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Matrices and DeterminantsQuestion and Answers: Page 11

Question Number 11353 Answers: 0 Comments: 0

reduce the matrix below to echelon form and then to row canonical form A = [((2 4 2 −2 5 1)),((3 6 2 2 0 4)),((4 8 2 6 −5 7)) ]

$reduce the matrix below to echelon form and then to row canonical form$ $A = [\begin{matrix} 2 4 2 - 2 5 1 \\ 3 6 2 2 0 4 \\ 4 8 2 6 - 5 7 \end{matrix}]$

Question Number 11035 Answers: 1 Comments: 0

A= [(1,(−1)),((−4),(−2)) ] A^(2016) =.....?

$A = [\begin{matrix} 1 & - 1 \\ - 4 & - 2 \end{matrix}]$ $A^{2016} = . . . . . ?$

Question Number 11034 Answers: 0 Comments: 1

u= [(1,(−1),2) ] A= [(3,2),(1,3),(0,1) ] v= [(2,(−1)) ] uAv^t =....?

$u = [\begin{matrix} 1 & - 1 & 2 \end{matrix}]$ $A = [\begin{matrix} 3 & 2 \\ 1 & 3 \\ 0 & 1 \end{matrix}]$ $v = [\begin{matrix} 2 & - 1 \end{matrix}]$ ${u A v}^{t} = . . . . ?$

Question Number 10268 Answers: 2 Comments: 0

Question Number 10245 Answers: 1 Comments: 0

prove that determinant ((1,1,1),(x,y,z),(x^2 ,y^2 ,z^2 ))=(x−y)(y−z)(z−y)

$prove that$ $| \begin{matrix} 1 & 1 & 1 \\ x & y & z \\ x^{2} & y^{2} & z^{2} \end{matrix} | = (x - y) (y - z) (z - y)$

Question Number 10243 Answers: 0 Comments: 0

solve the eqution determinant (((x−3),1,(−1)),((−7),(x+5),(−1)),((−6),6,(x−1)))=0

$solve the eqution$ $| \begin{matrix} x - 3 & 1 & - 1 \\ - 7 & x + 5 & - 1 \\ - 6 & 6 & x - 1 \end{matrix} | = 0$

Question Number 9387 Answers: 0 Comments: 2

Question Number 9287 Answers: 2 Comments: 0

find the determinant of the matrix below determinant ((0,4,0,0,0),(0,0,0,2,0),(0,0,3,0,0),(0,0,0,0,1),(5,0,0,0,0))

$f i n d t h e d e t e r m i n a n t o f t h e m a t r i x b e l o w$ $| \begin{matrix} 0 & 4 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 5 & 0 & 0 & 0 & 0 \end{matrix} |$

Question Number 9286 Answers: 1 Comments: 0

find the determinant of the matrix below determinant ((0,0,0,0,1),(0,0,0,2,0),(0,0,3,0,0),(0,4,0,0,0),(5,0,0,0,0))

$f i n d t h e d e t e r m i n a n t o f t h e m a t r i x b e l o w$ $| \begin{matrix} 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 3 & 0 & 0 \\ 0 & 4 & 0 & 0 & 0 \\ 5 & 0 & 0 & 0 & 0 \end{matrix} |$

Question Number 9285 Answers: 1 Comments: 0

find the determinant of the matrix below [(1,4,(−3),1),(2,0,6,3),(4,(−1),2,5),(1,0,(−2),4) ]

$f i n d t h e d e t e r m i n a n t o f t h e m a t r i x b e l o w$ $[\begin{matrix} 1 & 4 & - 3 & 1 \\ 2 & 0 & 6 & 3 \\ 4 & - 1 & 2 & 5 \\ 1 & 0 & - 2 & 4 \end{matrix}]$

Question Number 9271 Answers: 1 Comments: 0

Find the determinant of the matrix below. determinant (((3 1 5 3)),((4 3 8 5)),((6 2 1 7)),((8 5 8 1)))

