Exercises: Multiple Systems and Matrix Inversion

Solution. Taking \(R_1 \,\leftarrow\, R_1 - 2R_2 ,\) we obtain \[\left( \begin{array}{cc|cc} 1 & 2 & 10 & 2 \\ 0 & 1 & 4 & 2 \end{array}\right) \,\rightarrow\, \left( \begin{array}{cc|cr} 1 & 0 & 2 & -2 \\ 0 & 1 & 4 & 2 \end{array}\right). \] Thus \[\begin{pmatrix} x&y \\ z&w \end{pmatrix} = \left(\begin{array}{cr} 2 & -2 \\ 4 & 2 \end{array}\right).\]

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

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. \[ \begin{align} A \cdot \,^t \! A &= \begin{pmatrix} 2 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 5 & 1 \\ 1 & 2 \end{pmatrix}, \\[8pt] \,^t \! A \cdot A &= \begin{pmatrix} 2 & 0 \\ 0 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 2 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix} = \begin{pmatrix} 4 & 0 & 2 \\ 0 & 1 & 1 \\ 2 & 1 & 2 \end{pmatrix} . \end{align} \]

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

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

Solution. Suppose both \(A\) and \(B\) belong to \(M_n (K)\) and invertible. Take \(C = B^{-1} A^{-1}.\) Then \[\begin{gather} (AB)C = (AB)(B^{-1} A^{-1}) = I_n , \\[7pt] C(AB) = (B^{-1} A^{-1})(AB) = I_n . \end{gather}\] Therefore \(AB\) is invertible and the inverse matrix is \(C = B^{-1} A^{-1}.\)

Solution. Suppose that \(ad-bc \ne 0.\) A simple calculation shows that \[\begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot \frac{1}{ad-bc} \left(\begin{array}{rr} d & -b \\ -c & a \end{array}\right) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},\] therefore \[ \frac{1}{ad-bc} \left(\begin{array}{rr} d&-b\\-c&a\end{array}\right)\] is the inverse matrix of \(A,\) and \(A\) is invertible.

Now suppose that \(ad-bc = 0.\) Suppose contrarily that \(A\) is invertible and the inverse matrix is \[A^{-1} = \begin{pmatrix} w & x \\ y & z \end{pmatrix}.\] By definition of an inverse matrix, \[\begin{pmatrix} w & x \\ y & z \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} aw+cx & bw+dx \\ ay+cz & by+dz \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.\] Therefore we have \(d=0\), and \(b=0\) or \(c=0.\)

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

Solution. Let \[\begin{align} A &= \begin{pmatrix} a_1 & b_1 \\ 0 & c_1 \end{pmatrix}, \\[7pt] B &= \begin{pmatrix} a_2 & b_2 \\ 0 & c_2 \end{pmatrix}, \end{align}\] where \(a_1 c_1 \ne 0,\) \(a_2 c_2 \ne 0.\) Then \(B\) is invertible and \[\begin{align} AB^{-1} &= \frac{1}{a_2 c_2} \begin{pmatrix} a_1 & b_1 \\ 0 & c_1 \end{pmatrix} \left(\begin{array}{cr} c_2 & -b_2 \\ 0 & a_2 \end{array}\right) \\[7pt] &= \frac{1}{a_2 c_2} \begin{pmatrix} a_1 c_2 & a_2 b_1 - a_1 b_2 \\ 0 & a_2 c_1 \end{pmatrix} \end{align}\] which obviously belongs to \(\operatorname{GL}_2 (K).\)

Solution. Take \(A = \left( a_{ij} \right)_{n\times n}\) where \[a_{ij} = \begin{cases} 1 & \quad \text{if} \,\, i+1 = j \\[8pt] 0 & \quad \text{otherwise.} \end{cases}\] As an example, for \(n=4,\) we take \[A = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}. \]

