Exercises: Multiple Systems and Matrix Inversion

This set of exercises is retrieved from the fifth chapter of Linear Algebra by Robert J. Valenza. Note that these solutions are not fully elaborated; You have to fill the descriptions by yourself.

Problem 5.1
Solve the following matrix equation for $x,$ $y,$ $z$ and $w .$ $(\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}) (\begin{matrix} x & y \\ z & w \end{matrix}) = (\begin{matrix} 10 & 2 \\ 4 & 2 \end{matrix})$

Solution. Taking $R_{1} \leftarrow R_{1} - 2 R_{2},$ we obtain $(\begin{array}{cccc} 1 & 2 & 10 & 2 \\ 0 & 1 & 4 & 2 \end{array}) \to (\begin{array}{cccr} 1 & 0 & 2 & - 2 \\ 0 & 1 & 4 & 2 \end{array}) .$ Thus $(\begin{matrix} x & y \\ z & w \end{matrix}) = (\begin{array}{cr} 2 & - 2 \\ 4 & 2 \end{array}) .$

Problem 5.2
Suppose that $(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}) (\begin{matrix} x & y \\ z & w \end{matrix}) = (\begin{matrix} x & y \\ z & w \end{matrix}) (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}) .$ Show that $x = w,$ $z = 0$ and there is no constraint on $y .$

Solution. Evaluating each multiplication gives $(\begin{matrix} x + z & y + w \\ z & w \end{matrix}) = (\begin{matrix} x & x + y \\ z & z + w \end{matrix}) .$ From this, we obtain $(\begin{matrix} z & w - x \\ 0 & - z \end{matrix}) = (\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix})$ or $z = 0, w - x = 0.$ Therefore the solution is $x = w, z = 0$ and $y$ is any real number.

Problem 5.3
Let $A$ be an $m \times n$ matrix. Show that the matrix products $A \cdot^{t} A$ and $^{t} A \cdot A$ are both defined. Compute these products for the matrix $A = (\begin{matrix} 2 & 0 & 1 \\ 0 & 1 & 1 \end{matrix}) .$

Solution. $A$ is an $m \times n$ matrix and $^{t} A$ is an $n \times m$ matrix, thus both $A \cdot^{t} A$ and $^{t} A \cdot A$ are defined. $\begin{aligned} A \cdot^{t} A & = (\begin{array}{c} 2 & 0 & 1 \\ 0 & 1 & 1 \end{array}) (\begin{array}{c} 2 & 0 \\ 0 & 1 \\ 1 & 1 \end{array}) = (\begin{array}{c} 5 & 1 \\ 1 & 2 \end{array}), \\ ^{t} A \cdot A & = (\begin{array}{c} 2 & 0 \\ 0 & 1 \\ 1 & 1 \end{array}) (\begin{array}{c} 2 & 0 & 1 \\ 0 & 1 & 1 \end{array}) = (\begin{array}{c} 4 & 0 & 2 \\ 0 & 1 & 1 \\ 2 & 1 & 2 \end{array}) . \end{aligned}$

Problem 5.4
What is the dimension of the space of $4 \times 4$ symmetric matrices defined over a field $K ?$ Of all $n \times n$ symmetric matrices?

Solution. $\frac{n (n + 1)}{2} .$

Problem 5.5
What is the dimension of the space of all $5 \times 5$ upper triangular matrices in $M_{n} (K) ?$ Of all $n \times n$ upper triangular matrices?

Solution. $\frac{n (n + 1)}{2} .$

Problem 5.6
Show that the product of two invertible matrices in $M_{n} (K)$ is invertible.

Solution. Suppose both $A$ and $B$ belong to $M_{n} (K)$ and invertible. Take $C = B^{- 1} A^{- 1} .$ Then $\begin{matrix} (A B) C = (A B) (B^{- 1} A^{- 1}) = I_{n}, \\ C (A B) = (B^{- 1} A^{- 1}) (A B) = I_{n} . \end{matrix}$ Therefore $A B$ is invertible and the inverse matrix is $C = B^{- 1} A^{- 1} .$

Problem 5.7
Let $A = (\begin{matrix} a & b \\ c & d \end{matrix})$ lie in $M_{2} (K)$ and assume that $a d - b c \neq 0.$ Show that $A^{- 1}$ exists and that in fact $A^{- 1} = \frac{1}{a d - b c} (\begin{array}{rr} d & - b \\ - c & a \end{array}) .$ Conversely, show that if $a d - b c = 0,$ then $A$ is not invertible.

