{"id":5177,"date":"2020-09-08T15:06:56","date_gmt":"2020-09-08T06:06:56","guid":{"rendered":"https:\/\/sasamath.com\/blog\/?p=5177"},"modified":"2021-05-30T23:25:14","modified_gmt":"2021-05-30T14:25:14","slug":"exercises-multiple-systems-and-matrix-inversion","status":"publish","type":"post","link":"https:\/\/sasamath.com\/blog\/articles\/exercises-multiple-systems-and-matrix-inversion\/","title":{"rendered":"Exercises: Multiple Systems and Matrix Inversion"},"content":{"rendered":"
\n

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.\n<\/p>\n<\/div>\n

<\/p>\n

Problem 5.1<\/span>
\nSolve the following matrix equation for \\(x,\\) \\(y,\\) \\(z\\) and \\(w.\\)
\n\\[
\n\\begin{pmatrix} 1&2 \\\\ 0&1 \\end{pmatrix}
\n\\begin{pmatrix} x&y \\\\ z&w \\end{pmatrix}
\n=
\n\\begin{pmatrix} 10&2 \\\\ 4&2 \\end{pmatrix}
\n\\]\n<\/p>\n

\n

Solution.<\/span>
\nTaking \\(R_1 \\,\\leftarrow\\, R_1 – 2R_2 ,\\) we obtain
\n\\[\\left( \\begin{array}{cc|cc}
\n1 & 2 & 10 & 2 \\\\
\n0 & 1 & 4 & 2
\n\\end{array}\\right)
\n\\,\\rightarrow\\,
\n\\left( \\begin{array}{cc|cr}
\n1 & 0 & 2 & -2 \\\\
\n0 & 1 & 4 & 2
\n\\end{array}\\right).
\n\\]
\nThus
\n\\[\\begin{pmatrix} x&y \\\\ z&w \\end{pmatrix}
\n=
\n\\left(\\begin{array}{cr} 2 & -2 \\\\ 4 & 2 \\end{array}\\right).\\]\n<\/p>\n<\/div>\n

<\/p>\n

Problem 5.2<\/span>
\nSuppose that
\n\\[
\n\\begin{pmatrix} 1&1 \\\\ 0&1 \\end{pmatrix}
\n\\begin{pmatrix} x&y \\\\ z&w \\end{pmatrix}
\n=
\n\\begin{pmatrix} x&y \\\\ z&w \\end{pmatrix}
\n\\begin{pmatrix} 1&1 \\\\ 0&1 \\end{pmatrix}
\n.
\n\\]
\nShow that \\(x=w,\\) \\(z=0\\) and there is no constraint on \\(y.\\)\n<\/p>\n

\n

Solution.<\/span>
\nEvaluating each multiplication gives
\n\\[\\begin{pmatrix} x+z & y+w \\\\ z & w \\end{pmatrix} =
\n\\begin{pmatrix} x & x+y \\\\ z & z+w \\end{pmatrix}.\\]
\nFrom this, we obtain
\n\\[ \\begin{pmatrix} z & w-x \\\\ 0 & -z \\end{pmatrix}
\n= \\begin{pmatrix} 0 & 0 \\\\ 0 & 0 \\end{pmatrix}\\]
\nor
\n\\[z=0 ,\\,\\, w-x=0.\\]
\nTherefore the solution is
\n\\[x=w ,\\,\\, z=0 \\]
\nand \\(y\\) is any real number.\n<\/p>\n<\/div>\n

<\/p>\n

Problem 5.3<\/span>
\nLet \\(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
\n\\[A=\\begin{pmatrix}
\n2&0&1 \\\\ 0&1&1
\n\\end{pmatrix} .\\]\n<\/p>\n

\n

Solution.<\/span>
\n\\(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.
\n\\[ \\begin{align}
\nA \\cdot \\,^t \\! A
\n&= \\begin{pmatrix} 2 & 0 & 1 \\\\ 0 & 1 & 1 \\end{pmatrix}
\n\\begin{pmatrix} 2 & 0 \\\\ 0 & 1 \\\\ 1 & 1 \\end{pmatrix}
\n= \\begin{pmatrix} 5 & 1 \\\\ 1 & 2 \\end{pmatrix}, \\\\[8pt]<\/p>\n

