system is represented in matrix form \(A \uvec{x} = \uvec{b} \) where

\begin{align*} A \amp = \begin{abmatrix}{rr} 2 \amp -3 \\ -2 \amp 4 \end{abmatrix} \text{,} \amp \uvec{x} \amp = \begin{bmatrix} s \\ t \end{bmatrix} \text{,} \amp \uvec{b} \amp = \begin{abmatrix}{r} -7 \\ 3 \end{abmatrix} \text{.} \end{align*}

The coefficient matrix is identical to the matrix of Exercise 16 and has inverse

\begin{equation*} \begin{bmatrix} 2 \amp \frac{3}{2} \\ 1 \amp 1 \end{bmatrix}\text{.} \end{equation*}

The system therefore has unique solution

\begin{equation*} \uvec{x} = \inv{A} \uvec{b} = \begin{bmatrix} 2 \amp \frac{3}{2} \\ 1 \amp 1 \end{bmatrix} \begin{abmatrix}{r} -7 \\ 3 \end{abmatrix} = \begin{abmatrix}{r} -\frac{19}{2} \\ -4 \end{abmatrix}\text{.} \end{equation*}

In this unique solution, \(s = -19/2\) and \(t = -4\text{.}\)

21.

\(\displaystyle \begin{sysofeqns}{rcrcr} 2 x_1 \amp - \amp 3 x_2 \amp = \amp 3 \\ -2 x_1 \amp + \amp 4 x_2 \amp = \amp 4 \end{sysofeqns}\)

Solution.

This system has the same coefficient matrix as Exercise 20, but has variable and constant columns

\begin{align*} \uvec{x} \amp = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \text{,} \amp \uvec{b} \amp = \begin{bmatrix} 3 \\ 4 \end{bmatrix} \text{.} \end{align*}

We can use the same inverse matrix computed in Exercise 16 to compute the unique solution

\begin{equation*} \uvec{x} = \inv{A} \uvec{b} = \begin{abmatrix}{rr} -2 \amp 1 \\ -3 \amp 1 \end{abmatrix} \begin{abmatrix}{r} 3 \\ 4 \end{abmatrix} = \begin{bmatrix} 12 \\ 7 \end{bmatrix}\text{.} \end{equation*}

In this unique solution, \(x_1 = 1\) and \(x_2 = 7\text{.}\)

22.

\(\displaystyle \begin{sysofeqns}{rcrcrcr} x \amp + \amp y \amp - \amp 3 z \amp = 3 \\ \amp \amp y \amp - \amp 3 z \amp = 8 \\ -x \amp + \amp y \amp - \amp 2 z \amp = -1 \end{sysofeqns}\)

Hint.

First see Exercise 11.

Solution.

This system is represented in matrix form \(A \uvec{x} = \uvec{b} \) where

\begin{align*} A \amp = \begin{abmatrix}{rrr} 1 \amp 1 \amp -3 \\ 0 \amp 1 \amp -3 \\ -1 \amp 1 \amp -2 \end{abmatrix} \text{,} \amp \uvec{x} \amp = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \text{,} \amp \uvec{b} \amp = \begin{abmatrix}{r} 3 \\ 8 \\ -1 \end{abmatrix} \text{.} \end{align*}

The coefficient matrix is identical to the matrix of Exercise 11 and has inverse

\begin{equation*} \begin{abmatrix}{rrr} 1 \amp -1 \amp 0 \\ 3 \amp -5 \amp 3 \\ 1 \amp -2 \amp 1 \end{abmatrix}\text{.} \end{equation*}

The system therefore has unique solution

\begin{equation*} \uvec{x} = \inv{A} \uvec{b} = \begin{abmatrix}{rrr} 1 \amp -1 \amp 0 \\ 3 \amp -5 \amp 3 \\ 1 \amp -2 \amp 1 \end{abmatrix} \begin{abmatrix}{r} 3 \\ 8 \\ -1 \end{abmatrix} = \begin{abmatrix}{r} -5 \\ -34 \\ -14 \end{abmatrix}\text{.} \end{equation*}

In this unique solution, \(x = -5\text{,}\) \(y = -34\text{,}\) and \(z = -14\text{.}\)

23.