$Find the determinant of the matrix below .$ $| \begin{matrix} 3 1 5 3 \\ 4 3 8 5 \\ 6 2 1 7 \\ 8 5 8 1 \end{matrix} |$

Question Number 8650 Answers: 0 Comments: 3

Question Number 8647 Answers: 0 Comments: 1

Question Number 8082 Answers: 0 Comments: 0

Question Number 8080 Answers: 0 Comments: 0

Question Number 7196 Answers: 1 Comments: 0

Question Number 6856 Answers: 0 Comments: 0

Prove that the set of all m × n matrices having entries in a field is a vector space over F.

$P r o v e t h a t t h e s e t o f a l l m \times n m a t r i c e s h a v i n g e n t r i e s i n a$ $f i e l d i s a v e c t o r s p a c e o v e r F .$

Question Number 6251 Answers: 1 Comments: 0

3x−2y−4z=2 3y−4z=−2 2y+6z=−1

$3 x - 2 y - 4 z = 2$ $3 y - 4 z = - 2$ $2 y + 6 z = - 1$

Question Number 5963 Answers: 1 Comments: 2

for what value of k the system of equation has no solution x+2y+3z=1 2x+ky+5z=1 3x+4y+7z=1

$for what value of k the system of equation has no solution$ $x + 2 y + 3 z = 1$ $2 x + ky + 5 z = 1$ $3 x + 4 y + 7 z = 1$

Question Number 5465 Answers: 1 Comments: 0

$i$

Question Number 4812 Answers: 0 Comments: 6

Question Number 4638 Answers: 0 Comments: 0

Let P= (((1−p_1 ),p_2 ),(p_1 ,(1−p_2 )) )= ((a_(1,1) ,a_(1,2) ),(a_(2,1) ,a_(2,2) ) ) and that P^n be the nth power of P evaluated as P^n =P×P^(n−1) ;i.e by successive pre−multiplication of P to P^2 ,P^3 ,P^4 ,... up to P^(n−1) . Show that the element of P^n in the second row and first column is a_(2,1) =((p_1 (1−(1−p_1 −p_2 )^n ))/(p_1 +p_2 )). {The columns of P^n represent probability vectors for all n∈N. Hence, a_(1,1) +a_(2,1) =1 and a_(1,2) +a_(2,2) =1 for example.}

$L e t P = (\begin{matrix} 1 - p_{1} & p_{2} \\ p_{1} & 1 - p_{2} \end{matrix}) = (\begin{matrix} a_{1, 1} & a_{1, 2} \\ a_{2, 1} & a_{2, 2} \end{matrix})$ $a n d t h a t P^{n} b e t h e n t h p o w e r o f P e v a l u a t e d$ $a s P^{n} = P \times P^{n - 1}; i . e b y s u c c e s s i v e$ $p r e - m u l t i p l i c a t i o n o f P t o P^{2}, P^{3}, P^{4}, . . .$ $u p t o P^{n - 1} . S h o w t h a t t h e e l e m e n t o f$ $P^{n} i n t h e s e c o n d r o w a n d f i r s t c o l u m n$ $i s a_{2, 1} = \frac{p_{1} (1 - {(1 - p_{1} - p_{2})}^{n})}{p_{1} + p_{2}} .$ ${T h e c o l u m n s o f P^{n} r e p r e s e n t$ $p r o b a b i l i t y v e c t o r s f o r a l l n \in N . H e n c e,$ $a_{1, 1} + a_{2, 1} = 1 a n d a_{1, 2} + a_{2, 2} = 1 f o r e x a m p l e .}$

Question Number 4389 Answers: 0 Comments: 3

∫_(−∞) ^∞ ∫_(−∞) ^∞ (√(x^2 +y^2 ))exp(((−1)/2)(x^2 +y^2 ))dydx=?

