\n
Problema Sejam:<\/p>\n Considerando [tex]\\boxed{s=\\sqrt{D(X)}}[\/tex] e [tex]\\boxed{t=\\sqrt{t(X)+2s}} \\, [\/tex], prove que:<\/p>\n \n(a)<\/strong> Se [tex]t\\neq 0[\/tex] ent\u00e3o a matriz [tex]R[\/tex], dada por (b)<\/strong> Seja [tex]C[\/tex] a matriz quadrada dada por \n<\/div>\n \n Solu\u00e7\u00e3o <\/p>\n (a)<\/strong> Para que [tex]R[\/tex] seja uma raiz quadrada de [tex]X[\/tex] temos que mostrar a seguinte igualdade matricial [tex]\\boxed{R^2=R\\cdot R=X}.[\/tex] (Neste caso, diremos que [tex]R \\, [\/tex] \u00e9 a raiz quadrada matricial<\/strong> de [tex]X \\, [\/tex] e indicaremos assim: [tex]R=\\sqrt{X} \\, .[\/tex])<\/p>\n \n (b)<\/strong> Vamos imitar o processo de obten\u00e7\u00e3o da f\u00f3rmula usada na resolu\u00e7\u00f5es de equa\u00e7\u00f5es do 2\u00ba grau: [tex]\\qquad \\sqrt{Y}=\\dfrac{1}{t}\\left(\\begin{array}{cc} e uma raiz [tex]X[\/tex] da equa\u00e7\u00e3o matricial [tex] X^2+ X+C=\\left(\\begin{array}{cc} [tex]\\quad X=\\dfrac{-I+\\sqrt{I-4C}}{2}[\/tex]<\/p>\n [tex]\\quad X=\\dfrac{-I+\\sqrt{Y}}{2}[\/tex]<\/p>\n [tex]\\quad X=-\\dfrac{1}{2}I+\\dfrac{1}{2}\\sqrt{Y}[\/tex]<\/p>\n [tex]\\quad X=-\\dfrac{1}{2}\\left(\\begin{array}{cc} [tex]\\quad \\, \\fcolorbox{black}{#eee0e5}{$X=\\left(\\begin{array}{cc} Solu\u00e7\u00e3o elaborada pelos Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n \n","protected":false},"excerpt":{"rendered":" Problema (Indicado a partir do 2\u00ba ano do E. M.) Sejam: [tex]X=\\left(\\begin{array}{cc} x_1 & x_2 \\\\ x_3 & x_4 \\end{array}\\right)[\/tex] uma matriz quadrada de tamanho [tex]2\\times 2[\/tex], [tex]t(X)=x_1+x_4[\/tex], o tra\u00e7o de [tex]X[\/tex] e [tex]D(X)=x_1x_4-x_3x_2[\/tex], o determinante de [tex]X[\/tex]. Considerando [tex]\\boxed{s=\\sqrt{D(X)}}[\/tex] e [tex]\\boxed{t=\\sqrt{t(X)+2s}} \\, [\/tex], prove que: (a) Se [tex]t\\neq 0[\/tex] ent\u00e3o a matriz [tex]R[\/tex], dada … <\/p>\n
\n (Indicado a partir do 2\u00ba ano do E. M.)<\/span><\/p>\n
\n\n
\nx_1 & x_2 \\\\
\nx_3 & x_4
\n\\end{array}\\right)[\/tex] uma matriz quadrada de tamanho [tex]2\\times 2[\/tex], <\/li>\n
\n[tex]\\qquad \\qquad R=\\dfrac{1}{t}\\left(\\begin{array}{cc}
\nx_1+s & x_2 \\\\
\nx_3 & x_4+s
\n\\end{array}\\right)\\\\
\n[\/tex]
\n\u00e9 uma raiz quadrada de [tex]X[\/tex], isto \u00e9, [tex]R^2=R\\cdot R=X[\/tex].<\/p>\n
\n[tex]\\qquad \\qquad C=\\left(\\begin{array}{cc}
\n-2 & -1 \\\\
\n0 & 0
\n\\end{array}\\right)[\/tex].
