Problem 1 Find the indicated entry of the matrix, if it is defined. A = ( 1 3 1 2 − 1 4 ) {\displaystyle A={\begin{pmatrix}1&3&1\\2&-1&4\end{pmatrix}}} a 2 , 1 {\displaystyle a_{2,1}} a 1 , 2 {\displaystyle a_{1,2}} a 2 , 2 {\displaystyle a_{2,2}} a 3 , 1 {\displaystyle a_{3,1}} Answer 2 {\displaystyle 2} 3 {\displaystyle 3} − 1 {\displaystyle -1} Not defined. Problem 2 Give the size of each matrix. ( 1 0 4 2 1 5 ) {\displaystyle {\begin{pmatrix}1&0&4\\2&1&5\end{pmatrix}}} ( 1 1 − 1 1 3 − 1 ) {\displaystyle {\begin{pmatrix}1&1\\-1&1\\3&-1\end{pmatrix}}} ( 5 10 10 5 ) {\displaystyle {\begin{pmatrix}5&10\\10&5\end{pmatrix}}} Answer 2 × 3 {\displaystyle 2\!\times \!3} 3 × 2 {\displaystyle 3\!\times \!2} 2 × 2 {\displaystyle 2\!\times \!2} This exercise is recommended for all readers. Problem 3 Do the indicated vector operation, if it is defined. ( 2 1 1 ) + ( 3 0 4 ) {\displaystyle {\begin{pmatrix}2\\1\\1\end{pmatrix}}+{\begin{pmatrix}3\\0\\4\end{pmatrix}}} 5 ( 4 − 1 ) {\displaystyle 5{\begin{pmatrix}4\\-1\end{pmatrix}}} ( 1 5 1 ) − ( 3 1 1 ) {\displaystyle {\begin{pmatrix}1\\5\\1\end{pmatrix}}-{\begin{pmatrix}3\\1\\1\end{pmatrix}}} 7 ( 2 1 ) + 9 ( 3 5 ) {\displaystyle 7{\begin{pmatrix}2\\1\end{pmatrix}}+9{\begin{pmatrix}3\\5\end{pmatrix}}} ( 1 2 ) + ( 1 2 3 ) {\displaystyle {\begin{pmatrix}1\\2\end{pmatrix}}+{\begin{pmatrix}1\\2\\3\end{pmatrix}}} 6 ( 3 1 1 ) − 4 ( 2 0 3 ) + 2 ( 1 1 5 ) {\displaystyle 6{\begin{pmatrix}3\\1\\1\end{pmatrix}}-4{\begin{pmatrix}2\\0\\3\end{pmatrix}}+2{\begin{pmatrix}1\\1\\5\end{pmatrix}}} Answer ( 5 1 5 ) {\displaystyle {\begin{pmatrix}5\\1\\5\end{pmatrix}}} ( 20 − 5 ) {\displaystyle {\begin{pmatrix}20\\-5\end{pmatrix}}} ( − 2 4 0 ) {\displaystyle {\begin{pmatrix}-2\\4\\0\end{pmatrix}}} ( 41 52 ) {\displaystyle {\begin{pmatrix}41\\52\end{pmatrix}}} Not defined. ( 12 8 4 ) {\displaystyle {\begin{pmatrix}12\\8\\4\end{pmatrix}}} This exercise is recommended for all readers. Problem 4 Solve each system using matrix notation. Express the solution using vectors. 3 x + 6 y = 18 x + 2 y = 6 {\displaystyle {\begin{array}{*{2}{rc}r}3x&+&6y&=&18\\x&+&2y&=&6\end{array}}} x + y = 1 x − y = − 1 {\displaystyle {\begin{array}{*{2}{rc}r}x&+&y&=&1\\x&-&y&=&-1\end{array}}} x 1 + x 3 = 4 x 1 − x 2 + 2 x 3 = 5 4 x 1 − x 2 + 5 x 3 = 17 {\displaystyle {\begin{array}{*{3}{rc}r}x_{1}&&&+&x_{3}&=&4\\x_{1}&-&x_{2}&+&2x_{3}&=&5\\4x_{1}&-&x_{2}&+&5x_{3}&=&17\end{array}}} 2 a + b − c = 2 2 a + c = 3 a − b = 0 {\displaystyle {\begin{array}{*{3}{rc}r}2a&+&b&-&c&=&2\\2a&&&+&c&=&3\\a&-&b&&&=&0\end{array}}} x + 2 y − z = 3 2 x + y + w = 4 x − y + z + w = 1 {\displaystyle {\begin{array}{*{4}{rc}r}x&+&2y&-&z&&&=&3\\2x&+&y&&&+&w&=&4\\x&-&y&+&z&+&w&=&1\end{array}}} x + z + w = 4 2 x + y − w = 2 3 x + y + z = 7 {\displaystyle {\begin{array}{*{4}{rc}r}x&&&+&z&+&w&=&4\\2x&+&y&&&-&w&=&2\\3x&+&y&+&z&&&=&7\end{array}}} Answer This reduction ( 3 6 18 1 2 6 ) → ( − 1 / 3 ) ρ 1 + ρ 2 ( 3 6 18 0 0 0 ) {\displaystyle {\begin{array}{rcl}\left({\begin{array}{*{2}{c}|c}3&6&18\\1&2&6\end{array}}\right)&{\xrightarrow[{}]{(-1/3)\rho _{1}+\rho _{2}}}&\left({\begin{array}{*{2}{c}|c}3&6&18\\0&0&0\end{array}}\right)\end{array}}} leaves x {\displaystyle x} leading and y {\displaystyle y} free. Making y {\displaystyle y} the parameter, we have x = 6 − 2 y {\displaystyle x=6-2y} so the solution set is { ( 6 0 ) + ( − 2 1 ) y | y ∈ R } . {\displaystyle \{{\begin{pmatrix}6\\0\end{pmatrix}}+{\begin{pmatrix}-2\\1\end{pmatrix}}y\,{\big |}\,y\in \mathbb {R} \}.