Choose the correct statement related to the matrices A- 1 0- 0 1 - and B- 0 1- 1 0 -

The correct option is C A^3=A, B^3 = B Clearly, A is an identity matrix #160;Now, A ^ 3 = #160; [ 1 amp; 0; 0 amp; 1 ][ ...

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

Byju's Answer

Standard XII

Mathematics

Addition and Subtraction of a Matrix

Choose the co...

Question

Choose the correct statement related to the matrices $A = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}]$ and $B = [\begin{matrix} 0 & 1 1 & 0 \end{matrix}]$

A^{3} = A, B^{3} \neq B

No worries! We‘ve got your back. Try BYJU‘S free classes today!

A^{3} \neq A, B^{3} = B

No worries! We‘ve got your back. Try BYJU‘S free classes today!

A^{3} = A, B^{3} = B

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

A^{3} \neq A, B^{3} \neq B

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $A^{3} = A, B^{3} = B$
Clearly, $A$ is an identity matrix

Now, $A^{3} =$ $[\begin{matrix} 1 & 0 0 & 1 \end{matrix}] [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] [\begin{matrix} 1 & 0 0 & 1 \end{matrix}]$

$= [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] = A$

$⟹ A^{3} = A$

Also, $B^{3} = [\begin{matrix} 0 & 1 1 & 0 \end{matrix}] [\begin{matrix} 0 & 1 1 & 0 \end{matrix}] [\begin{matrix} 0 & 1 1 & 0 \end{matrix}] = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] [\begin{matrix} 0 & 1 1 & 0 \end{matrix}] = [\begin{matrix} 0 & 1 1 & 0 \end{matrix}] = B$

$⟹ B^{3} = B$

Hence, option C is correct.

Suggest Corrections

Similar questions

Q. If

\frac{s i n^{4} A}{a} + \frac{c o s^{4} A}{b} = \frac{1}{a + b}, t h e n \frac{s i n^{8} A}{a^{3}} + \frac{c o s^{8} A}{b^{3}} =

Q. Prove that:

(a + b)^{3} (b + c)^{3} + (c + a)^{3} - 3 (a + b) (b + c) (c + a) = 2 (a^{3} + b^{3} + c^{3} - 3 a b c)

Q. If

\frac{s i n^{4} A}{a} + \frac{c o s^{4} A}{b} = \frac{1}{a + b},

then

\frac{s i n^{8} A}{a^{3}} + \frac{c o s^{8} A}{b^{3}} =

If $α$ and $β$ are imaginary cube roots of unity and x = a+b,y = a $α$ + b $β$ , z= a $β$ + b $α$ , then xyz =

$I f \frac{s i n^{4} A}{a} + \frac{c o s^{4} A}{b} = \frac{1}{a + b}, t h e n t h e v a l u e o f \frac{s i n^{8} A}{a^{3}} + \frac{c o s^{8} A}{b^{3}} i s e q u a l t o$

Join BYJU'S Learning Program

Choose the correct statement related to the matrices A=[1001] and B=[0110]

The correct option is C A3=A,B3=BClearly, A is an identity matrix Now, A3= [1001][1001][1001] =[1001][1001]=[1001]=A⟹A3=AAlso, B3=[0110][0110][0110]=[1001][0110]=[0110]=B⟹B3=BHence, option C is correct.

Choose the correct statement related to the matrices $A = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}]$ and $B = [\begin{matrix} 0 & 1 1 & 0 \end{matrix}]$