\nEncontre uma matriz [tex]X[\/tex] de tamanho [tex]2\\times 2[\/tex] que seja solu\u00e7\u00e3o da seguinte equa\u00e7\u00e3o matricial de segundo grau
\n[tex]\\qquad \\qquad X^2+ X+C=\\left(\\begin{array}{cc}
\n0 & 0 \\\\
\n0 & 0
\n\\end{array}\\right).[\/tex]<\/p>\n
\n
\nAssim, para solucionar este item, basta realizar a multiplica\u00e7\u00e3o matricial:
\n[tex]\\qquad \\begin{align*} R^2=R\\cdot R&=\\dfrac{1}{t}\\left(\\begin{array}{cc}
\nx_1+s & x_2 \\\\
\nx_3 & x_4+s
\n\\end{array}\\right)\\dfrac{1}{t}\\left(\\begin{array}{cc}
\nx_1+s & x_2 \\\\
\nx_3 & x_4+s
\n\\end{array}\\right)\\\\
\n\\\\
\n&=\\dfrac{1}{t^2}\\left(\\begin{array}{cc}
\n(x_1+s)^2+x_2x_3 & (x_1+s)x_2+x_2(x_4+s) \\\\
\nx_3(x_1+s)+(x_4+s)x_3 & x_3x_2+(x_4+s)^2
\n\\end{array}\\right)
\n\\end{align*}[\/tex]
\ne observar que
\n[tex]\\qquad \\begin{align*} \\boxed{(x_1+s)^2+x_2x_3}&=x_1^2+2x_1s+s^2+x_2x_3=x_1^2+2x_1s+(x_1x_4-x_2x_3)+x_2x_3\\\\
\n&=x_1(x_1+2s+x_4)=x_1(t(X)+2s)=\\boxed{x_1t^2} \\, ; \\end{align*}[\/tex]
\n[tex]\\qquad \\boxed{(x_1+s)x_2+x_2(x_4+s)} =x_2(x_1+2s+x_4)=x_2(t(X)+2s)=\\boxed{x_2t^2} \\, ;[\/tex]
\n[tex]\\qquad \\boxed{x_3(x_1+s)+(x_4+s)x_3}=x_3(x_1+2s+x_4)=x_3(t(X)+2s)=\\boxed{x_3t^2} \\, ;[\/tex]
\n[tex]\\qquad \\begin{align*} \\boxed{x_3x_2+(x_4+s)^2}&=x_3x_2+x_4^2+2x_4s+s^2=x_3x_2+x_4^2+2x_4s+(x_1x_4-x_2x_3)\\\\
\n&=x_4(x_4+2s+x_1)=x_4(t(X)+2s)=\\boxed{x_4t^2} \\, . \\end{align*}[\/tex]
\nAssim,
\n[tex]\\qquad \\, \\fcolorbox{black}{#eee0e5}{$R^2$}=R\\cdot R=\\dfrac{1}{t^2}\\left(\\begin{array}{cc}
\nt^2x_1 & t^2x_2 \\\\
\nt^2x_3 & t^2x_4
\n\\end{array}\\right)=\\left(\\begin{array}{cc}
\nx_1 & x_2 \\\\
\nx_3 & x_4
\n\\end{array}\\right)= \\, \\fcolorbox{black}{#eee0e5}{$X$} \\, .[\/tex]<\/p>\n
\nSe [tex]I[\/tex] denota a matriz identidade de tamanho [tex]2\\times 2[\/tex], ent\u00e3o temos que:
\n[tex]\\qquad \\qquad X^2+ X=-C \\, \\, \\, \\Longleftrightarrow \\, \\, \\, \\boxed{X^2+ X+\\dfrac{1}{4}I}=\\dfrac{1}{4}I-C=\\boxed{\\dfrac{I-4C}{4}} \\, . \\qquad \\textcolor{#800000}{(i)}[\/tex]
\nDo fato de a multiplica\u00e7\u00e3o matricial ser distributiva com rela\u00e7\u00e3o \u00e0 adi\u00e7\u00e3o matricial obtemos que:
\n[tex]\\qquad \\begin{align*}\\boxed{\\left(X+\\dfrac{1}{2}I\\right)^2}&=\\left(X+\\dfrac{1}{2}I\\right)\\cdot \\left(X+\\dfrac{1}{2}I\\right)=X^2+\\dfrac{1}{2}X+\\dfrac{1}{2}X+\\dfrac{1}{4}I\\\\
\n&=\\boxed{X^2+ X+\\dfrac{1}{4}I} \\, .\\qquad \\textcolor{#800000}{(ii)}\\end{align*}[\/tex]
\nDesta maneira,
\n[tex]\\qquad \\left(X+\\dfrac{1}{2}I\\right)^2\\stackrel{\\textcolor{#800000}{(ii)}}{=}X^2+ X+\\dfrac{1}{4}I\\stackrel{\\textcolor{#800000}{(i)}}{=}\\dfrac{I-4C}{4}.[\/tex]
\nAssim,
\n[tex]\\qquad X=\\dfrac{-I+\\sqrt{I-4C}}{2},[\/tex]
\nem que [tex]\\sqrt{ \\, }[\/tex] denota, neste caso, a raiz matricial introduzida no item anterior.
\nSeja [tex]Y[\/tex] a matriz
\n[tex]\\qquad Y=I-4C=\\left(\\begin{array}{cc}
\n1 & 0 \\\\
\n0 & 1
\n\\end{array}\\right)+\\left(\\begin{array}{cc}
\n8 & 4 \\\\
\n0 & 0
\n\\end{array}\\right)=\\left(\\begin{array}{cc}
\n9 & 4 \\\\
\n0 & 1
\n\\end{array}\\right)[\/tex]
\nPara determinarmos [tex]X[\/tex], precisaremos encontrar uma raiz quadrada matricial para [tex]Y[\/tex] e para isso vamos utilizar a f\u00f3rmula do item (a)<\/strong>.
\nNeste caso, note que [tex]\\boxed{t(Y)=9+1=10} \\, , \\, \\boxed{D(Y)=9\\times 1+0\\times 4=9} \\, , \\, \\boxed{s=\\sqrt{9}=3} \\, , \\, \\boxed{t=\\sqrt{10+2\\times 3}=4}[\/tex] e, com isso, uma raiz quadrada de [tex]Y[\/tex] \u00e9 dada por<\/p>\n
\n9+s & 4 \\\\
\n0 & 1+s
\n\\end{array}\\right)=\\dfrac{1}{4}\\left(\\begin{array}{cc}
\n9+3 & 4 \\\\
\n0 & 1+3
\n\\end{array}\\right)=\\left(\\begin{array}{cc}
\n3 & 1 \\\\
\n0 & 1
\n\\end{array}\\right)[\/tex]<\/p>\n
\n0 & 0 \\\\
\n0 & 0
\n\\end{array}\\right)[\/tex] pode finalmente ser obtida:<\/p>\n
\n1 & 0 \\\\
\n0 & 1
\n\\end{array}\\right)+\\dfrac{1}{2}\\left(\\begin{array}{cc}
\n3 & 1 \\\\
\n0 & 1
\n\\end{array}\\right)[\/tex]<\/p>\n
\n1 & \\frac{1}{2} \\\\
\n0 & 0
\n\\end{array}\\right)$} \\, .[\/tex]<\/p>\n
\n