} This reduction ( 1 1 1 1 − 1 − 1 ) → − ρ 1 + ρ 2 ( 1 1 1 0 − 2 − 2 ) {\displaystyle {\begin{array}{rcl}\left({\begin{array}{*{2}{c}|c}1&1&1\\1&-1&-1\end{array}}\right)&{\xrightarrow[{}]{-\rho _{1}+\rho _{2}}}&\left({\begin{array}{*{2}{c}|c}1&1&1\\0&-2&-2\end{array}}\right)\end{array}}} gives the unique solution y = 1 {\displaystyle y=1} , x = 0 {\displaystyle x=0} . The solution set is { ( 0 1 ) } . {\displaystyle \{{\begin{pmatrix}0\\1\end{pmatrix}}\}.} This use of Gauss' method ( 1 0 1 4 1 − 1 2 5 4 − 1 5 17 ) → − 4 ρ 1 + ρ 3 − ρ 1 + ρ 2 ( 1 0 1 4 0 − 1 1 1 0 − 1 1 1 ) → − ρ 2 + ρ 3 ( 1 0 1 4 0 − 1 1 1 0 0 0 0 ) {\displaystyle \left({\begin{array}{*{3}{c}|c}1&0&1&4\\1&-1&2&5\\4&-1&5&17\end{array}}\right)\;{\xrightarrow[{-4\rho _{1}+\rho _{3}}]{-\rho _{1}+\rho _{2}}}\;\left({\begin{array}{*{3}{c}|c}1&0&1&4\\0&-1&1&1\\0&-1&1&1\end{array}}\right)\;{\xrightarrow[{}]{-\rho _{2}+\rho _{3}}}\;\left({\begin{array}{*{3}{c}|c}1&0&1&4\\0&-1&1&1\\0&0&0&0\end{array}}\right)} leaves x 1 {\displaystyle x_{1}} and x 2 {\displaystyle x_{2}} leading with x 3 {\displaystyle x_{3}} free. The solution set is { ( 4 − 1 0 ) + ( − 1 1 1 ) x 3 | x 3 ∈ R } . {\displaystyle \{{\begin{pmatrix}4\\-1\\0\end{pmatrix}}+{\begin{pmatrix}-1\\1\\1\end{pmatrix}}x_{3}\,{\big |}\,x_{3}\in \mathbb {R} \}.} This reduction ( 2 1 − 1 2 2 0 1 3 1 − 1 0 0 ) → − ( 1 / 2 ) ρ 1 + ρ 3 − ρ 1 + ρ 2 ( 2 1 − 1 2 0 − 1 2 1 0 − 3 / 2 1 / 2 − 1 ) → ( − 3 / 2 ) ρ 2 + ρ 3 ( 2 1 − 1 2 0 − 1 2 1 0 0 − 5 / 2 − 5 / 2 ) {\displaystyle \left({\begin{array}{*{3}{c}|c}2&1&-1&2\\2&0&1&3\\1&-1&0&0\end{array}}\right)\;{\xrightarrow[{-(1/2)\rho _{1}+\rho _{3}}]{-\rho _{1}+\rho _{2}}}\;\left({\begin{array}{*{3}{c}|c}2&1&-1&2\\0&-1&2&1\\0&-3/2&1/2&-1\end{array}}\right)\;{\xrightarrow[{}]{(-3/2)\rho _{2}+\rho _{3}}}\;\left({\begin{array}{*{3}{c}|c}2&1&-1&2\\0&-1&2&1\\0&0&-5/2&-5/2\end{array}}\right)} shows that the solution set is a singleton set. { ( 1 1 1 ) } {\displaystyle \{{\begin{pmatrix}1\\1\\1\end{pmatrix}}\}} This reduction is easy ( 1 2 − 1 0 3 2 1 0 1 4 1 − 1 1 1 1 ) → − ρ 1 + ρ 3 − 2 ρ 1 + ρ 2 ( 1 2 − 1 0 3 0 − 3 2 1 − 2 0 − 3 2 1 − 2 ) → − ρ 2 + ρ 3 ( 1 2 − 1 0 3 0 − 3 2 1 − 2 0 0 0 0 0 ) {\displaystyle \left({\begin{array}{*{4}{c}|c}1&2&-1&0&3\\2&1&0&1&4\\1&-1&1&1&1\end{array}}\right)\;{\xrightarrow[{-\rho _{1}+\rho _{3}}]{-2\rho _{1}+\rho _{2}}}\;\left({\begin{array}{*{4}{c}|c}1&2&-1&0&3\\0&-3&2&1&-2\\0&-3&2&1&-2\end{array}}\right)\;{\xrightarrow[{}]{-\rho _{2}+\rho _{3}}}\;\left({\begin{array}{*{4}{c}|c}1&2&-1&0&3\\0&-3&2&1&-2\\0&0&0&0&0\end{array}}\right)} and ends with x {\displaystyle x} and y {\displaystyle y} leading, while z {\displaystyle z} and w {\displaystyle w} are free. Solving for y {\displaystyle y} gives y = ( 2 + 2 z + w ) / 3 {\displaystyle y=(2+2z+w)/3} and substitution shows that x + 2 ( 2 + 2 z + w ) / 3 − z = 3 {\displaystyle x+2(2+2z+w)/3-z=3} so x = ( 5 / 3 ) − ( 1 / 3 ) z − ( 2 / 3 ) w {\displaystyle x=(5/3)-(1/3)z-(2/3)w} , making the solution set { ( 5 / 3 2 / 3 0 0 ) + ( − 1 / 3 2 / 3 1 0 ) z + ( − 2 / 3 1 / 3 0 1 ) w | z , w ∈ R } . {\displaystyle \{{\begin{pmatrix}5/3\\2/3\\0\\0\end{pmatrix}}+{\begin{pmatrix}-1/3\\2/3\\1\\0\end{pmatrix}}z+{\begin{pmatrix}-2/3\\1/3\\0\\1\end{pmatrix}}w\,{\big |}\,z,w\in \mathbb {R} \}.