Exercise 1

Prove, without developing, that the following determinants are zero:

A = \begin {vmatrix} 1 & a & b + c  \\ 1 & b & a + c \\  1 & c & a + b \\     \end {vmatrix}

B = \begin {vmatrix} a & 3a & 4a  \\ a & 5a & 6a\\  a & 7a & 8a \\     \end {vmatrix}

 

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Exercise 2

Knowing that |A|=5, calculate the value of the other determinants.

A = \begin {vmatrix} x & y & z  \\ 3 & 0 & 2 \\  1 & 1 & 1 \\     \end {vmatrix} = 5

B = \begin {vmatrix} 2x & 2y & 2z  \\ \frac{3}{2} && 0 & 1 \\  1 & 1 & 1 \\     \end {vmatrix}

C = \begin {bmatrix} x & y & z  \\ 3x + 3 & 3y & 3z + 2\\  x + 1 & y + 1 & z + 1 \\     \end {vmatrix}

 

 

Exercise 3

Prove that the following determinants are a multiple of 5 and 4 respectively, without developing them.

A = \begin {vmatrix} 2 & 4& 1  \\ 3 & 2 & 0 \\  7 & 1& 3 \\     \end {vmatrix}

B = \begin {vmatrix} -1 & 2& 3 \\ 2 & -1 & 3 \\  3 & -1& 2 \\     \end {vmatrix}

 

 

Exercise 4

Prove, without developing, that the following determinant is a multiple of 15:

A = \begin {vmatrix} 1 & 5& 0  \\ 2 & 2 & 5 \\  2 & 5& 5  \\     \end {vmatrix}

 

 

Exercise 5

Prove, without developing.

\begin {vmatrix}  </div> a^2 & a & bc  \\ b^2 &b & ca \\  c ^ 2&  c& ab \\  \end {vmatrix} =  \begin {vmatrix} a ^3 &a ^2&1 \\ b ^3 & b ^2&1 \\  c ^3 &c^2&1 \\  \end {vmatrix}

 

 

Exercise 6

Solve the following equations without developing the determinants.

\begin {vmatrix} 1 & 1& 1 \\ 1 & x & 1 \\  1 & 1& x^2\\  \end {vmatrix} = 0

 

\begin {vmatrix} a& b& c \\ a& x & c \\  a& b& x \\     \end {vmatrix} = 0

 

Exercise 7

Prove that the following determinant is divisible by 21:

A =\begin {vmatrix} 1 & 2& 3& 4& 5 &6 \\ 2&3 & 4&5&6&1 \\ 3 & 4& 5&6&1&2 \\4&5 & 6&1&2&3 \\ 5&6& 1&2&3&4 \\6&1 &2&3&4&5 \\\end {vmatrix} = 21,

 

Solution of exercise 1

Prove, without developing, that the following determinants are zero:

A = \begin {vmatrix} 1 & a & b + c  \\ 1 & b & a + c \\  1 & c & a + b \\     \end {vmatrix}

 

  = \begin {vmatrix} 1 & a & a + b + c  \\ 1 & b & a + b + c \\  1 & c & a + b + c \\     \end {vmatrix}

 

  = (a + b + c) \begin {vmatrix} 1 & a & 1  \\ 1 & b & 1\\  1 & c & 1 \\     \end {vmatrix} = 0

2.  B = \begin {vmatrix} a & 3a & 4a  \\ a & 5a & 6a\\a & 7a & 8a \\\end {vmatrix} = 0

It has two proportional lines. The third column equals the sum of the first two.

 

Solution of exercise 2

Knowing that |A|=5, calculate the value of the other determinants.

A = \begin {vmatrix} x & y & z  \\ 3 & 0 & 2 \\  1 & 1 & 1 \\  \end {vmatrix} = 5

 