Solution. Suppose that $a d - b c \neq 0.$ A simple calculation shows that $(\begin{matrix} a & b \\ c & d \end{matrix}) \cdot \frac{1}{a d - b c} (\begin{array}{rr} d & - b \\ - c & a \end{array}) = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}),$ therefore $\frac{1}{a d - b c} (\begin{array}{rr} d & - b \\ - c & a \end{array})$ is the inverse matrix of $A,$ and $A$ is invertible.

Now suppose that $a d - b c = 0.$ Suppose contrarily that $A$ is invertible and the inverse matrix is $A^{- 1} = (\begin{matrix} w & x \\ y & z \end{matrix}) .$ By definition of an inverse matrix, $(\begin{matrix} w & x \\ y & z \end{matrix}) (\begin{matrix} a & b \\ c & d \end{matrix}) = (\begin{matrix} a w + c x & b w + d x \\ a y + c z & b y + d z \end{matrix}) = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) .$ Therefore we have $d = 0$ , and $b = 0$ or $c = 0.$

If $d = 0$ and $b = 0,$ then $(\begin{matrix} * & * \\ * & * \end{matrix}) (\begin{matrix} a & b \\ c & d \end{matrix}) = (\begin{matrix} * & 0 \\ * & 0 \end{matrix}),$ which cannot be the identity matrix. If $d = 0$ and $c = 0,$ then $(\begin{matrix} a & b \\ c & d \end{matrix}) (\begin{matrix} * & * \\ * & * \end{matrix}) = (\begin{matrix} * & * \\ 0 & 0 \end{matrix}),$ which neither cannot be the identity matrix.

Problem 5.8
Let $B$ be the subset of ${GL}_{2} (K)$ consisting of all matrices of the form $A = (\begin{matrix} a & b \\ 0 & c \end{matrix}), a c \neq 0.$ Show that $B$ is a subgroup of ${GL}_{2} (K) .$

Solution. Let $\begin{aligned} A & = (\begin{array}{c} a_{1} & b_{1} \\ 0 & c_{1} \end{array}), \\ B & = (\begin{array}{c} a_{2} & b_{2} \\ 0 & c_{2} \end{array}), \end{aligned}$ where $a_{1} c_{1} \neq 0,$ $a_{2} c_{2} \neq 0.$ Then $B$ is invertible and $\begin{aligned} A B^{- 1} & = \frac{1}{a_{2} c_{2}} (\begin{array}{c} a_{1} & b_{1} \\ 0 & c_{1} \end{array}) (\begin{array}{cr} c_{2} & - b_{2} \\ 0 & a_{2} \end{array}) \\ = \frac{1}{a_{2} c_{2}} (\begin{array}{c} a_{1} c_{2} & a_{2} b_{1} - a_{1} b_{2} \\ 0 & a_{2} c_{1} \end{array}) \end{aligned}$ which obviously belongs to ${GL}_{2} (K) .$

For this problem and next(9-10), superscripts on matrices indicate exponents rather than columns.

Problem 5.9
Find a $2 \times 2$ matrix $A$ such that $A \neq 0$ but nonetheless $A^{2} = 0.$ Next, find a $3 \times 3$ matrix such that $A \neq 0,$ $A^{2} \neq 0$ but $A^{3} = 0.$ Finally, for arbitrary $n$ find an $n \times n$ matrix $A$ such that none of $A,$ $A^{2},$ $\dots,$ $A^{n - 1}$ equals $0,$ but $A^{n} = 0.$

Solution. Take $A = {(a_{i j})}_{n \times n}$ where $a_{i j} = {\begin{cases} 1 & if i + 1 = j \\ 0 & otherwise. \end{cases}$ As an example, for $n = 4,$ we take $A = (\begin{matrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{matrix}) .$

Problem 5.10
Let $A \in M_{n} (K)$ be such that $A^{r} = 0$ for some integer $r \geq 1.$ Show that $I_{n} - A$ is invertible.