\\,^t \\! A \\cdot A
\n&= \\begin{pmatrix} 2 & 0 \\\\ 0 & 1 \\\\ 1 & 1 \\end{pmatrix}
\n \\begin{pmatrix} 2 & 0 & 1 \\\\ 0 & 1 & 1 \\end{pmatrix}
\n= \\begin{pmatrix} 4 & 0 & 2 \\\\ 0 & 1 & 1 \\\\ 2 & 1 & 2 \\end{pmatrix} .
\n\\end{align}
\n\\]\n<\/p>\n<\/div>\n

<\/p>\n

Problem 5.4<\/span>
\nWhat is the dimension of the space of \\(4\\times 4\\) symmetric matrices defined over a field \\(K?\\) Of all \\(n \\times n\\) symmetric matrices?\n<\/p>\n

\n

Solution.<\/span>
\n\\[\\frac{n(n+1)}{2}.\\]\n<\/p>\n<\/div>\n

<\/p>\n

Problem 5.5<\/span>
\nWhat 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?\n<\/p>\n

\n

Solution.<\/span>
\n\\[\\frac{n(n+1)}{2}.\\]\n<\/p>\n<\/div>\n

<\/p>\n

Problem 5.6<\/span>
\nShow that the product of two invertible matrices in \\(M_n (K)\\) is invertible.\n<\/p>\n

\n

Solution.<\/span>
\nSuppose both \\(A\\) and \\(B\\) belong to \\(M_n (K)\\) and invertible. Take \\(C = B^{-1} A^{-1}.\\) Then
\n\\[\\begin{gather}
\n(AB)C = (AB)(B^{-1} A^{-1}) = I_n , \\\\[7pt]
\nC(AB) = (B^{-1} A^{-1})(AB) = I_n .
\n\\end{gather}\\]
\nTherefore \\(AB\\) is invertible and the inverse matrix is \\(C = B^{-1} A^{-1}.\\)\n<\/p>\n<\/div>\n

<\/p>\n

Problem 5.7<\/span>
\nLet
\n\\[A=\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}\\]
\nlie in \\(M_2 (K)\\) and assume that \\(ad-bc \\ne 0.\\) Show that \\(A^{-1}\\) exists and that in fact
\n\\[A^{-1} = \\frac{1}{ad-bc}
\n\\left(\\begin{array}{rr} d&-b\\\\-c&a\\end{array}\\right).\\]
\nConversely, show that if \\(ad-bc=0,\\) then \\(A\\) is not invertible.\n<\/p>\n

\n

Solution.<\/span>
\nSuppose that \\(ad-bc \\ne 0.\\) A simple calculation shows that
\n\\[\\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},\\]
\ntherefore
\n\\[ \\frac{1}{ad-bc}
\n\\left(\\begin{array}{rr} d&-b\\\\-c&a\\end{array}\\right)\\]
\nis the inverse matrix of \\(A,\\) and \\(A\\) is invertible.<\/p>\n

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

If \\(d=0\\) and \\(b=0,\\) then
\n\\[\\begin{pmatrix} \\ast & \\ast \\\\ \\ast & \\ast \\end{pmatrix}
\n\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} =
\n\\begin{pmatrix} \\ast & 0 \\\\ \\ast & 0 \\end{pmatrix},\\]
\nwhich cannot be the identity matrix. If \\(d=0\\) and \\(c=0,\\) then
\n\\[\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix}
\n\\begin{pmatrix} \\ast & \\ast \\\\ \\ast & \\ast \\end{pmatrix}
\n=
\n\\begin{pmatrix} \\ast & \\ast \\\\ 0 & 0 \\end{pmatrix},\\]
\nwhich neither cannot be the identity matrix.\n<\/p>\n<\/div>\n