$\int_{- \infty}^{\infty} \int_{- \infty}^{\infty} \sqrt{x^{2} + y^{2}} e x p (\frac{- 1}{2} (x^{2} + y^{2})) d y d x = ?$

Question Number 3965 Answers: 2 Comments: 1

Find det(A) where A is an n×n matrix of the form A= ((λ,1,1,…,…,…,…,1),(1,λ,1,…,…,…,…,1),(1,1,λ,…,…,…,…,1),(⋮,⋮,⋮,⋱,…,…,…,1),(⋮,⋮,⋮,⋮,⋱,…,…,1),(⋮,⋮,⋮,⋮,⋮,⋱,…,1),(⋮,⋮,⋮,⋮,⋮,⋮,⋱,1),(1,1,1,1,1,1,1,λ) ) λ=constant for leading diagonal elements 1s for all other elements.

$F i n d d e t (A) w h e r e A i s a n n \times n m a t r i x$ $o f t h e f o r m$ $A = (\begin{matrix} λ & 1 & 1 & \dots & \dots & \dots & \dots & 1 \\ 1 & λ & 1 & \dots & \dots & \dots & \dots & 1 \\ 1 & 1 & λ & \dots & \dots & \dots & \dots & 1 \\ ⋮ & ⋮ & ⋮ & ⋱ & \dots & \dots & \dots & 1 \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋱ & \dots & \dots & 1 \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋱ & \dots & 1 \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋱ & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & λ \end{matrix})$ $λ = c o n s t a n t f o r l e a d i n g d i a g o n a l e l e m e n t s$ $1 s f o r a l l o t h e r e l e m e n t s .$

Question Number 3566 Answers: 2 Comments: 1

Let A= ((a,b),(c,d) ). Find a condition on a,b,c,d so that A^(n+1) −A^n =nA, n∈N. A^(n+1) −A^n =nA A^n −A^(n−1) =(n−1)A A^(n−1) −A^(n−2) =(n−2)A A^(n−2) −A^(n−3) =(n−3)A ... A^4 −A^3 =3A A^3 −A^2 =2A A^2 −A=A So, A^(n+1) −A=(n+n−1+n−2+...+3+2+1)A A^(n+1) =A+((n(n+1))/2)A A^(n+1) =((n^2 +n+2)/2)A A^n =(((n−1)^2 +n+1)/2)A A^n =((n^2 −n+2)/2)A (n≥1), A= ((a,b),(c,d) ) .

$L e t A = (\begin{matrix} a & b \\ c & d \end{matrix}) . F i n d a c o n d i t i o n o n$ $a, b, c, d s o t h a t A^{n + 1} - A^{n} = n A, n \in N .$ $A^{n + 1} - A^{n} = n A$ $A^{n} - A^{n - 1} = (n - 1) A$ $A^{n - 1} - A^{n - 2} = (n - 2) A$ $A^{n - 2} - A^{n - 3} = (n - 3) A$ $. . .$ $A^{4} - A^{3} = 3 A$ $A^{3} - A^{2} = 2 A$ $A^{2} - A = A$ $S o, A^{n + 1} - A = (n + n - 1 + n - 2 + . . . + 3 + 2 + 1) A$ $A^{n + 1} = A + \frac{n (n + 1)}{2} A$ $A^{n + 1} = \frac{n^{2} + n + 2}{2} A$ $A^{n} = \frac{{(n - 1)}^{2} + n + 1}{2} A$ $A^{n} = \frac{n^{2} - n + 2}{2} A (n ⩾ 1), A = (\begin{matrix} a & b \\ c & d \end{matrix}) .$

Question Number 2564 Answers: 1 Comments: 0

A= [(1,2),(2,5) ],B= [(3,(−1)),(4,(+2)) ] determine A^B

$A = [\begin{matrix} 1 & 2 \\ 2 & 5 \end{matrix}], B = [\begin{matrix} 3 & - 1 \\ 4 & + 2 \end{matrix}]$ $d e t e r m i n e A^{B}$

Pg 6 Pg 7 Pg 8 Pg 9 Pg 10 Pg 11 Pg 12

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