Solution. Take \(B = I_n + A+ A^2 + \cdots + A^{r-1} ,\) then \[\begin{align} (I_n - A)B &= (I_n -A)(I_n +A +A^2 + \cdots + A^{r-1})\\[7pt] &= (I_n + A+A^2 + \cdots + A^{r-1}) - (A+A^2 + A^3 + \cdots + A^r )\\[7pt] &= I_n - A^r \\[7pt] &= I_n - O = I_n. \end{align}\]

Solution. Suppose \[\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \operatorname{GL}_2 (F_2 )\] then \(ad-bc \ne 0,\) that is, \(ad \ne bc.\) Since each of \(a,\) \(b,\) \(c\) and \(d\) is \(0\) or \(1,\) \(ad \ne bc\) arises as one of the following form: \[\begin{gather} 0 \times 0 \ne 1 \times 1, \\[7pt] 0 \times 1 \ne 1 \times 1, \\[7pt] 1 \times 0 \ne 1 \times 1, \\[7pt] 1 \times 1 \ne 0 \times 0, \\[7pt] 1 \times 1 \ne 0 \times 1, \\[7pt] 1 \times 1 \ne 1 \times 0. \end{gather}\] Thus all the desired matrices are \[\begin{gather} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \,\, \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}, \,\, \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}, \\[7pt] \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \,\, \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}, \,\, \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}. \end{gather}\] In fact the multiplicative group consisting of these matrices are isomorphic to \(S_3.\)

Solution. Let a system be given as follows. \[\left\{ \begin{align} a_{11} x_1 + a_{12} x_2 + \cdots + a_{1n} x_n &= 0 \\[7pt] a_{21} x_1 + a_{22} x_2 + \cdots + a_{2n} x_n &= 0 \\[7pt] \vdots \quad\quad \\[7pt] a_{m1} x_1 + a_{m2} x_2 + \cdots + a_{mn} x_n &= 0 \end{align} \right.\] Let \(V\) be the subset of \(K^n\) consisting of the vectors \((x_1 ,\, \cdots ,\, x_n )\) that satisfy the above system. Let \[\begin{align} \mathbf{v} &= (v_1 ,\, v_2 ,\, \cdots ,\, v_n) \in V, \\[7pt] \mathbf{w} &= (w_1 ,\, w_2 ,\, \cdots ,\, w_n) \in V, \end{align}\] and let \(\lambda\) be a scalar. Since \[\begin{pmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{m1} & \cdots & a_{mn} \end{pmatrix} \begin{pmatrix} v_1 \\ \vdots \\ v_n \end{pmatrix} = \begin{pmatrix} 0 \\ \vdots \\ 0 \end{pmatrix}\] and \[\begin{pmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{m1} & \cdots & a_{mn} \end{pmatrix} \begin{pmatrix} w_1 \\ \vdots \\ w_n \end{pmatrix} = \begin{pmatrix} 0 \\ \vdots \\ 0 \end{pmatrix},\] we have \[\begin{pmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{m1} & \cdots & a_{mn} \end{pmatrix} \left[ \begin{pmatrix} v_1 \\ \vdots \\ v_n \end{pmatrix} + \begin{pmatrix} w_1 \\ \vdots \\ w_n \end{pmatrix} \right] = \begin{pmatrix} 0 \\ \vdots \\ 0 \end{pmatrix}\] and \[\begin{pmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{m1} & \cdots & a_{mn} \end{pmatrix} \begin{pmatrix} \lambda v_1 \\ \vdots \\ \lambda v_n \end{pmatrix} = \begin{pmatrix} 0 \\ \vdots \\ 0 \end{pmatrix},\] that is \[\mathbf{v} + \mathbf{w} \in V ,\,\, \lambda \mathbf{v} \in V.\] Hence \(V\) is a subspace of \(F^n.\)

Solution. \[ \left(\begin{array}{cr} 2 & -4 \\ 5 & 9 \end{array}\right) \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 7 \\ 4 \end{pmatrix}. \]

