|1 \u00a00 \u00a01| \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0 |3 \u00a01 \u00a0a|
|2 \u00a04 \u00a03| = 5 \u00a0e \u00a0 \u00a0|2 \u00a0b \u00a01| \u00a0= 1
|a \u00a0b \u00a05| \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0 |0 \u00a03 \u00a05| <\/p>\n\n\n\n\n\n\n\n
Para calcularmos a e b, temos que calcular o determinante, como se a e b fossem n\u00fameros mesmo. Voc\u00ea j\u00e1 deve saber calcular o determinante de uma matriz 3×3, ent\u00e3o vamos calcular primeiro o determinante da 1\u00aa matriz, que sabemos que \u00e9 igual a 5:<\/p>\n\n\n\n
1.4.5 + 0.3.a + 1.2.b – 1.4.a – 3.b.1 – 5.0.2 = 5<\/p>\n\n\n\n
20 + 2b – 4a – 3b = 5<\/p>\n\n\n\n
20 – 5 = 4a + 3b – 2b<\/p>\n\n\n\n
15 = 4a + b<\/p>\n\n\n\n
Agora o outro determinante:<\/p>\n\n\n\n
3.b.5 + 1.1.0 + a.2.3 – a.b.0 – 1.3.3 – 5.1.2 = 1<\/p>\n\n\n\n
15b + 6a – 9 – 10 = 1<\/p>\n\n\n\n
15b + 6a – 19 = 1<\/p>\n\n\n\n
15b + 6a = 1 + 19<\/p>\n\n\n\n
15b + 6a = 20<\/p>\n\n\n\n
E temos que resolver o sistema:<\/p>\n\n\n\n
15 = 4a + b ( i )<\/p>\n\n\n\n
15b + 6a = 20 ( ii )<\/p>\n\n\n\n
Da equa\u00e7\u00e3o ( i ), podemos achar b em fun\u00e7\u00e3o de a:<\/p>\n\n\n\n
15 = 4a + b<\/p>\n\n\n\n
15 – 4a = b<\/p>\n\n\n\n
b = 15 – 4a<\/p>\n\n\n\n
E substituindo na equa\u00e7\u00e3o ( ii ):<\/p>\n\n\n\n
15b + 6a = 20<\/p>\n\n\n\n
15.(15 – 4a) + 6a = 20<\/p>\n\n\n\n
225 – 60a + 6a = 20<\/p>\n\n\n\n
225 – 20 = 60a – 6a <\/p>\n\n\n\n
205 = 54a<\/p>\n\n\n\n
54a = 205<\/p>\n\n\n\n
a = 205\/54<\/p>\n\n\n\n
E agora para achar b, colocamos o valor de a na equa\u00e7\u00e3o:<\/p>\n\n\n\n
b = 15 – 4a<\/p>\n\n\n\n
b = 15 – 4.(205\/54)<\/p>\n\n\n\n
b = 15 – 2.(205\/27)<\/p>\n\n\n\n
b = 15 – 410\/27, tirando o m\u00ednimo,<\/p>\n\n\n\n
b = (405 – 410)\/27<\/p>\n\n\n\n
b = – 5\/27<\/p>\n\n\n\n
Resposta: a = 205\/54; b = – 5\/27<\/p>\n","protected":false},"excerpt":{"rendered":"
|1 \u00a00 \u00a01| \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0 |3 \u00a01 \u00a0a||2 \u00a04 \u00a03| = 5 \u00a0e \u00a0 \u00a0|2 \u00a0b \u00a01| \u00a0= 1|a \u00a0b \u00a05| \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0 |0 \u00a03 \u00a05|<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[15],"tags":[28],"class_list":["post-371","post","type-post","status-publish","format-standard","hentry","category-matrizes","tag-facil"],"_links":{"self":[{"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/posts\/371","targetHints":{"allow":["GET"]}}],"collection":[{"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/comments?post=371"}],"version-history":[{"count":1,"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/posts\/371\/revisions"}],"predecessor-version":[{"id":372,"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/posts\/371\/revisions\/372"}],"wp:attachment":[{"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/media?parent=371"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/categories?post=371"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/tags?post=371"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}