} The reduction ( 1 0 1 1 4 2 1 0 − 1 2 3 1 1 0 7 ) → − 3 ρ 1 + ρ 3 − 2 ρ 1 + ρ 2 ( 1 0 1 1 4 0 1 − 2 − 3 − 6 0 1 − 2 − 3 − 5 ) → − ρ 2 + ρ 3 ( 1 0 1 1 4 0 1 − 2 − 3 − 6 0 0 0 0 1 ) {\displaystyle \left({\begin{array}{*{4}{c}|c}1&0&1&1&4\\2&1&0&-1&2\\3&1&1&0&7\end{array}}\right)\;{\xrightarrow[{-3\rho _{1}+\rho _{3}}]{-2\rho _{1}+\rho _{2}}}\;\left({\begin{array}{*{4}{c}|c}1&0&1&1&4\\0&1&-2&-3&-6\\0&1&-2&-3&-5\end{array}}\right)\;{\xrightarrow[{}]{-\rho _{2}+\rho _{3}}}\;\left({\begin{array}{*{4}{c}|c}1&0&1&1&4\\0&1&-2&-3&-6\\0&0&0&0&1\end{array}}\right)} shows that there is no solution— the solution set is empty. Problem 5 Solve each system using matrix notation. Give each solution set in vector notation. 2 x + y − z = 1 4 x − y = 3 {\displaystyle {\begin{array}{*{3}{rc}r}2x&+&y&-&z&=&1\\4x&-&y&&&=&3\end{array}}} x − z = 1 y + 2 z − w = 3 x + 2 y + 3 z − w = 7 {\displaystyle {\begin{array}{*{4}{rc}r}x&&&-&z&&&=&1\\&&y&+&2z&-&w&=&3\\x&+&2y&+&3z&-&w&=&7\end{array}}} x − y + z = 0 y + w = 0 3 x − 2 y + 3 z + w = 0 − y − w = 0 {\displaystyle {\begin{array}{*{4}{rc}r}x&-&y&+&z&&&=&0\\&&y&&&+&w&=&0\\3x&-&2y&+&3z&+&w&=&0\\&&-y&&&-&w&=&0\end{array}}} a + 2 b + 3 c + d − e = 1 3 a − b + c + d + e = 3 {\displaystyle {\begin{array}{*{5}{rc}r}a&+&2b&+&3c&+&d&-&e&=&1\\3a&-&b&+&c&+&d&+&e&=&3\end{array}}} Answer This reduction ( 2 1 − 1 1 4 − 1 0 3 ) → − 2 ρ 1 + ρ 2 ( 2 1 − 1 1 0 − 3 2 1 ) {\displaystyle {\begin{array}{rcl}\left({\begin{array}{*{3}{c}|c}2&1&-1&1\\4&-1&0&3\end{array}}\right)&{\xrightarrow[{}]{-2\rho _{1}+\rho _{2}}}&\left({\begin{array}{*{3}{c}|c}2&1&-1&1\\0&-3&2&1\end{array}}\right)\end{array}}} ends with x {\displaystyle x} and y {\displaystyle y} leading while z {\displaystyle z} is free. Solving for y {\displaystyle y} gives y = ( 1 − 2 z ) / ( − 3 ) {\displaystyle y=(1-2z)/(-3)} , and then substitution 2 x + ( 1 − 2 z ) / ( − 3 ) − z = 1 {\displaystyle 2x+(1-2z)/(-3)-z=1} shows that x = ( ( 4 / 3 ) + ( 1 / 3 ) z ) / 2 {\displaystyle x=((4/3)+(1/3)z)/2} . Hence the solution set is { ( 2 / 3 − 1 / 3 0 ) + ( 1 / 6 2 / 3 1 ) z | z ∈ R } . {\displaystyle \{{\begin{pmatrix}2/3\\-1/3\\0\end{pmatrix}}+{\begin{pmatrix}1/6\\2/3\\1\end{pmatrix}}z\,{\big |}\,z\in \mathbb {R} \}.} This application of Gauss' method ( 1 0 − 1 0 1 0 1 2 − 1 3 1 2 3 − 1 7 ) → − ρ 1 + ρ 3 ( 1 0 − 1 0 1 0 1 2 − 1 3 0 2 4 − 1 6 ) → − 2 ρ 2 + ρ 3 ( 1 0 − 1 0 1 0 1 2 − 1 3 0 0 0 1 0 ) {\displaystyle \left({\begin{array}{*{4}{c}|c}1&0&-1&0&1\\0&1&2&-1&3\\1&2&3&-1&7\end{array}}\right)\;{\xrightarrow[{}]{-\rho _{1}+\rho _{3}}}\;\left({\begin{array}{*{4}{c}|c}1&0&-1&0&1\\0&1&2&-1&3\\0&2&4&-1&6\end{array}}\right)\;{\xrightarrow[{}]{-2\rho _{2}+\rho _{3}}}\;\left({\begin{array}{*{4}{c}|c}1&0&-1&0&1\\0&1&2&-1&3\\0&0&0&1&0\end{array}}\right)} leaves x {\displaystyle x} , y {\displaystyle y} , and w {\displaystyle w} leading. The solution set is { ( 1 3 0 0 ) + ( 1 − 2 1 0 ) z | z ∈ R } . {\displaystyle \{{\begin{pmatrix}1\\3\\0\\0\end{pmatrix}}+{\begin{pmatrix}1\\-2\\1\\0\end{pmatrix}}z\,{\big |}\,z\in \mathbb {R} \}.