Let A- 1 -1 0- 0 1 -1- 0 0 1 - and B-7A20-20A7-2I- where I is an identity matrix of order 3- 3- If B-bij- then b13 is equal to

Let A = I + P where P=[ 0 amp; 1 nbsp; amp;0; 0 nbsp; nbsp; amp;0 nbsp; amp; 1; 0 nbsp; nbsp; amp;0 nbsp; amp;0 nbsp; ] P^2=[ 0 a ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Symmetric Matrix

Let A=[ 1 -1 ...

Question

Let $A = ⎛ ⎜ ⎝ \begin{matrix} 1 & - 1 & 0 0 & 1 & - 1 0 & 0 & 1 \end{matrix} ⎞ ⎟ ⎠$ and $B = 7 A^{20} - 20 A^{7} + 2 I,$ where $I$ is an identity matrix of order $3 \times 3$ . If $B = [b_{i j}],$ then $b_{13}$ is equal to

Open in App

Solution

Let $A = I + P$ where $P = ⎡ ⎢ ⎣ \begin{matrix} 0 & - 1 & 0 0 & 0 & - 1 0 & 0 & 0 \end{matrix} ⎤ ⎥ ⎦$
$P^{2} = ⎡ ⎢ ⎣ \begin{matrix} 0 & 0 & 1 0 & 0 & 0 0 & 0 & 0 \end{matrix} ⎤ ⎥ ⎦$ and $P^{n} = 0$ for $n \geq 3$
So, $A^{n} = (I + P)^{n} = I +^{n} C_{1} \cdot P +^{n} C_{2} \cdot P^{2} + \dots + P^{n}$
So, $(b_{13})_{n} = 0 + 0 +^{n} C_{2} \cdot 1 =^{n} C_{2}$

Now $b_{13}$ element of $7 \cdot A^{20} - 20 \cdot A^{7} + 2 I$
$= 7 (^{20} C_{2}) - 20 (^{7} C_{2}) + 0$
$= 7 \times 190 - 20 \times 21$
$= 70 [19 - 6] = 910$

Suggest Corrections

Similar questions

Q. Let

P = [a_{i j}]

be a 3

\times

3 matrix and let

Q = [b_{i j}]

, where

b_{i j} = 2^{i + j} a_{i j}

for

1 \leq i, j \leq 3

. If the determinant of P is 2, then the determinant of the matrix Q is

Q. Let

P = [a_{i j}]

be a

3 \times 3

matrix and let

Q = (b_{i j})

where

b_{i j} = 2^{i + j} a_{i j}

for

1 \leq i, j \leq 3

. If the determinant of P is

2

then the determinant of the matrix Q is?

Q. Let

P = [a_{i j}]

be a

3 \times 3

matrix and let

Q = [b_{i j}]

, where

b_{i j} = 2^{i + j} \cdot a_{i j}

for

1 \leq i, j \leq 3

. If the determinant of

P

2,

then the determinant of the matrix

Q

Q. Let

A = ⎡ ⎢ ⎣ \begin{matrix} 2 & 0 & 7 0 & 1 & 0 1 & - 2 & 1 \end{matrix} ⎤ ⎥ ⎦

and

⎡ ⎢ ⎣ \begin{matrix} - α & 14 α & 7 α 0 & 1 & 0 α & - 4 α & - 2 α \end{matrix} ⎤ ⎥ ⎦

. If

A B = I

, where

I

is an identity matrix of order

3

then trace

B

has value equal to

Let P=[a_ij] be a 3×3 matrix and let Q=[b_ij], where b_ij=2 ^i+ja_ij for 1≤i,j≤3. If the determinant of P is 2, then the determinant of the matrix Q is

Join BYJU'S Learning Program

Let A=⎛⎜⎝1−1001−1001⎞⎟⎠ and B=7A20−20A7+2I, where I is an identity matrix of order 3×3. If B=[bij], then b13 is equal to

Let A=I+P where P=⎡⎢⎣0−1000−1000⎤⎥⎦ P2=⎡⎢⎣001000000⎤⎥⎦ and Pn=0 for n≥3 So, An=(I+P)n=I+ nC1⋅P+ nC2⋅P2+⋯+Pn So, (b13)n=0+0+ nC2⋅1= nC2 Now b13 element of 7⋅A20−20⋅A7+2I =7(20C2)−20(7C2)+0 =7×190−20×21 =70[19−6]=910

Let $A = ⎛ ⎜ ⎝ \begin{matrix} 1 & - 1 & 0 0 & 1 & - 1 0 & 0 & 1 \end{matrix} ⎞ ⎟ ⎠$ and $B = 7 A^{20} - 20 A^{7} + 2 I,$ where $I$ is an identity matrix of order $3 \times 3$ . If $B = [b_{i j}],$ then $b_{13}$ is equal to