Solution. Take $B = I_{n} + A + A^{2} + \dots + A^{r - 1},$ then $\begin{aligned} (I_{n} - A) B & = (I_{n} - A) (I_{n} + A + A^{2} + \dots + A^{r - 1}) \\ = (I_{n} + A + A^{2} + \dots + A^{r - 1}) - (A + A^{2} + A^{3} + \dots + A^{r}) \\ = I_{n} - A^{r} \\ = I_{n} - O = I_{n} . \end{aligned}$

Problem 5.11
List the elements in ${GL}_{2} (F_{2})$ where $F_{2}$ is the finite field of two elements $0$ and $1.$

Solution. Suppose $(\begin{matrix} a & b \\ c & d \end{matrix}) \in {GL}_{2} (F_{2})$ then $a d - b c \neq 0,$ that is, $a d \neq b c .$ Since each of $a,$ $b,$ $c$ and $d$ is $0$ or $1,$ $a d \neq b c$ arises as one of the following form: $\begin{matrix} 0 \times 0 \neq 1 \times 1, \\ 0 \times 1 \neq 1 \times 1, \\ 1 \times 0 \neq 1 \times 1, \\ 1 \times 1 \neq 0 \times 0, \\ 1 \times 1 \neq 0 \times 1, \\ 1 \times 1 \neq 1 \times 0. \end{matrix}$ Thus all the desired matrices are $\begin{matrix} (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}), (\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix}), (\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}), \\ (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}), (\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}), (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}) . \end{matrix}$ In fact the multiplicative group consisting of these matrices are isomorphic to $S_{3} .$

Problem 5.12
Explain succinctly why the solution space to a homogeneous system of $m$ linear equations in $n$ unknowns defined over a field $K$ is a subspace of $K^{n} .$

Solution. Let a system be given as follows. ${\begin{aligned} a_{11} x_{1} + a_{12} x_{2} + \dots + a_{1 n} x_{n} & = 0 \\ a_{21} x_{1} + a_{22} x_{2} + \dots + a_{2 n} x_{n} & = 0 \\ ⋮ \\ a_{m 1} x_{1} + a_{m 2} x_{2} + \dots + a_{m n} x_{n} & = 0 \end{aligned}$ Let $V$ be the subset of $K^{n}$ consisting of the vectors $(x_{1}, \dots, x_{n})$ that satisfy the above system. Let $\begin{aligned} v & = (v_{1}, v_{2}, \dots, v_{n}) \in V, \\ w & = (w_{1}, w_{2}, \dots, w_{n}) \in V, \end{aligned}$ and let $λ$ be a scalar. Since $(\begin{matrix} a_{11} & \dots & a_{1 n} \\ ⋮ & ⋱ & ⋮ \\ a_{m 1} & \dots & a_{m n} \end{matrix}) (\begin{matrix} v_{1} \\ ⋮ \\ v_{n} \end{matrix}) = (\begin{matrix} 0 \\ ⋮ \\ 0 \end{matrix})$ and $(\begin{matrix} a_{11} & \dots & a_{1 n} \\ ⋮ & ⋱ & ⋮ \\ a_{m 1} & \dots & a_{m n} \end{matrix}) (\begin{matrix} w_{1} \\ ⋮ \\ w_{n} \end{matrix}) = (\begin{matrix} 0 \\ ⋮ \\ 0 \end{matrix}),$ we have $(\begin{matrix} a_{11} & \dots & a_{1 n} \\ ⋮ & ⋱ & ⋮ \\ a_{m 1} & \dots & a_{m n} \end{matrix}) [(\begin{matrix} v_{1} \\ ⋮ \\ v_{n} \end{matrix}) + (\begin{matrix} w_{1} \\ ⋮ \\ w_{n} \end{matrix})] = (\begin{matrix} 0 \\ ⋮ \\ 0 \end{matrix})$ and $(\begin{matrix} a_{11} & \dots & a_{1 n} \\ ⋮ & ⋱ & ⋮ \\ a_{m 1} & \dots & a_{m n} \end{matrix}) (\begin{matrix} λ v_{1} \\ ⋮ \\ λ v_{n} \end{matrix}) = (\begin{matrix} 0 \\ ⋮ \\ 0 \end{matrix}),$ that is $v + w \in V, λ v \in V .$ Hence $V$ is a subspace of $F^{n} .$