} This row reduction ( 1 − 1 1 0 0 0 1 0 1 0 3 − 2 3 1 0 0 − 1 0 − 1 0 ) → − 3 ρ 1 + ρ 3 ( 1 − 1 1 0 0 0 1 0 1 0 0 1 0 1 0 0 − 1 0 − 1 0 ) → ρ 2 + ρ 4 − ρ 2 + ρ 3 ( 1 − 1 1 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 ) {\displaystyle \left({\begin{array}{*{4}{c}|c}1&-1&1&0&0\\0&1&0&1&0\\3&-2&3&1&0\\0&-1&0&-1&0\end{array}}\right)\;{\xrightarrow[{}]{-3\rho _{1}+\rho _{3}}}\;\left({\begin{array}{*{4}{c}|c}1&-1&1&0&0\\0&1&0&1&0\\0&1&0&1&0\\0&-1&0&-1&0\end{array}}\right)\;{\xrightarrow[{\rho _{2}+\rho _{4}}]{-\rho _{2}+\rho _{3}}}\;\left({\begin{array}{*{4}{c}|c}1&-1&1&0&0\\0&1&0&1&0\\0&0&0&0&0\\0&0&0&0&0\end{array}}\right)} ends with z {\displaystyle z} and w {\displaystyle w} free. The solution set is { ( 0 0 0 0 ) + ( − 1 0 1 0 ) z + ( − 1 − 1 0 1 ) w | z , w ∈ R } . {\displaystyle \{{\begin{pmatrix}0\\0\\0\\0\end{pmatrix}}+{\begin{pmatrix}-1\\0\\1\\0\end{pmatrix}}z+{\begin{pmatrix}-1\\-1\\0\\1\end{pmatrix}}w\,{\big |}\,z,w\in \mathbb {R} \}.} Gauss' method done in this way ( 1 2 3 1 − 1 1 3 − 1 1 1 1 3 ) → − 3 ρ 1 + ρ 2 ( 1 2 3 1 − 1 1 0 − 7 − 8 − 2 4 0 ) {\displaystyle {\begin{array}{rcl}\left({\begin{array}{*{5}{c}|c}1&2&3&1&-1&1\\3&-1&1&1&1&3\end{array}}\right)&{\xrightarrow[{}]{-3\rho _{1}+\rho _{2}}}&\left({\begin{array}{*{5}{c}|c}1&2&3&1&-1&1\\0&-7&-8&-2&4&0\end{array}}\right)\end{array}}} ends with c {\displaystyle c} , d {\displaystyle d} , and e {\displaystyle e} free. Solving for b {\displaystyle b} shows that b = ( 8 c + 2 d − 4 e ) / ( − 7 ) {\displaystyle b=(8c+2d-4e)/(-7)} and then substitution a + 2 ( 8 c + 2 d − 4 e ) / ( − 7 ) + 3 c + 1 d − 1 e = 1 {\displaystyle a+2(8c+2d-4e)/(-7)+3c+1d-1e=1} shows that a = 1 − ( 5 / 7 ) c − ( 3 / 7 ) d − ( 1 / 7 ) e {\displaystyle a=1-(5/7)c-(3/7)d-(1/7)e} and so the solution set is { ( 1 0 0 0 0 ) + ( − 5 / 7 − 8 / 7 1 0 0 ) c + ( − 3 / 7 − 2 / 7 0 1 0 ) d + ( − 1 / 7 4 / 7 0 0 1 ) e | c , d , e ∈ R } . {\displaystyle \{{\begin{pmatrix}1\\0\\0\\0\\0\end{pmatrix}}+{\begin{pmatrix}-5/7\\-8/7\\1\\0\\0\end{pmatrix}}c+{\begin{pmatrix}-3/7\\-2/7\\0\\1\\0\end{pmatrix}}d+{\begin{pmatrix}-1/7\\4/7\\0\\0\\1\end{pmatrix}}e\,{\big |}\,c,d,e\in \mathbb {R} \}.} Problem 6 The vector is in the set. What value of the parameters produces that vector? ( 5 − 5 ) {\displaystyle {\begin{pmatrix}5\\-5\end{pmatrix}}} , { ( 1 − 1 ) k | k ∈ R } {\displaystyle \{{\begin{pmatrix}1\\-1\end{pmatrix}}k\,{\big |}\,k\in \mathbb {R} \}} ( − 1 2 1 ) {\displaystyle {\begin{pmatrix}-1\\2\\1\end{pmatrix}}} , { ( − 2 1 0 ) i + ( 3 0 1 ) j | i , j ∈ R } {\displaystyle \{{\begin{pmatrix}-2\\1\\0\end{pmatrix}}i+{\begin{pmatrix}3\\0\\1\end{pmatrix}}j\,{\big |}\,i,j\in \mathbb {R} \}} ( 0 − 4 2 ) {\displaystyle {\begin{pmatrix}0\\-4\\2\end{pmatrix}}} , { ( 1 1 0 ) m + ( 2 0 1 ) n | m , n ∈ R } {\displaystyle \{{\begin{pmatrix}1\\1\\0\end{pmatrix}}m+{\begin{pmatrix}2\\0\\1\end{pmatrix}}n\,{\big |}\,m,n\in \mathbb {R} \}} Answer For each problem we get a system of linear equations by looking at the equations of components. k = 5 {\displaystyle k=5} The second components show that i = 2 {\displaystyle i=2} , the third components show that j = 1 {\displaystyle j=1} . m = − 4 {\displaystyle m=-4} , n = 2 {\displaystyle n=2} Problem 7 Decide if the vector is in the set. ( 3 − 1 ) {\displaystyle {\begin{pmatrix}3\\-1\end{pmatrix}}} , { ( − 6 2 ) k | k ∈ R } {\displaystyle \{{\begin{pmatrix}-6\\2\end{pmatrix}}k\,{\big |}\,k\in \mathbb {R} \}} ( 5 4 ) {\displaystyle {\begin{pmatrix}5\\4\end{pmatrix}}} , { ( 5 − 4 ) j | j ∈ R } {\displaystyle \{{\begin{pmatrix}5\\-4\end{pmatrix}}j\,{\big |}\,j\in \mathbb {R} \}} ( 2 1 − 1 ) {\displaystyle {\begin{pmatrix}2\\1\\-1\end{pmatrix}}} , { ( 0 3 − 7 ) + ( 1 − 1 3 ) r | r ∈ R } {\displaystyle \{{\begin{pmatrix}0\\3\\-7\end{pmatrix}}+{\begin{pmatrix}1\\-1\\3\end{pmatrix}}r\,{\big |}\,r\in \mathbb {R} \}} ( 1 0 1 ) {\displaystyle {\begin{pmatrix}1\\0\\1\end{pmatrix}}} , { ( 2 0 1 ) j + ( − 3 − 1 1 ) k | j , k ∈ R } {\displaystyle \{{\begin{pmatrix}2\\0\\1\end{pmatrix}}j+{\begin{pmatrix}-3\\-1\\1\end{pmatrix}}k\,{\big |}\,j,k\in \mathbb {R} \}} Answer For each problem we get a system of linear equations by looking at the equations of components. Yes; take k = − 1 / 2 {\displaystyle k=-1/2} . No; the system with equations 5 = 5 ⋅ j {\displaystyle 5=5\cdot j} and 4 = − 4 ⋅ j {\displaystyle 4=-4\cdot j} has no solution. Yes; take r = 2 {\displaystyle r=2} . No. The second components give k = 0 {\displaystyle k=0} . Then the third components give j = 1 {\displaystyle j=1} . But the first components don't check. Problem 8 Parametrize the solution set of this one-equation system. x 1 + x 2 + ⋯ + x n = 0 {\displaystyle x_{1}+x_{2}+\cdots +x_{n}=0} Answer This system has 1 {\displaystyle 1} equation. The leading variable is x 1 {\displaystyle x_{1}} , the other variables are free. { ( − 1 1 ⋮ 0 ) x 2 + ⋯ + ( − 1 0 ⋮ 1 ) x n | x 2 , ... , x n ∈ R } {\displaystyle \{{\begin{pmatrix}-1\\1\\\vdots \\0\end{pmatrix}}x_{2}+\cdots +{\begin{pmatrix}-1\\0\\\vdots \\1\end{pmatrix}}x_{n}\,{\big |}\,x_{2},\ldots ,x_{n}\in \mathbb {R} \}} This exercise is recommended for all readers. Problem 9 Apply Gauss' method to the left-hand side to solve x + 2 y − w = a 2 x + z = b x + y + 2 w = c {\displaystyle {\begin{array}{*{4}{rc}r}x&+&2y&&&-&w&=&a\\2x&&&+&z&&&=&b\\x&+&y&&&+&2w&=&c\end{array}}} for x {\displaystyle x} , y {\displaystyle y} , z {\displaystyle z} , and w {\displaystyle w} , in terms of the constants a {\displaystyle a} , b {\displaystyle b} , and c {\displaystyle c} . Note that w {\displaystyle w} will be a free variable. Use your answer from the prior part to solve this. x + 2 y − w = 3 2 x + z = 1 x + y + 2 w = − 2 {\displaystyle {\begin{array}{*{4}{rc}r}x&+&2y&&&-&w&=&3\\2x&&&+&z&&&=&1\\x&+&y&&&+&2w&=&-2\end{array}}} Answer Gauss' method here gives ( 1 2 0 − 1 a 2 0 1 0 b 1 1 0 2 c ) → − ρ 1 + ρ 3 − 2 ρ 1 + ρ 2 ( 1 2 0 − 1 a 0 − 4 1 2 − 2 a + b 0 − 1 0 3 − a + c ) → − ( 1 / 4 ) ρ 2 + ρ 3 ( 1 2 0 − 1 a 0 − 4 1 2 − 2 a + b 0 0 − 1 / 4 5 / 2 − ( 1 / 2 ) a − ( 1 / 4 ) b + c ) , {\displaystyle {\begin{array}{rcl}\left({\begin{array}{*{4}{c}|c}1&2&0&-1&a\\2&0&1&0&b\\1&1&0&2&c\end{array}}\right)&{\xrightarrow[{-\rho _{1}+\rho _{3}}]{-2\rho _{1}+\rho _{2}}}&\left({\begin{array}{*{4}{c}|c}1&2&0&-1&a\\0&-4&1&2&-2a+b\\0&-1&0&3&-a+c\end{array}}\right)\\[3em]&{\xrightarrow[{}]{-(1/4)\rho _{2}+\rho _{3}}}&\left({\begin{array}{*{4}{c}|c}1&2&0&-1&a\\0&-4&1&2&-2a+b\\0&0&-1/4&5/2&-(1/2)a-(1/4)b+c\end{array}}\right),\end{array}}} leaving w {\displaystyle w} free. Solve: z = 2 a + b − 4 c + 10 w {\displaystyle z=2a+b-4c+10w} , and − 4 y = − 2 a + b − ( 2 a + b − 4 c + 10 w ) − 2 w {\displaystyle -4y=-2a+b-(2a+b-4c+10w)-2w} so y = a − c + 3 w {\displaystyle y=a-c+3w} , and x = a − 2 ( a − c + 3 w ) + w = − a + 2 c − 5 w . {\displaystyle x=a-2(a-c+3w)+w=-a+2c-5w.