<\/p>\n

Problem 5.8<\/span>
\nLet \\(B\\) be the subset of \\(\\mathrm{GL}_2 (K)\\) consisting of all matrices of the form
\n\\[A = \\begin{pmatrix}a&b\\\\0&c\\end{pmatrix} ,\\,\\, ac \\ne 0.\\]
\nShow that \\(B\\) is a subgroup of \\(\\mathrm{GL}_2 (K).\\)\n<\/p>\n

\n

Solution.<\/span>
\nLet
\n\\[\\begin{align}
\nA &= \\begin{pmatrix} a_1 & b_1 \\\\ 0 & c_1 \\end{pmatrix}, \\\\[7pt]
\nB &= \\begin{pmatrix} a_2 & b_2 \\\\ 0 & c_2 \\end{pmatrix},
\n\\end{align}\\]
\nwhere \\(a_1 c_1 \\ne 0,\\) \\(a_2 c_2 \\ne 0.\\) Then \\(B\\) is invertible and
\n\\[\\begin{align}
\nAB^{-1}
\n&= \\frac{1}{a_2 c_2} \\begin{pmatrix} a_1 & b_1 \\\\ 0 & c_1 \\end{pmatrix}
\n\\left(\\begin{array}{cr} c_2 & -b_2 \\\\ 0 & a_2 \\end{array}\\right) \\\\[7pt]
\n&= \\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}
\n\\end{align}\\]
\nwhich obviously belongs to \\(\\operatorname{GL}_2 (K).\\)\n<\/p>\n<\/div>\n

<\/p>\n

For this problem and next(9-10), superscripts on matrices indicate exponents rather than columns.<\/div>\n

Problem 5.9<\/span>
\nFind a \\(2\\times 2\\) matrix \\(A\\) such that \\(A\\ne 0\\) but nonetheless \\(A^2 =0.\\) Next, find a \\(3\\times 3\\) matrix such that \\(A \\ne 0,\\) \\(A^2 \\ne 0\\) but \\(A^3 =0.\\) Finally, for arbitrary \\(n\\) find an \\(n\\times n\\) matrix \\(A\\) such that none of \\(A,\\) \\(A^2,\\) \\(\\cdots,\\) \\(A^{n-1}\\) equals \\(0,\\) but \\(A^n =0.\\)<\/p>\n

\n

Solution.<\/span>
\nTake \\(A = \\left( a_{ij} \\right)_{n\\times n}\\) where
\n\\[a_{ij} = \\begin{cases}
\n1 & \\quad \\text{if} \\,\\, i+1 = j \\\\[8pt]
\n0 & \\quad \\text{otherwise.}
\n\\end{cases}\\]
\nAs an example, for \\(n=4,\\) we take
\n\\[A = \\begin{pmatrix}
\n0 & 1 & 0 & 0 \\\\
\n0 & 0 & 1 & 0 \\\\
\n0 & 0 & 0 & 1 \\\\
\n0 & 0 & 0 & 0
\n\\end{pmatrix}.
\n\\]\n<\/p>\n<\/div>\n

<\/p>\n

Problem 5.10<\/span>
\nLet \\(A\\in M_n(K)\\) be such that \\(A^r = 0\\) for some integer \\(r \\ge 1.\\) Show that \\(I_n – A\\) is invertible.\n<\/p>\n

\n

Solution.<\/span>
\nTake \\(B = I_n + A+ A^2 + \\cdots + A^{r-1} ,\\) then
\n\\[\\begin{align}
\n(I_n – A)B &= (I_n -A)(I_n +A +A^2 + \\cdots + A^{r-1})\\\\[7pt]
\n&= (I_n + A+A^2 + \\cdots + A^{r-1}) – (A+A^2 + A^3 + \\cdots + A^r )\\\\[7pt]
\n&= I_n – A^r \\\\[7pt]
\n&= I_n – O = I_n.
\n\\end{align}\\]\n<\/p>\n<\/div>\n

<\/p>\n

Problem 5.11<\/span>
\nList the elements in \\(\\mathrm{GL}_2 ( F_2 )\\) where \\(F_2 \\) is the finite field of two elements \\(0\\) and \\(1.\\)\n<\/p>\n