Problem 5.13
Express the following linear system as a single matrix equation and as a single vector equation. $\begin{aligned} 2 x_{1} - 4 x_{2} & = 7 \\ 5 x_{1} + 9 x_{2} & = 4 \end{aligned}$

Solution. $(\begin{array}{cr} 2 & - 4 \\ 5 & 9 \end{array}) (\begin{matrix} x_{1} \\ x_{2} \end{matrix}) = (\begin{matrix} 7 \\ 4 \end{matrix}) .$

Problem 5.14
Without explicitly solving, show that the system above has a unique solution.

Solution. Since $det (\begin{array}{cr} 2 & - 4 \\ 5 & 9 \end{array}) = 38 \neq 0,$ by Problem 7, given matrix is invertible.

Problem 5.15
Without explicitly solving, show that the system $\begin{aligned} 2 x_{1} + x_{2} & = y_{1} \\ - x_{1} + 2 x_{2} & = y_{2} \end{aligned}$ has a unique solution for all $y_{1}, y_{2} \in R .$

Solution. $det (\begin{array}{rc} 2 & 1 \\ - 1 & 2 \end{array}) = 5 \neq 0.$

Problem 5.16
Find the rank of the matrix $A = (\begin{matrix} 1 & 2 & 1 \\ 2 & 0 & 2 \\ 0 & 0 & 1 \end{matrix}) .$

Solution. Begin with the given matrix. $(\begin{matrix} 1 & 2 & 1 \\ 2 & 0 & 2 \\ 0 & 0 & 1 \end{matrix})$ Take $R_{2} \leftarrow R_{2} \times \frac{1}{2},$ then we obtain $(\begin{matrix} 1 & 2 & 1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{matrix}) .$ Take $R_{2} \leftarrow R_{2} - R_{3},$ then we obtain $(\begin{matrix} 1 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}) .$ Take $R_{1} \leftarrow R_{1} - R_{2} - R_{3},$ then we obtain $(\begin{matrix} 0 & 2 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}) .$ Take $R_{1} \leftrightarrow R_{2},$ then we obtain $(\begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{matrix}) .$ Take $R_{2} \leftarrow R_{2} \times \frac{1}{2},$ then we obtain $(\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}),$ which equals $I_{3} .$ Therefore $A$ is invertible, and $rk (A) = 3.$

Problem 5.17
Let $A$ be a $4 \times 7$ matrix (over any field) with at least one nonzero entry. What are the possible values for the rank of $A ?$

Solution. $1 \leq rk (A) \leq 4.$

Problem 5.18
Using Gauss-Jordan elimination, solve the system given in Problem 13.

Solution. $\begin{aligned} (\begin{array}{crc} 2 & - 4 & 7 \\ 5 & 9 & 4 \end{array}) & \to (\begin{array}{crr} 2 & - 4 & 7 \\ 0 & 19 & - \frac{27}{2} \end{array}) \\ \to (\begin{array}{ccr} 2 & 0 & \frac{79}{19} \\ 0 & 19 & - \frac{27}{2} \end{array}) \\ \to (\begin{array}{ccr} 1 & 0 & \frac{79}{38} \\ 0 & 19 & - \frac{27}{2} \end{array}) . \end{aligned}$ Therefore $x_{1} = \frac{79}{38}, x_{2} = - \frac{27}{38} .$

Problem 5.19
Suppose that we are solving a $2 \times 3$ homogeneous linear system $A x = 0$ by Gauss-Jordan elimination and reach the following augmented matrix in reduced row-echelon form: $(\begin{array}{ccrc} 1 & 0 & - 5 & 0 \\ 0 & 1 & 4 & 0 \end{array}) .$ What is the general solution to the original system? What is the dimension of the solution space?