} Therefore the solution set is this. { ( − a + 2 c a − c 2 a + b − 4 c 0 ) + ( − 5 3 10 1 ) w | w ∈ R } {\displaystyle \{{\begin{pmatrix}-a+2c\\a-c\\2a+b-4c\\0\end{pmatrix}}+{\begin{pmatrix}-5\\3\\10\\1\end{pmatrix}}w\,{\big |}\,w\in \mathbb {R} \}} Plug in with a = 3 {\displaystyle a=3} , b = 1 {\displaystyle b=1} , and c = − 2 {\displaystyle c=-2} . { ( − 7 5 15 0 ) + ( − 5 3 10 1 ) w | w ∈ R } {\displaystyle \{{\begin{pmatrix}-7\\5\\15\\0\end{pmatrix}}+{\begin{pmatrix}-5\\3\\10\\1\end{pmatrix}}w\,{\big |}\,w\in \mathbb {R} \}} This exercise is recommended for all readers. Problem 10 Why is the comma needed in the notation " a i , j {\displaystyle a_{i,j}} " for matrix entries? Answer Leaving the comma out, say by writing a 123 {\displaystyle a_{123}} , is ambiguous because it could mean a 1 , 23 {\displaystyle a_{1,23}} or a 12 , 3 {\displaystyle a_{12,3}} . Problem 11 Give the 4 × 4 {\displaystyle 4\!\times \!4} matrix whose i , j {\displaystyle i,j} -th entry is i + j {\displaystyle i+j} ; − 1 {\displaystyle -1} to the i + j {\displaystyle i+j} power. Answer ( 2 3 4 5 3 4 5 6 4 5 6 7 5 6 7 8 ) {\displaystyle {\begin{pmatrix}2&3&4&5\\3&4&5&6\\4&5&6&7\\5&6&7&8\end{pmatrix}}} ( 1 − 1 1 − 1 − 1 1 − 1 1 1 − 1 1 − 1 − 1 1 − 1 1 ) {\displaystyle {\begin{pmatrix}1&-1&1&-1\\-1&1&-1&1\\1&-1&1&-1\\-1&1&-1&1\end{pmatrix}}} Problem 12 For any matrix A {\displaystyle A} , the transpose of A {\displaystyle A} , written A t r a n s {\displaystyle {{A}^{\rm {trans}}}} , is the matrix whose columns are the rows of A {\displaystyle A} . Find the transpose of each of these. ( 1 2 3 4 5 6 ) {\displaystyle {\begin{pmatrix}1&2&3\\4&5&6\end{pmatrix}}} ( 2 − 3 1 1 ) {\displaystyle {\begin{pmatrix}2&-3\\1&1\end{pmatrix}}} ( 5 10 10 5 ) {\displaystyle {\begin{pmatrix}5&10\\10&5\end{pmatrix}}} ( 1 1 0 ) {\displaystyle {\begin{pmatrix}1\\1\\0\end{pmatrix}}} Answer ( 1 4 2 5 3 6 ) {\displaystyle {\begin{pmatrix}1&4\\2&5\\3&6\end{pmatrix}}} ( 2 1 − 3 1 ) {\displaystyle {\begin{pmatrix}2&1\\-3&1\end{pmatrix}}} ( 5 10 10 5 ) {\displaystyle {\begin{pmatrix}5&10\\10&5\end{pmatrix}}} ( 1 1 0 ) {\displaystyle {\begin{pmatrix}1&1&0\end{pmatrix}}} This exercise is recommended for all readers. Problem 13 Describe all functions f ( x ) = a x 2 + b x + c {\displaystyle f(x)=ax^{2}+bx+c} such that f ( 1 ) = 2 {\displaystyle f(1)=2} and f ( − 1 ) = 6 {\displaystyle f(-1)=6} . Describe all functions f ( x ) = a x 2 + b x + c {\displaystyle f(x)=ax^{2}+bx+c} such that f ( 1 ) = 2 {\displaystyle f(1)=2} . Answer Plugging in x = 1 {\displaystyle x=1} and x = − 1 {\displaystyle x=-1} gives a + b + c = 2 a − b + c = 6 → − ρ 1 + ρ 2 a + b + c = 2 − 2 b = 4 {\displaystyle {\begin{array}{rcl}{\begin{array}{*{3}{rc}r}a&+&b&+&c&=&2\\a&-&b&+&c&=&6\end{array}}&{\xrightarrow[{}]{-\rho _{1}+\rho _{2}}}&{\begin{array}{*{3}{rc}r}a&+&b&+&c&=&2\\&&-2b&&&=&4\end{array}}\end{array}}} so the set of functions is { f ( x ) = ( 4 − c ) x 2 − 2 x + c | c ∈ R } {\displaystyle \{f(x)=(4-c)x^{2}-2x+c\,{\big |}\,c\in \mathbb {R} \}} . Putting in x = 1 {\displaystyle x=1} gives a + b + c = 2 {\displaystyle {\begin{array}{*{3}{rc}r}a&+&b&+&c&=&2\end{array}}} so the set of functions is { f ( x ) = ( 2 − b − c ) x 2 + b x + c | b , c ∈ R } {\displaystyle \{f(x)=(2-b-c)x^{2}+bx+c\,{\big |}\,b,c\in \mathbb {R} \}} . Problem 14 Show that any set of five points from the plane R 2 {\displaystyle \mathbb {R} ^{2}} lie on a common conic section, that is, they all satisfy some equation of the form a x 2 + b y 2 + c x y + d x + e y + f = 0 {\displaystyle ax^{2}+by^{2}+cxy+dx+ey+f=0} where some of a , ... , f {\displaystyle a,\,\ldots \,,f} are nonzero. Answer On plugging in the five pairs ( x , y ) {\displaystyle (x,y)} we get a system with the five equations and six unknowns a {\displaystyle a} , ..., f {\displaystyle f} . Because there are more unknowns than equations, if no inconsistency exists among the equations then there are infinitely many solutions (at least one variable will end up free). But no inconsistency can exist because a = 0 {\displaystyle a=0} , ..., f = 0 {\displaystyle f=0} is a solution (we are only using this zero solution to show that the system is consistent— the prior paragraph shows that there are nonzero solutions). Problem 15 Make up a four equations/four unknowns system having a one-parameter solution set; a two-parameter solution set; a three-parameter solution set. Answer Here is one— the fourth equation is redundant but still OK. x + y − z + w = 0 y − z = 0 2 z + 2 w = 0 z + w = 0 {\displaystyle {\begin{array}{*{4}{rc}r}x&+&y&-&z&+&w&=&0\\&&y&-&z&&&=&0\\&&&&2z&+&2w&=&0\\&&&&z&+&w&=&0\end{array}}} Here is one. x + y − z + w = 0 w = 0 w = 0 w = 0 {\displaystyle {\begin{array}{*{4}{rc}r}x&+&y&-&z&+&w&=&0\\&&&&&&w&=&0\\&&&&&&w&=&0\\&&&&&&w&=&0\end{array}}} This is one. x + y − z + w = 0 x + y − z + w = 0 x + y − z + w = 0 x + y − z + w = 0 {\displaystyle {\begin{array}{*{4}{rc}r}x&+&y&-&z&+&w&=&0\\x&+&y&-&z&+&w&=&0\\x&+&y&-&z&+&w&=&0\\x&+&y&-&z&+&w&=&0\end{array}}} ? Problem 16 Solve the system of equations. a x + y = a 2 x + a y = 1 {\displaystyle {\begin{array}{*{2}{rc}r}ax&+&y&=&a^{2}\\x&+&ay&=&1\end{array}}} For what values of a {\displaystyle a} does the system fail to have solutions, and for what values of a {\displaystyle a} are there infinitely many solutions? Answer the above question for the system. a x + y = a 3 x + a y = 1 {\displaystyle {\begin{array}{*{2}{rc}r}ax&+&y&=&a^{3}\\x&+&ay&=&1\end{array}}} (USSR Olympiad #174) Answer This is how the answer was given in the cited source. Formal solution of the system yields x = a 3 − 1 a 2 − 1 y = − a 2 + a a 2 − 1 . {\displaystyle x={\frac {a^{3}-1}{a^{2}-1}}\qquad y={\frac {-a^{2}+a}{a^{2}-1}}.} If a + 1 ≠ 0 {\displaystyle a+1\neq 0} and a − 1 ≠ 0 {\displaystyle a-1\neq 0} , then the system has the single solution x = a 2 + a + 1 a + 1 y = − a a + 1 . {\displaystyle x={\frac {a^{2}+a+1}{a+1}}\qquad y={\frac {-a}{a+1}}.} If a = − 1 {\displaystyle a=-1} , or if a = + 1 {\displaystyle a=+1} , then the formulas are meaningless; in the first instance we arrive at the system { − x + y = 1 x − y = 1 {\displaystyle \left\{{\begin{array}{*{2}{rc}r}-x&+&y&=&1\\x&-&y&=&1\end{array}}\right.