\n

Solution.<\/span>
\nSuppose
\n\\[\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} \\in \\operatorname{GL}_2 (F_2 )\\]
\nthen \\(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:
\n\\[\\begin{gather}
\n0 \\times 0 \\ne 1 \\times 1, \\\\[7pt]
\n0 \\times 1 \\ne 1 \\times 1, \\\\[7pt]
\n1 \\times 0 \\ne 1 \\times 1, \\\\[7pt]
\n1 \\times 1 \\ne 0 \\times 0, \\\\[7pt]
\n1 \\times 1 \\ne 0 \\times 1, \\\\[7pt]
\n1 \\times 1 \\ne 1 \\times 0.
\n\\end{gather}\\]
\nThus all the desired matrices are
\n\\[\\begin{gather}
\n\\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \\end{pmatrix}, \\,\\,
\n\\begin{pmatrix} 0 & 1 \\\\ 1 & 1 \\end{pmatrix}, \\,\\,
\n\\begin{pmatrix} 1 & 1 \\\\ 1 & 0 \\end{pmatrix}, \\\\[7pt]
\n\\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix}, \\,\\,
\n\\begin{pmatrix} 1 & 0 \\\\ 1 & 1 \\end{pmatrix}, \\,\\,
\n\\begin{pmatrix} 1 & 1 \\\\ 0 & 1 \\end{pmatrix}.
\n\\end{gather}\\]
\nIn fact the multiplicative group consisting of these matrices are isomorphic to \\(S_3.\\)\n<\/p>\n<\/div>\n

<\/p>\n

Problem 5.12<\/span>
\nExplain 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.\\)\n<\/p>\n

\n

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

<\/p>\n

Problem 5.13<\/span>
\nExpress the following linear system as a single matrix equation and as a single vector equation.
\n\\[ \\begin{align}
\n2x_1 – 4x_2 &=7 \\\\[7pt]
\n5x_1 + 9x_2 &=4
\n\\end{align}\\]\n<\/p>\n

\n

Solution.<\/span>
\n\\[
\n\\left(\\begin{array}{cr} 2 & -4 \\\\ 5 & 9 \\end{array}\\right)
\n\\begin{pmatrix} x_1 \\\\ x_2 \\end{pmatrix} =
\n\\begin{pmatrix} 7 \\\\ 4 \\end{pmatrix}.
\n\\]\n<\/p>\n<\/div>\n

<\/p>\n

Problem 5.14<\/span>
\nWithout explicitly solving, show that the system above has a unique solution.\n<\/p>\n

\n

Solution.<\/span>
\nSince
\n\\[\\det \\left( \\begin{array}{cr} 2 & -4 \\\\ 5 & 9 \\end{array}\\right) = 38 \\ne 0,\\]
\nby Problem 7, given matrix is invertible.\n<\/p>\n<\/div>\n

<\/p>\n

Problem 5.15<\/span>
\nWithout explicitly solving, show that the system
\n\\[\\begin{align}
\n2x_1 + x_2 &= y_1 \\\\[7pt]
\n-x_1 + 2x_2 &= y_2
\n\\end{align}\\]
\nhas a unique solution for all \\(y_1 ,\\, y_2 \\in \\mathbb{R}.\\)\n<\/p>\n

\n

Solution.<\/span>
\n\\[\\det\\left(\\begin{array}{rc} 2 & 1 \\\\ -1 & 2 \\end{array}\\right) =5 \\ne 0.\\]\n<\/p>\n<\/div>\n

<\/p>\n

Problem 5.16<\/span>
\nFind the rank of the matrix
\n\\[A = \\begin{pmatrix} 1&2&1 \\\\ 2&0&2 \\\\ 0&0&1 \\end{pmatrix}.\\]\n<\/p>\n