Solution. Let $A = (\begin{array}{ccr} 1 & 0 & - 5 \\ 0 & 1 & 4 \end{array}) .$ Then $A x = 0$ if and only if $x \in Ker (T_{A}) .$ From a simple calculations we obtain $(5, - 4, 1)$ belongs to the kernel. Therefore all the solutions for the given system have the form $x = t (5, - 4, 1)$ where $t \in R .$

Problem 5.20
Suppose that we are solving a $3 \times 4$ linear system $A x = y$ by Gauss-Jordan elimination and reach the following augmented matrix in reduced row-echelon form: $(\begin{array}{ccccc} 1 & 2 & 0 & 1 & 5 \\ 0 & 0 & 1 & 4 & 2 \\ 0 & 0 & 0 & 0 & 4 \end{array}) .$ What can one say about solutions to the original system?

Solution. There are no solutions.

Problem 5.21
Find all solutions to the following system by Gauss-Jordan elimination: $\begin{aligned} x_{1} + x_{2} - x_{3} & = 0 \\ x_{1} + 2 x_{2} + 4 x_{3} & = 0 \end{aligned}$ What is the dimension of the solution space?

Solution. $\begin{aligned} (\begin{array}{ccrc} 1 & 1 & - 1 & 0 \\ 1 & 2 & 4 & 0 \end{array}) \\ \to & (\begin{array}{ccrc} 1 & 1 & - 1 & 0 \\ 0 & 1 & 5 & 0 \end{array}) \\ \to & (\begin{array}{ccrc} 1 & 0 & - 6 & 0 \\ 0 & 1 & 5 & 0 \end{array}) . \end{aligned}$ Hence the solutions is $x = t (6, - 5, 1)$ where $t \in R .$ The dimension of the solution space is $1.$

Problem 5.22
Does every linear system for which there are more variables than equations have a solution? If not, what additional condition is needed?

Solution. No. The rank of augmented matrix has to equal to the rank of coefficient matrix.

Problem 5.23
Summarize in your own words why reduced row-echelon form is an effective device for solving linear systems of equation.

Solution. First, it is easy to determine whether the system has any solutions or not; Second, it is easy to find the unknowns step by step.

Problem 5.24
Factor the following matrix into the product of a lower triangular and an upper triangular triangular matrices: $A = (\begin{matrix} 2 & 5 \\ 1 & 2 \end{matrix}) .$

Solution. Suppose $(\begin{matrix} a & 0 \\ b & c \end{matrix}) (\begin{matrix} x & y \\ 0 & z \end{matrix}) = (\begin{matrix} 2 & 5 \\ 1 & 2 \end{matrix}) .$ There are $6$ unknowns while the number of equations is $2.$ Therefore some of the unknowns can be taken arbitrarily. Taking $a = 1$ and $x = 2,$ we obtain $(\begin{matrix} 2 & y \\ 2 b & b y + c z \end{matrix}) = (\begin{matrix} 2 & 5 \\ 1 & 2 \end{matrix}) .$ In this equation, we inevitably have $y = 5$ and $b = \frac{1}{2},$ and we obtain $(\begin{matrix} 2 & 5 \\ 1 & \frac{5}{2} + c z \end{matrix}) = (\begin{matrix} 2 & 5 \\ 1 & 2 \end{matrix}) .$ Taking $c = \frac{1}{2},$ we have $z = - 1.$ Therefore $A$ can be factored as follows: $(\begin{matrix} 1 & 0 \\ \frac{1}{2} & \frac{1}{2} \end{matrix}) (\begin{array}{cr} 2 & 5 \\ 0 & - 1 \end{array}) = (\begin{matrix} 2 & 5 \\ 1 & 2 \end{matrix}) .$

Problem 5.25
Given the matrix factorization $(\begin{matrix} 1 & 1 & 1 \\ 2 & 4 & 8 \\ 1 & 5 & 11 \end{matrix}) = (\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 1 & 2 & 1 \end{matrix}) (\begin{matrix} 1 & 1 & 2 \\ 0 & 2 & 4 \\ 0 & 0 & 1 \end{matrix})$ solve the linear system $\begin{aligned} x_{1} + x_{2} + 2 x_{3} & = 11 \\ 2 x_{1} + 4 x_{2} + 8 x_{3} & = 30 \\ x_{1} + 5 x_{2} + 11 x_{3} & = 28 \end{aligned}$ by the method of $L U$ decomposition.