\(\displaystyle \begin{sysofeqns}{rcrcrcr} x_1 \amp + \amp x_2 \amp - \amp 3 x_3 \amp = 3 \\ \amp \amp x_2 \amp - \amp 3 x_3 \amp = 5 \\ -x_1 \amp + \amp x_2 \amp - \amp 2 x_3 \amp = 4 \end{sysofeqns}\)

Hint.

First see Exercise 11.

Solution.

This system has the same coefficient matrix as Exercise 22, but has variable and constant columns

\begin{align*} \uvec{x} \amp = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \text{,} \amp \uvec{b} \amp = \begin{bmatrix} 3 \\ 5 \\ 4 \end{bmatrix} \text{.} \end{align*}

We can use the same inverse matrix computed in Exercise 11 to compute the unique solution

\begin{equation*} \uvec{x} = \inv{A} \uvec{b} = \begin{abmatrix}{rrr} 1 \amp -1 \amp 0 \\ 3 \amp -5 \amp 3 \\ 1 \amp -2 \amp 1 \end{abmatrix} \begin{abmatrix}{r} 3 \\ 5 \\ 4 \end{abmatrix} = \begin{bmatrix} -2 \\ -4 \\ -3 \end{bmatrix}\text{.} \end{equation*}

In this unique solution, \(x_1 = -2\text{,}\) \(x_2 = -4\text{,}\) and \(x_3 = -3\text{.}\)

Isolating via matrix algebra.

In each case, use matrix algebra to isolate and determine the unknown variable matrix.

24.

\(\displaystyle \begin{abmatrix}{rr} 4 \amp 1 \\ 7 \amp -8 \end{abmatrix} - \begin{bmatrix} 1 \amp 3 \\ 1 \amp 2 \end{bmatrix} X = \begin{abmatrix}{rr} 0 \amp 6 \\ -6 \amp 1 \end{abmatrix}\)

Solution.

It’s less messy to perform algebraic manipulations with letters, so assign

\begin{align*} A \amp = \begin{abmatrix}{rr} 4 \amp 1 \\ 7 \amp -8 \end{abmatrix} \text{,} \amp B \amp = \begin{bmatrix} 1 \amp 3 \\ 1 \amp 2 \end{bmatrix} \text{,} \amp C \amp = \begin{abmatrix}{rr} 0 \amp 6 \\ -6 \amp 1 \end{abmatrix} \end{align*}

Then we have

\begin{gather*} A - B X = C \\ A - C = B X \\ \inv{B} (A - C) = \inv{B} B X \\ \inv{B} (A - C) = I X \\ X = \inv{B} (A - C) \text{,} \end{gather*}

assuming that \(B\) is invertible. Applying Proposition 5.5.4, for \(B\) we have

\begin{equation*} a d - b c = 1 \cdot 2 - 3 \cdot 1 = -1 \text{,} \end{equation*}

and so \(B\) is indeed invertible, with inverse

\begin{equation*} \inv{B} = \frac{1}{-1} \begin{abmatrix}{rr} 2 \amp -3 \\ -1 \amp 1 \end{abmatrix} = \begin{abmatrix}{rr} -2 \amp 3 \\ 1 \amp -1 \end{abmatrix}\text{.} \end{equation*}

With this inverse in hand, we may calculate

\begin{align*} A - C \amp = \begin{abmatrix}{rr} 4 \amp 1 \\ 7 \amp -8 \end{abmatrix} - \begin{abmatrix}{rr} 0 \amp 6 \\ -6 \amp 1 \end{abmatrix}\\ \amp = \begin{abmatrix}{rr} 4 \amp -5 \\ 13 \amp -9 \end{abmatrix}\text{,} \end{align*}

\begin{align*} X \amp = \inv{B} (A - C) \\ \amp = \begin{abmatrix}{rr} -2 \amp 3 \\ 1 \amp -1 \end{abmatrix} \begin{abmatrix}{rr} 4 \amp -5 \\ 13 \amp -9 \end{abmatrix}\\ \amp = \begin{abmatrix}{rr} 31 \amp -17 \\ -9 \amp 4 \end{abmatrix}\text{.} \end{align*}

25.