\n

Solution.<\/span>
\nBegin with the given matrix.
\n\\[\\begin{pmatrix} 1&2&1 \\\\ 2&0&2 \\\\ 0&0&1 \\end{pmatrix}\\]
\nTake \\(R_2 \\,\\leftarrow\\, R_2 \\times \\frac{1}{2},\\) then we obtain
\n\\[\\begin{pmatrix} 1&2&1 \\\\ 1&0&1 \\\\ 0&0&1 \\end{pmatrix}.\\]
\nTake \\(R_2 \\,\\leftarrow\\,R_2 – R_3,\\) then we obtain
\n\\[\\begin{pmatrix} 1&2&1 \\\\ 1&0&0 \\\\ 0&0&1 \\end{pmatrix}.\\]
\nTake \\(R_1 \\,\\leftarrow\\, R_1 – R_2 – R_3,\\) then we obtain
\n\\[\\begin{pmatrix} 0&2&0 \\\\ 1&0&0 \\\\ 0&0&1 \\end{pmatrix}.\\]
\nTake \\(R_1 \\,\\leftrightarrow\\, R_2,\\) then we obtain
\n\\[\\begin{pmatrix} 1&0&0 \\\\ 0&2&0 \\\\ 0&0&1 \\end{pmatrix}.\\]
\nTake \\(R_2 \\,\\leftarrow\\, R_2 \\times\\frac{1}{2},\\) then we obtain
\n\\[\\begin{pmatrix} 1&0&0 \\\\ 0&1&0 \\\\ 0&0&1 \\end{pmatrix} ,\\]
\nwhich equals \\(I_3.\\)
\nTherefore \\(A\\) is invertible, and \\(\\operatorname{rk}(A) = 3.\\)<\/p>\n<\/div>\n

<\/p>\n

Problem 5.17<\/span>
\nLet \\(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?\\)\n<\/p>\n

\n

Solution.<\/span>
\n\\(1\\le \\operatorname{rk}(A) \\le 4.\\)\n<\/p>\n<\/div>\n

<\/p>\n

Problem 5.18<\/span>
\nUsing Gauss-Jordan elimination, solve the system given in Problem 13.\n<\/p>\n

\n

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

<\/p>\n

Problem 5.19<\/span>
\nSuppose that we are solving a \\(2\\times 3\\) homogeneous linear system \\(A\\mathbf{x} = \\mathbf{0}\\) by Gauss-Jordan elimination and reach the following augmented matrix in reduced row-echelon form:
\n\\[
\n\\left(
\n\\begin{array}{ccr|c}
\n1&0&-5&0 \\\\ 0&1&4&0
\n\\end{array}
\n\\right).
\n\\]
\nWhat is the general solution to the original system? What is the dimension of the solution space?\n<\/p>\n

\n

Solution.<\/span>
\nLet
\n\\[A = \\left(\\begin{array}{ccr} 1 & 0 & -5 \\\\ 0 & 1 & 4 \\end{array}\\right).\\]
\nThen \\(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}.\\)\n<\/p>\n<\/div>\n

<\/p>\n

Problem 5.20<\/span>
\nSuppose that we are solving a \\(3\\times 4\\) linear system \\(A\\mathbf{x}=\\mathbf{y}\\) by Gauss-Jordan elimination and reach the following augmented matrix in reduced row-echelon form:
\n\\[
\n\\left(
\n\\begin{array}{cccc|c}
\n1&2&0&1&5 \\\\
\n0&0&1&4&2 \\\\
\n0&0&0&0&4
\n\\end{array}
\n\\right).
\n\\]
\nWhat can one say about solutions to the original system?\n<\/p>\n

\n

Solution.<\/span>
\nThere are no solutions.\n<\/p>\n<\/div>\n

<\/p>\n

Problem 5.21<\/span>
\nFind all solutions to the following system by Gauss-Jordan elimination:
\n\\[\\begin{align} x_1 + x_2 – x_3 &=0 \\\\[7pt] x_1 + 2x_2 +4x_3 &=0 \\end{align}\\]
\nWhat is the dimension of the solution space?\n<\/p>\n

\n

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

<\/p>\n

Problem 5.22<\/span>
\nDoes every linear system for which there are more variables than equations have a solution? If not, what additional condition is needed?\n<\/p>\n

\n

Solution.<\/span>
\nNo. The rank of augmented matrix has to equal to the rank of coefficient matrix.\n<\/p>\n<\/div>\n

<\/p>\n

Problem 5.23<\/span>
\nSummarize in your own words why reduced row-echelon form is an effective device for solving linear systems of equation.\n<\/p>\n