Solution. Express the given matrix factorization as $A = L U$ respectively, and let $\begin{aligned} x & = (x_{1}, x_{2}, x_{3}), \\ y & = (11, 30, 28), \\ z & = (z_{1}, z_{2}, z_{3}) . \end{aligned}$ From the equation $L z = y$ we have $z = (11, 8, 1) .$ Next, solving the equation $U x = z$ we have $x = (7, 2, 1) .$ Therefore the solution is $x_{1} = 7, x_{2} = 2, x_{3} = 1.$

Problem 5.26
Summarize in your own words why $L U$ decomposition in an effective device for solving linear systems of equations.

Solution. There exists straightforward obvious algorithm to find the solutions.

Problem 5.27
Use the technique of finding solutions of multiple systems to derive the general formula for the inversion of $2 \times 2$ matrices.

Solution. Assume that $a d - b c \neq 0.$ $\begin{aligned} (\begin{array}{cccc} a & b & 1 & 0 \\ c & d & 0 & 1 \end{array}) \\ \to & (\begin{array}{cccc} a c & b c & c & 0 \\ a c & a d & 0 & a \end{array}) \\ \to & (\begin{array}{ccrc} a c & b c & c & 0 \\ 0 & a d - b c & - c & a \end{array}) \\ \to & (\begin{array}{cccc} a c & b c & c & 0 \\ 0 & 1 & \frac{- c}{a d - b c} & \frac{a}{a d - b c} \end{array}) \\ \to & (\begin{array}{cccc} a c & 0 & \frac{a c d}{a d - b c} & \frac{- a b c}{a d - b c} \\ 0 & 1 & \frac{- c}{a d - b c} & \frac{a}{a d - b c} \end{array}) \\ \to & (\begin{array}{cccc} 1 & 0 & \frac{d}{a d - b c} & \frac{- b}{a d - b c} \\ 0 & 1 & \frac{- c}{a d - b c} & \frac{a}{a d - b c} \end{array}) . \end{aligned}$

Problem 5.28
Using the technique of finding solutions of multiple systems, invert the following carefully contrived matrix: $A = (\begin{matrix} 1 & 2 & 2 \\ 2 & 4 & 3 \\ 3 & 5 & 4 \end{matrix}) .$ This should go rather smoothly.

Solution. $\begin{aligned} (\begin{array}{cccccc} 1 & 2 & 2 & 1 & 0 & 0 \\ 2 & 4 & 3 & 0 & 1 & 0 \\ 3 & 5 & 4 & 0 & 0 & 1 \end{array}) \\ \to & (\begin{array}{crrrcc} 1 & 2 & 2 & 1 & 0 & 0 \\ 0 & 0 & - 1 & - 2 & 1 & 0 \\ 0 & - 1 & - 2 & - 3 & 0 & 0 \end{array}) \\ \to & (\begin{array}{crrrrc} 1 & 2 & 0 & - 3 & 2 & 0 \\ 0 & 0 & - 1 & - 2 & 1 & 0 \\ 0 & - 1 & 0 & 1 & - 2 & 0 \end{array}) \\ \to & (\begin{array}{cccrrc} 1 & 0 & 0 & - 1 & - 2 & 0 \\ 0 & 0 & 1 & 2 & - 1 & 0 \\ 0 & 1 & 0 & - 1 & 2 & 0 \end{array}) \\ \to & (\begin{array}{cccrrc} 1 & 0 & 0 & - 1 & 02 & 0 \\ 0 & 1 & 0 & - 1 & 2 & 0 \\ 0 & 0 & 1 & 2 & - 1 & 0 \end{array}) . \end{aligned}$

Problem 5.29
To the sound of the rain and the chamber music of Claude Debussy, the author reaches for his calculator, an old but serviceable hp-11C. Punching the random number key nine times and recording the first digit to the right of the decimal point, he produces the following matrix: $A = (\begin{matrix} 9 & 2 & 0 \\ 2 & 4 & 3 \\ 8 & 6 & 1 \end{matrix}) .$ He ponders; he frowns; he leaves it to the student to find $A^{- 1} .$

Solution. $A^{- 1} = \frac{1}{82} (\begin{array}{rrr} 14 & 2 & - 6 \\ - 22 & 9 & 27 \\ 20 & 38 & - 32 \end{array}) .$

Exercises: Multiple Systems and Matrix Inversion

Exercises: Dimension

Exercises: Representation of Linear Transformations