    B = \begin {vmatrix} 2x & 2y & 2z  \\ \frac{3}{2} & 0 & 1 \\1 & 1 & 1 \\\end {vmatrix}<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://www.superprof.co.uk/resources/wp-content/ql-cache/quicklatex.com-ae40d3724186f3ddd585c6688499d6c8_l3.png" height="934" width="774" class="ql-img-displayed-equation " alt="\[B = 2 \cdot \frac {1} {2} \begin {vmatrix} x & y & z  \\ 3 & 0 & 2 \\1 & 1 & 1 \\\end {vmatrix} = 5$</section>  $ C = \begin {bmatrix} x & y & z  \\ 3x + 3 & 3y & 3z + 2\\ x + 1 & y + 1 & z + 1 \\ \end {vmatrix}$ $r_2 - 3r_1$ $r_3 - r_1$ $ C = \begin {bmatrix} x & y & z  \\ 3 & 0 &  2\\ 1 &  1 & 1 \\ \end {vmatrix} = 5$ <section id="am"> <h2>Solution of exercise 3</h2> Prove that the following determinants are a multiple of 5 and 4 respectively, without developing them. $ A = \begin {bmatrix} 2& 4& 1 \\ 3 & 2& 0\\ 7&  1 & 3\\ \end {vmatrix} $ $c_1 = c_1 + c_2 - c_3$ $  = \begin {bmatrix} 5& 4& 1 \\ 5 & 2& 0\\ 5&  1 & 3\\ \end {vmatrix}  = 5 \cdot \begin {bmatrix} 1& 4& 1 \\ 1& 2& 0\\ 1&  1 & 3\\ \end {vmatrix} = 5 $ $ B = \begin {bmatrix} -1 & 2 & 3  \\ 2 & -1 &  3\\ 3& -1 & 2 \\ \end {vmatrix} $ $c_1 = c_1 + c_2 + c_3$ $ C = \begin {bmatrix} 4 & 2 & 3 \\ 4& -1 &  3\\ 4 & -1 & 2 \\ \end {vmatrix} $ $  = 4 \cdot \begin {bmatrix} 1 & 2& 3  \\ 1 & -1 &  3\\ 1 &  -1 & 2 \\ \end {vmatrix} = 4$ </section>  <section id="am"> <h2>Solution of exercise 4</h2> Prove, without developing, that the following determinant is a multiple of 15: $ \begin {bmatrix} 1 & 5 & 0  \\ 2 & 2 & 5\\ 2 &  5 & 5 \\ \end {vmatrix} $ $100 c_1 + 10c_2 + c_3$ $  = \begin {bmatrix} 1 & 5 & 150  \\ 2 & 2 &  225\\ 2&  5& 225 \\ \end {vmatrix}$ $ = 15 \cdot \begin {bmatrix} 1 & 5 & 10  \\ 2 & 2 & 15\\ 2 &  5& 17 \\ \end {vmatrix} = 15$ <h2>Solution of exercise 5</h2> Prove, without developing. $ C = \begin {bmatrix} a^2 & a & bc  \\ b^2 & b & ca\\ c^2& c & ab \\ \end {vmatrix} =\begin {bmatrix} a^3 & a^2 & abc  \\ b^3 & b^2 &  abc\\ c^3 & c^2 & abc \\ \end {vmatrix} $ </section>  <section id="am">$  = \frac {1} {abc} \begin {bmatrix} a^3 & a^2 & abc  \\ b^3 & b^2 &  abc\\c^3 &  c^2 & abc \\\end {vmatrix} $</section>  $ =\frac {abc} {abc} \begin {bmatrix} a^3& a^2 & 1  \\ b^3 & b^2 &  1\\ c^3 & c^2 & 1 \\ \end {vmatrix} $ $ = \begin {bmatrix} a^ 3& a^2 & 1  \\ b^3 & b^2 & 1\\ c^3 &  c^2 & 1 \\ \end {vmatrix} $ <section id="am"> <h2>Solution of exercise 6</h2> Solve the following equations without developing the determinants. $ \begin {bmatrix} 1 & 1 & 1  \\ 1& x & 1\\ 1 &  1 & x^2 \\ \end {vmatrix} = 0$ $r_2 - r_1$ $r_3 - r_1$ </section>$ \begin {bmatrix} 1 & 1 & 1  \\ 0& x - 1 & 0\\ 0 &  0& x^2 - 1 \\ \end {vmatrix} = 0$ <section id="am">$(x - 1) (x ^2 - 1) = 0$, $x = 1$ and $x = -1\]" title="Rendered by QuickLaTeX.com"/> \begin {bmatrix} a & b& c  \\ a& x & c\\a& b & x \\ \end {vmatrix}

 

= a \cdot \begin {bmatrix} 1 & b& c  \\ 1& x & c\\  1& b & x \\  \end {vmatrix}

 

= a \cdot \begin {bmatrix} 1 & b& c  \\ 0& b - x & 0\\  0& 0 & c - x \\  \end {vmatrix}

 

= a (b - x) (c - x) a (b - x) (c - x) = 0

 

Solution of exercise 7

Prove that the following determinant is divisible by 21:

A =\begin {vmatrix} 1 & 2& 3& 4& 5 &6 \\ 2&3 & 4&5&6&1 \\ 3 & 4& 5&6&1&2 \\4&5 & 6&1&2&3 \\ 5&6& 1&2&3&4 \\6&1 &2&3&4&5 \\\end {vmatrix} = 21

 

c_6 = c_1 + c_2 + c_3 + c_4 + c_5 + c_6

 

  =\begin {vmatrix} 1 & 2& 3& 4& 5 &21\\ 2&3 & 4&5&6&21 \\ 3 & 4& 5&6&1&21 \\4&5 & 6&1&2&21 \\ 5&6& 1&2&3&21 \\6&1 &2&3&4&21\\\end {vmatrix}

 

21 \cdot \begin {vmatrix} 1 & 2& 3& 4& 5 &1\\ 2&3 & 4&5&6&1 \\ 3 & 4& 5&6&1&1 \\4&5 & 6&1&2&1 \\ 5&6& 1&2&3&1 \\6&1 &2&3&4&1\\\end {vmatrix} = 21

 

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.