\n

Solution.<\/span>
\nFirst, it is easy to determine whether the system has any solutions or not; Second, it is easy to find the unknowns step by step.\n<\/p>\n<\/div>\n

<\/p>\n

Problem 5.24<\/span>
\nFactor the following matrix into the product of a lower triangular and an upper triangular triangular matrices:
\n\\[A=\\begin{pmatrix} 2&5 \\\\ 1&2 \\end{pmatrix}.\\]\n<\/p>\n

\n

Solution.<\/span>
\nSuppose
\n\\[
\n\\begin{pmatrix} a & 0 \\\\ b & c \\end{pmatrix}
\n\\begin{pmatrix} x & y \\\\ 0 & z \\end{pmatrix}
\n=
\n\\begin{pmatrix} 2 & 5 \\\\ 1 & 2 \\end{pmatrix}.
\n\\]
\nThere 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
\n\\[\\begin{pmatrix} 2 & y \\\\ 2b & by+cz \\end{pmatrix} = \\begin{pmatrix} 2 & 5 \\\\ 1 & 2 \\end{pmatrix}.\\]
\nIn this equation, we inevitably have \\(y=5\\) and \\(b = \\frac{1}{2},\\) and we obtain
\n\\[\\begin{pmatrix} 2 & 5 \\\\ 1 & \\frac{5}{2}+cz \\end{pmatrix} = \\begin{pmatrix} 2 & 5 \\\\ 1 & 2 \\end{pmatrix}.\\]
\nTaking \\(c=\\frac{1}{2},\\) we have \\(z=-1.\\) Therefore \\(A\\) can be factored as follows:
\n\\[
\n\\begin{pmatrix} 1 & 0 \\\\ \\frac{1}{2} & \\frac{1}{2} \\end{pmatrix}
\n\\left(\\begin{array}{cr} 2 & 5 \\\\ 0 & -1 \\end{array}\\right)
\n=
\n\\begin{pmatrix} 2 & 5 \\\\ 1 & 2 \\end{pmatrix}.
\n\\]<\/p>\n<\/div>\n

<\/p>\n

Problem 5.25<\/span>
\nGiven the matrix factorization
\n\\[
\n\\begin{pmatrix} 1&1&1 \\\\ 2&4&8 \\\\ 1&5&11 \\end{pmatrix}
\n=
\n\\begin{pmatrix} 1&0&0 \\\\ 2&1&0 \\\\ 1&2&1 \\end{pmatrix}
\n\\begin{pmatrix} 1&1&2 \\\\ 0&2&4 \\\\ 0&0&1 \\end{pmatrix}
\n\\]
\nsolve the linear system
\n\\[\\begin{align}
\nx_1 + x_2 + 2x_3 &= 11 \\\\[7pt]
\n2x_1 + 4x_2 + 8x_3 &= 30 \\\\[7pt]
\nx_1 + 5x_2 + 11x_3 &= 28
\n\\end{align}\\]
\nby the method of \\(LU\\) decomposition.\n<\/p>\n

\n

Solution.<\/span>
\nExpress the given matrix factorization as \\(A = LU\\) respectively, and let
\n\\[\\begin{align}
\n\\mathbf{x} &= (x_1 ,\\,x_2 ,\\,x_3 ), \\\\[7pt]
\n\\mathbf{y} &= (11 ,\\,30 ,\\,28 ), \\\\[7pt]
\n\\mathbf{z} &= (z_1 ,\\,z_2 ,\\,z_3 ).
\n\\end{align}\\]
\nFrom the equation \\(L\\mathbf{z} = \\mathbf{y}\\) we have
\n\\[\\mathbf{z} = (11,\\,8,\\,1).\\]
\nNext, solving the equation \\(U \\mathbf{x} = \\mathbf{z}\\) we have
\n\\[\\mathbf{x} = (7,\\,2,\\,1).\\]
\nTherefore the solution is
\n\\[x_1 = 7,\\,\\, x_2 =2 ,\\,\\, x_3 = 1.\\]\n<\/p>\n<\/div>\n