} which is a contradictory system. In the second instance we have { x + y = 1 x + y = 1 {\displaystyle \left\{{\begin{array}{*{2}{rc}r}x&+&y&=&1\\x&+&y&=&1\end{array}}\right.} which has an infinite number of solutions (for example, for x {\displaystyle x} arbitrary, y = 1 − x {\displaystyle y=1-x} ). Solution of the system yields x = a 4 − 1 a 2 − 1 y = − a 3 + a a 2 − 1 . {\displaystyle x={\frac {a^{4}-1}{a^{2}-1}}\qquad y={\frac {-a^{3}+a}{a^{2}-1}}.} Here, is a 2 − 1 ≠ 0 {\displaystyle a^{2}-1\neq 0} , the system has the single solution x = a 2 + 1 {\displaystyle x=a^{2}+1} , y = − a {\displaystyle y=-a} . For a = − 1 {\displaystyle a=-1} and a = 1 {\displaystyle a=1} , we obtain the systems { − x + y = − 1 x − y = 1 { x + y = 1 x + y = 1 {\displaystyle \left\{{\begin{array}{*{2}{rc}r}-x&+&y&=&-1\\x&-&y&=&1\end{array}}\right.\qquad \left\{{\begin{array}{*{2}{rc}r}x&+&y&=&1\\x&+&y&=&1\end{array}}\right.} both of which have an infinite number of solutions. ? Problem 17 In air a gold-surfaced sphere weighs 7588 {\displaystyle 7588} grams. It is known that it may contain one or more of the metals aluminum, copper, silver, or lead. When weighed successively under standard conditions in water, benzene, alcohol, and glycerine its respective weights are 6588 {\displaystyle 6588} , 6688 {\displaystyle 6688} , 6778 {\displaystyle 6778} , and 6328 {\displaystyle 6328} grams. How much, if any, of the forenamed metals does it contain if the specific gravities of the designated substances are taken to be as follows? (Duncan & Quelch 1952) Answer This is how the answer was given in the cited source. Let u {\displaystyle u} , v {\displaystyle v} , x {\displaystyle x} , y {\displaystyle y} , z {\displaystyle z} be the volumes in c m 3 {\displaystyle {\rm {cm}}^{3}} of Al, Cu, Pb, Ag, and Au, respectively, contained in the sphere, which we assume to be not hollow. Since the loss of weight in water (specific gravity 1.00 {\displaystyle 1.00} ) is 1000 {\displaystyle 1000} grams, the volume of the sphere is 1000  cm 3 {\displaystyle 1000{\mbox{ cm}}^{3}} . Then the data, some of which is superfluous, though consistent, leads to only 2 {\displaystyle 2} independent equations, one relating volumes and the other, weights. u + v + x + y + z = 1000 2.7 u + 8.9 v + 11.3 x + 10.5 y + 19.3 z = 7558 {\displaystyle {\begin{array}{*{5}{rc}r}u&+&v&+&x&+&y&+&z&=&1000\\2.7u&+&8.9v&+&11.3x&+&10.5y&+&19.3z&=&7558\end{array}}} Clearly the sphere must contain some aluminum to bring its mean specific gravity below the specific gravities of all the other metals. There is no unique result to this part of the problem, for the amounts of three metals may be chosen arbitrarily, provided that the choices will not result in negative amounts of any metal. If the ball contains only aluminum and gold, there are 294.5  cm 3 {\displaystyle 294.5{\mbox{ cm}}^{3}} of gold and 705.5  cm 3 {\displaystyle 705.5{\mbox{ cm}}^{3}} of aluminum. Another possibility is 124.7  cm 3 {\displaystyle 124.7{\mbox{ cm}}^{3}} each of Cu, Au, Pb, and Ag and 501.2  cm 3 {\displaystyle 501.2{\mbox{ cm}}^{3}} of Al. The USSR Mathematics Olympiad, number 174. Duncan, Dewey (proposer); Quelch, W. H. (solver) (Sept.-Oct. 1952), Mathematics Magazine 26 (1): 48
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