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

Solution. \[\det\left(\begin{array}{rc} 2 & 1 \\ -1 & 2 \end{array}\right) =5 \ne 0.\]

Solution. Begin with the given matrix. \[\begin{pmatrix} 1&2&1 \\ 2&0&2 \\ 0&0&1 \end{pmatrix}\] Take \(R_2 \,\leftarrow\, R_2 \times \frac{1}{2},\) then we obtain \[\begin{pmatrix} 1&2&1 \\ 1&0&1 \\ 0&0&1 \end{pmatrix}.\] Take \(R_2 \,\leftarrow\,R_2 - R_3,\) then we obtain \[\begin{pmatrix} 1&2&1 \\ 1&0&0 \\ 0&0&1 \end{pmatrix}.\] Take \(R_1 \,\leftarrow\, R_1 - R_2 - R_3,\) then we obtain \[\begin{pmatrix} 0&2&0 \\ 1&0&0 \\ 0&0&1 \end{pmatrix}.\] Take \(R_1 \,\leftrightarrow\, R_2,\) then we obtain \[\begin{pmatrix} 1&0&0 \\ 0&2&0 \\ 0&0&1 \end{pmatrix}.\] Take \(R_2 \,\leftarrow\, R_2 \times\frac{1}{2},\) then we obtain \[\begin{pmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{pmatrix} ,\] which equals \(I_3.\) Therefore \(A\) is invertible, and \(\operatorname{rk}(A) = 3.\)

Solution. \(1\le \operatorname{rk}(A) \le 4.\)

Solution. \[\begin{align} \left(\begin{array}{cr|c} 2 & -4 & 7 \\ 5 & 9 & 4 \end{array}\right) \,&\rightarrow\, \left(\begin{array}{cr|r} 2 & -4 & 7\,\, \\ 0 & 19 & -\frac{27}{2} \end{array}\right) \\[7pt] &\rightarrow\, \left(\begin{array}{cc|r} 2 & 0 & \frac{79}{19} \\ 0 & 19 & -\frac{27}{2} \end{array}\right) \\[7pt] &\rightarrow\, \left(\begin{array}{cc|r} 1 & 0 & \frac{79}{38} \\ 0 & 19 & -\frac{27}{2} \end{array}\right) . \end{align}\] Therefore \[x_1 = \frac{79}{38} ,\,\,\, x_2 = - \frac{27}{38}.\]

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

Solution. There are no solutions.

Solution. \[\begin{align} & \left(\begin{array}{ccr|c} 1 & 1 & -1 & 0 \\ 1 & 2 & 4 & 0 \end{array}\right) \\[7pt] \rightarrow \, & \left(\begin{array}{ccr|c} 1 & 1 & -1 & 0 \\ 0 & 1 & 5 & 0 \end{array}\right) \\[7pt] \rightarrow \, & \left(\begin{array}{ccr|c} 1 & 0 & -6 & 0 \\ 0 & 1 & 5 & 0 \end{array}\right) . \end{align}\] Hence the solutions is \(\mathbf{x} = t(6,\,-5,\,1)\) where \(t\in\mathbb{R}.\) The dimension of the solution space is \(1.\)

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

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.

Solution. Suppose \[ \begin{pmatrix} a & 0 \\ b & c \end{pmatrix} \begin{pmatrix} x & y \\ 0 & z \end{pmatrix} = \begin{pmatrix} 2 & 5 \\ 1 & 2 \end{pmatrix}. \] 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 \[\begin{pmatrix} 2 & y \\ 2b & by+cz \end{pmatrix} = \begin{pmatrix} 2 & 5 \\ 1 & 2 \end{pmatrix}.\] In this equation, we inevitably have \(y=5\) and \(b = \frac{1}{2},\) and we obtain \[\begin{pmatrix} 2 & 5 \\ 1 & \frac{5}{2}+cz \end{pmatrix} = \begin{pmatrix} 2 & 5 \\ 1 & 2 \end{pmatrix}.\] Taking \(c=\frac{1}{2},\) we have \(z=-1.\) Therefore \(A\) can be factored as follows: \[ \begin{pmatrix} 1 & 0 \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} \left(\begin{array}{cr} 2 & 5 \\ 0 & -1 \end{array}\right) = \begin{pmatrix} 2 & 5 \\ 1 & 2 \end{pmatrix}. \]