<\/p>\n

Problem 5.26<\/span>
\nSummarize in your own words why \\(LU\\) decomposition in an effective device for solving linear systems of equations.\n<\/p>\n

\n

Solution.<\/span>
\nThere exists straightforward obvious algorithm to find the solutions.\n<\/p>\n<\/div>\n

<\/p>\n

Problem 5.27<\/span>
\nUse the technique of finding solutions of multiple systems to derive the general formula for the inversion of \\(2\\times 2\\) matrices.\n<\/p>\n

\n

Solution.<\/span>
\nAssume that \\(ad-bc \\ne 0.\\)
\n\\[\\begin{align}
\n& \\left(\\begin{array}{cc|cc} a & b & 1 & 0 \\\\ c & d & 0 & 1 \\end{array}\\right) \\\\[7pt]
\n\\rightarrow \\,
\n& \\left(\\begin{array}{cc|cc} ac & bc & c & 0 \\\\ ac & ad & 0 & a \\end{array}\\right) \\\\[7pt]
\n\\rightarrow \\,
\n& \\left(\\begin{array}{cc|rc} ac & bc & c & 0 \\\\ 0 & ad-bc & -c & a \\end{array}\\right) \\\\[7pt]
\n\\rightarrow \\,
\n& \\left(\\begin{array}{cc|cc} ac & bc & c & 0 \\\\ 0 & 1 & \\frac{-c}{ad-bc} & \\frac{a}{ad-bc} \\end{array}\\right) \\\\[7pt]
\n\\rightarrow \\,
\n& \\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]
\n\\rightarrow \\,
\n& \\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) .
\n\\end{align}\\]\n<\/p>\n<\/div>\n

<\/p>\n

Problem 5.28<\/span>
\nUsing the technique of finding solutions of multiple systems, invert the following carefully contrived matrix:
\n\\[A = \\begin{pmatrix} 1&2&2 \\\\ 2&4&3 \\\\ 3&5&4 \\end{pmatrix}.\\]
\nThis should go rather smoothly.\n<\/p>\n

\n

Solution.<\/span>
\n\\[\\begin{align}
\n&\\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]
\n\\rightarrow\\,
\n&\\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]
\n\\rightarrow\\,
\n&\\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]
\n\\rightarrow\\,
\n&\\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]
\n\\rightarrow\\,
\n&\\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) .
\n\\end{align}\\]\n<\/p>\n<\/div>\n

<\/p>\n

Problem 5.29<\/span>
\nTo 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:
\n\\[A = \\begin{pmatrix} 9&2&0 \\\\ 2&4&3 \\\\ 8&6&1 \\end{pmatrix}.\\]
\nHe ponders; he frowns; he leaves it to the student to find \\(A^{-1}.\\)\n<\/p>\n

\n

Solution.<\/span>
\n\\[A^{-1} = \\frac{1}{82}
\n\\left(\\begin{array}{rrr}
\n14 & 2 & -6 \\\\ -22 & 9 & 27 \\\\ 20 & 38 & -32
\n\\end{array}\\right).\\]\n<\/p>\n<\/div>\n

<\/p>\n

<\/p>\n

<\/p>\n","protected":false},"excerpt":{"rendered":"

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{pmatrix} 1&2 \\\\ 0&1 \\end{pmatrix} \\begin{pmatrix} x&y \\\\ z&w \\end{pmatrix} = \\begin{pmatrix} 10&2 \\\\ 4&2 \\end{pmatrix} \\)…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_lmt_disableupdate":"","_lmt_disable":"","footnotes":""},"categories":[57],"tags":[402,400,414,411],"class_list":["post-5177","post","type-post","status-publish","format-standard","hentry","category-linear-algebra","tag-exercises","tag-linear-algebra","tag-matrix","tag-vector-space"],"_links":{"self":[{"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/posts\/5177","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/comments?post=5177"}],"version-history":[{"count":88,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/posts\/5177\/revisions"}],"predecessor-version":[{"id":6513,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/posts\/5177\/revisions\/6513"}],"wp:attachment":[{"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/media?parent=5177"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/categories?post=5177"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/tags?post=5177"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}