\(\displaystyle \utrans{X} \begin{abmatrix}{rr} 1 \amp -3 \\ -3 \amp 10 \end{abmatrix} + \begin{abmatrix}{rr} 3 \amp 1 \\ 3 \amp -3 \end{abmatrix} = I\)

Solution.

It’s less messy to perform algebraic manipulations with letters, so assign

\begin{align*} A \amp = \begin{abmatrix}{rr} 1 \amp -3 \\ -3 \amp 10 \end{abmatrix} \text{,} \amp B \amp = \begin{abmatrix}{rr} 3 \amp 1 \\ 3 \amp -3 \end{abmatrix} \text{.} \end{align*}

Then we have

\begin{gather*} \utrans{X} A + B = I \\ \utrans{X} A = I - B \\ \utrans{X} A \inv{A} = (I - B) \inv{A} \\ \utrans{X} I = (I - B) \inv{A} \\ \utrans{X} = (I - B) \inv{A} \text{,} \end{gather*}

assuming that \(A\) is invertible.

Note how we have multiplied by \(\inv{A}\) on the right on both sides of the second equality above to obtain the third. Since order of matrix multiplication matters, multiplying by \(\inv{A}\) on the left would not cancel the \(A\) on the left-hand side of the equality.

The final algebraic step is to “undo” the transpose that has been applied to \(X\text{,}\) but by Rule 5.a of Proposition 4.5.1, transpose reverses itself:

\begin{gather*} \utrans{X} = (I - B) \inv{A} \\ \utrans{(\utrans{X})} = \utrans{\bbrac{(I - B) \inv{A}}} \\ X = \utrans{\bbrac{(I - B) \inv{A}}} \text{.} \end{gather*}

(We could apply more Rules involving the transpose from Proposition 4.5.1 to further “simplify” the right-hand side of the last line above, but it’s probably easier to just calculate one final transpose at the end instead of several intermediate transposes along the way.)

Our first calculation should be to confirm that \(A\) is invertible. Applying Proposition 5.5.4, for \(A\) we have

\begin{equation*} a d - b c = 1 \cdot 10 - (-3) \cdot (-3) = 1 \text{,} \end{equation*}

and so \(A\) is indeed invertible, with inverse

\begin{equation*} \inv{A} = \begin{bmatrix} 10 \amp 3 \\ 3 \amp 1 \end{bmatrix}\text{.} \end{equation*}

With this inverse in hand, and noting that the \(I\) in our equality should be the \(2 \times 2\) identity matrix to match the dimensions of the other matrices involved, we may calculate

\begin{align*} I - B \amp = \bidentmattwo - \begin{abmatrix}{rr} 3 \amp 1 \\ 3 \amp -3 \end{abmatrix}\\ \amp = \begin{abmatrix}{rr} -2 \amp -1 \\ -3 \amp 4 \end{abmatrix}\text{,} \end{align*}

\begin{align*} (I - B) \inv{A} \amp = \begin{abmatrix}{rr} -2 \amp -1 \\ -3 \amp 4 \end{abmatrix} \begin{bmatrix} 10 \amp 3 \\ 3 \amp 1 \end{bmatrix}\\ \amp = \begin{abmatrix}{rr} -23 \amp -7 \\ -18 \amp -5 \end{abmatrix}\text{,} \end{align*}

\begin{align*} X \amp = \utrans{\bbrac{(I - B) \inv{A}}} \\ \amp = \utrans{ \begin{abmatrix}{rr} -23 \amp -7 \\ -18 \amp -5 \end{abmatrix} }\\ \amp = \begin{abmatrix}{rr} -23 \amp -18 \\ - 7 \amp - 5 \end{abmatrix}\text{.} \end{align*}

26.