Solution. Express the given matrix factorization as \(A = LU\) respectively, and let \[\begin{align} \mathbf{x} &= (x_1 ,\,x_2 ,\,x_3 ), \\[7pt] \mathbf{y} &= (11 ,\,30 ,\,28 ), \\[7pt] \mathbf{z} &= (z_1 ,\,z_2 ,\,z_3 ). \end{align}\] From the equation \(L\mathbf{z} = \mathbf{y}\) we have \[\mathbf{z} = (11,\,8,\,1).\] Next, solving the equation \(U \mathbf{x} = \mathbf{z}\) we have \[\mathbf{x} = (7,\,2,\,1).\] Therefore the solution is \[x_1 = 7,\,\, x_2 =2 ,\,\, x_3 = 1.\]

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

Solution. Assume that \(ad-bc \ne 0.\) \[\begin{align} & \left(\begin{array}{cc|cc} a & b & 1 & 0 \\ c & d & 0 & 1 \end{array}\right) \\[7pt] \rightarrow \, & \left(\begin{array}{cc|cc} ac & bc & c & 0 \\ ac & ad & 0 & a \end{array}\right) \\[7pt] \rightarrow \, & \left(\begin{array}{cc|rc} ac & bc & c & 0 \\ 0 & ad-bc & -c & a \end{array}\right) \\[7pt] \rightarrow \, & \left(\begin{array}{cc|cc} ac & bc & c & 0 \\ 0 & 1 & \frac{-c}{ad-bc} & \frac{a}{ad-bc} \end{array}\right) \\[7pt] \rightarrow \, & \left(\begin{array}{cc|cc} ac & 0 & \frac{acd}{ad-bc} & \frac{-abc}{ad-bc} \\ 0 & 1 & \frac{-c}{ad-bc} & \frac{a}{ad-bc} \end{array}\right) \\[7pt] \rightarrow \, & \left(\begin{array}{cc|cc} 1 & 0 & \frac{d}{ad-bc} & \frac{-b}{ad-bc} \\ 0 & 1 & \frac{-c}{ad-bc} & \frac{a}{ad-bc} \end{array}\right) . \end{align}\]

Solution. \[\begin{align} &\left(\begin{array}{ccc|ccc} 1 & 2 & 2 & 1 & 0 & 0 \\ 2 & 4 & 3 & 0 & 1 & 0 \\ 3 & 5 & 4 & 0 & 0 & 1 \end{array}\right) \\[7pt] \rightarrow\, &\left(\begin{array}{crr|rcc} 1 & 2 & 2 & 1 & 0 & 0 \\ 0 & 0 & -1 & -2 & 1 & 0 \\ 0 & -1 & -2 & -3 & 0 & 0 \end{array}\right) \\[7pt] \rightarrow\, &\left(\begin{array}{crr|rrc} 1 & 2 & 0 & -3 & 2 & 0 \\ 0 & 0 & -1 & -2 & 1 & 0 \\ 0 & -1 & 0 & 1 & -2 & 0 \end{array}\right) \\[7pt] \rightarrow\, &\left(\begin{array}{ccc|rrc} 1 & 0 & 0 & -1 & -2 & 0 \\ 0 & 0 & 1 & 2 & -1 & 0 \\ 0 & 1 & 0 & -1 & 2 & 0 \end{array}\right) \\[7pt] \rightarrow\, &\left(\begin{array}{ccc|rrc} 1 & 0 & 0 & -1 & 02 & 0 \\ 0 & 1 & 0 & -1 & 2 & 0 \\ 0 & 0 & 1 & 2 & -1 & 0 \end{array}\right) . \end{align}\]

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