\(\displaystyle \begin{abmatrix}{rrrr} 1 \amp 3 \amp -2 \amp 2 \\ 0 \amp 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 1 \amp 0 \\ 1 \amp 4 \amp 2 \amp -1 \end{abmatrix} X \begin{abmatrix}{rrrr} 4 \amp -11 \amp 14 \amp -3 \\ -1 \amp 3 \amp -4 \amp 1 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \end{abmatrix} - 5 I = \begin{abmatrix}{rrrr} -8 \amp 2 \amp -9 \amp 7 \\ 7 \amp 1 \amp 3 \amp -5 \\ 2 \amp 6 \amp 6 \amp -4 \\ -8 \amp 1 \amp 9 \amp 3 \end{abmatrix}\)

Hint.

First see Exercise 7.

Solution.

It’s less messy to perform algebraic manipulations with letters, so assign

\begin{align*} A \amp = \begin{abmatrix}{rrrr} 1 \amp 3 \amp -2 \amp 2 \\ 0 \amp 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 1 \amp 0 \\ 1 \amp 4 \amp 2 \amp -1 \end{abmatrix} \text{,}\\ B \amp = \begin{abmatrix}{rrrr} 4 \amp -11 \amp 14 \amp -3 \\ -1 \amp 3 \amp -4 \amp 1 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \end{abmatrix} \text{,}\\ C \amp = \begin{abmatrix}{rrrr} -8 \amp 2 \amp -9 \amp 7 \\ 7 \amp 1 \amp 3 \amp -5 \\ 2 \amp 6 \amp 6 \amp -4 \\ -8 \amp 1 \amp 9 \amp 3 \end{abmatrix} \text{.} \end{align*}

Then we have

\begin{gather*} A X B - 5 I = C \\ A X B = C + 5 I \\ \inv{A} (A X B) \inv{B} = \inv{A} (C + 5 I) \inv{B} \\ (\inv{A} A) X (B \inv{B}) = \inv{A} (C + 5 I) \inv{B} \\ I X I = \inv{A} (C + 5 I) \inv{B} \\ X = \inv{A} (C + 5 I) \inv{B} \text{,} \end{gather*}

assuming that both \(A\) and \(B\) are invertible.

Remember that order of matrix multiplication matters. In particular, we cannot rearrange \(A X B\) to either \(A B X\) or \(X A B\text{,}\) and we have been careful to cancel \(A\) and \(B\) by multiplying by \(\inv{A}\) on the left and by \(\inv{B}\) on the right on both sides of the equality.

Now back to the question of the invertibility of \(A\) and \(B\text{.}\) Per the hint, matrices \(A\) and \(B\) in this exercise match the matrices in Exercise 5.6.7, where we verified that they are inverses of each other; that is,

\begin{align*} \inv{A} \amp = B \text{,} \amp \inv{B} \amp = A \text{.} \end{align*}

While normally there is no calculation advantage to expanding the brackets in our last algebraic expression above, in this case we might be able to further simplify our right-hand side a bit more:

\begin{gather*} X = \inv{A} (C + 5 I) \inv{B} \\ X = \inv{A} (C + 5 I) A \\ X = \inv{A} C A + 5 \inv{A} I A \\ X = \inv{A} C A + 5 \inv{A} A \\ X = \inv{A} C A + 5 I \\ X = B C A + 5 I \text{.} \end{gather*}

Once again, remember that order of matrix multiplication matters, and so while \(\inv{A} I A = I\) is true because first we may simplify \(I A = A\) before “cancelling” \(A\) with its inverse, we cannot similarly simplify \(\inv{A} C A\) because the \(A\) is not right next to its inverse to “cancel.”

Our final algebraic expression above now tells us what to calculate:

\begin{align*} B C \amp = \begin{abmatrix}{rrrr} 4 \amp -11 \amp 14 \amp -3 \\ -1 \amp 3 \amp -4 \amp 1 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \end{abmatrix} \begin{abmatrix}{rrrr} -8 \amp 2 \amp -9 \amp 7 \\ 7 \amp 1 \amp 3 \amp -5 \\ 2 \amp 6 \amp 6 \amp -4 \\ -8 \amp 1 \amp 9 \amp 3 \end{abmatrix}\\ \amp = \begin{abmatrix}{rrrr} -57 \amp 78 \amp -12 \amp 18 \\ 13 \amp -22 \amp 3 \amp -3 \\ 2 \amp 6 \amp 6 \amp -4 \\ 7 \amp 1 \amp 3 \amp -5 \end{abmatrix}\text{,} \end{align*}

\begin{align*} B C A \amp = (B C) A \\ \amp = \begin{abmatrix}{rrrr} -57 \amp 78 \amp -12 \amp 18 \\ 13 \amp -22 \amp 3 \amp -3 \\ 2 \amp 6 \amp 6 \amp -4 \\ 7 \amp 1 \amp 3 \amp -5 \end{abmatrix} \begin{abmatrix}{rrrr} 1 \amp 3 \amp -2 \amp 2 \\ 0 \amp 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 1 \amp 0 \\ 1 \amp 4 \amp 2 \amp -1 \end{abmatrix}\\ \amp = \begin{abmatrix}{rrrr} -39 \amp -99 \amp 138 \amp -54 \\ 10 \amp 27 \amp -29 \amp 7 \\ - 2 \amp -10 \amp - 6 \amp 14 \\ 2 \amp 1 \amp -21 \amp 20 \end{abmatrix}\text{.} \end{align*}

Taking the \(I\) in the equality to represent the \(4 \times 4\) identity matrix to match the dimensions of the other matrices, we also have

\begin{equation*} 5 I = 5 \bidentmatfour = \begin{bmatrix} 5 \amp 0 \amp 0 \amp 0 \\ 0 \amp 5 \amp 0 \amp 0 \\ 0 \amp 0 \amp 5 \amp 0 \\ 0 \amp 0 \amp 0 \amp 5 \\ \end{bmatrix}\text{.} \end{equation*}

Finally,

\begin{align*} X \amp = B C A + 5 I \\ \amp = \begin{abmatrix}{rrrr} -39 \amp -99 \amp 138 \amp -54 \\ 10 \amp 27 \amp -29 \amp 7 \\ - 2 \amp -10 \amp - 6 \amp 14 \\ 2 \amp 1 \amp -21 \amp 20 \end{abmatrix} + \begin{bmatrix} 5 \amp 0 \amp 0 \amp 0 \\ 0 \amp 5 \amp 0 \amp 0 \\ 0 \amp 0 \amp 5 \amp 0 \\ 0 \amp 0 \amp 0 \amp 5 \\ \end{bmatrix}\\ \amp = \begin{abmatrix}{rrrr} -34 \amp -99 \amp 138 \amp -54 \\ 10 \amp 32 \amp -29 \amp 7 \\ - 2 \amp -10 \amp - 1 \amp 14 \\ 2 \amp 1 \amp -21 \amp 25 \end{abmatrix}\text{.} \end{align*}

Guessing inversion patterns from multiplication patterns.

In each of the following, use the referenced multiplication pattern from Exercises 4.6 to answer the following about the given matrix.

Using the idea that an inverse should “reverse” multiplication, guess the inverse of matrix.
Verify that your guess is correct by multiplying it against the original matrix.
Determine whether there any values or combinations of values for the variable entries in the original matrix for which your inverse doesn’t work. In those situations, is the original matrix invertible?

27.

Multiplication pattern from Exercise 4.6.36; matrix

\begin{equation*} \begin{bmatrix} a \amp 0 \amp 0 \\ 0 \amp b \amp 0 \\ 0 \amp 0 \amp c \end{bmatrix}\text{.} \end{equation*}

28.

Multiplication pattern from Exercise 4.6.94; matrix

\begin{equation*} \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp a \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{bmatrix}\text{.} \end{equation*}

Singular patterns of zero entries.

Having nonzero entries does not guarantee that a matrix will be invertible. But, on the other hand, certain patterns of zero entries in a matrix guarantee that it will be singular.

29.

Argue that a matrix with a row that is all zeros must be singular.

Hint.

Consider what Exercise 4.6.101 says about the possibility that the definition of inverse matrix could be fulfilled for such a matrix.

30.

Argue that a matrix with a column that is all zeros must be singular.

Hint.

Consider what Exercise 4.6.102 says about the possibility that the definition of inverse matrix could be fulfilled for such a matrix.

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