If two matrices A- 0 2- 1 -1- - and B- x y- 0 0- - are such that AB-O- Then x-1-y-2 is

AB=[ 0 amp;2; 1 amp; 1; ][ x amp;y; 0 amp;0; ] =[ 0 amp;0; x amp;y; ]=[ 0 amp;0; 0 amp;0; ] ⇒ x=0,y ...

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Standard XII

Mathematics

Matrix Definition and Representation

If two matric...

Question

If two matrices $A = [\begin{matrix} 0 & 2 1 & - 1 \end{matrix}]$ and $B = [\begin{matrix} x & y 0 & 0 \end{matrix}]$ are such that $A B = O .$ Then $(x + 1, y - 2)$ is

(1, 2)

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(- 1, 2)

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(1, - 2)

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(2, - 1)

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Solution

The correct option is C $(1, - 2)$
$A B = [\begin{matrix} 0 & 2 1 & - 1 \end{matrix}] [\begin{matrix} x & y 0 & 0 \end{matrix}] = [\begin{matrix} 0 & 0 x & y \end{matrix}] = [\begin{matrix} 0 & 0 0 & 0 \end{matrix}]$
$\Rightarrow x = 0, y = 0$
$∴ (x + 1, y - 2) \equiv (1, - 2)$

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Similar questions

Q. Suppose the vectors

X_{1}

X_{2}

and

X_{3}

are the solutions of the system of linear equations,

A x = b

when the vector

b

on the right side is equal to

b_{1}

b_{2}

and

b_{3}

respectively. if

X_{1} = ⎡ ⎢ ⎣ \begin{matrix} 111 \end{matrix} ⎤ ⎥ ⎦, X_{2} = ⎡ ⎢ ⎣ \begin{matrix} 021 \end{matrix} ⎤ ⎥ ⎦, X_{3} = ⎡ ⎢ ⎣ \begin{matrix} 001 \end{matrix} ⎤ ⎥ ⎦, b_{1} = ⎡ ⎢ ⎣ \begin{matrix} 100 \end{matrix} ⎤ ⎥ ⎦, b_{2} = ⎡ ⎢ ⎣ \begin{matrix} 020 \end{matrix} ⎤ ⎥ ⎦

and

b_{3} = ⎡ ⎢ ⎣ \begin{matrix} 002 \end{matrix} ⎤ ⎥ ⎦

, then the determinant of

A

is equal to

Q. If two matrices

A = [\begin{matrix} 2 x & 2 1 & y \end{matrix}]

and

B = [\begin{matrix} 1 & - 2 2 & 1 \end{matrix}]

are such that

B A = O .

Then

(x, y)

Q. Given the matrices

A = [\begin{matrix} 1 & - 1 4 & - 1 \end{matrix}]

and

B = [\begin{matrix} 1 & - 1 2 & - 2 \end{matrix}]

. The two matrices

X

and

Y

are such that

X = B A^{- 1}

and

Y = A^{- 1} B,

then the matrix

3 (X + Y)

is:

If A and B are two matrices such that AB=B and BA=A, then $A^{2} + B^{2}$ =

Q. Assertion :If

A

and

B

are two

3 \times 3

matrices such that

A B = 0

, then

A = 0

B = 0

. Reason: If

A

B

and

X

are three

3 \times 3

matrices such that

A X = B

| A | \neq 0

, then

X = A^{- 1} B

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If two matrices A=[021−1] and B=[xy00] are such that AB=O. Then (x+1,y−2) is

The correct option is C (1,−2)AB=[021−1][xy00]=[00xy]=[0000] ⇒x=0,y=0 ∴(x+1,y−2)≡(1,−2)

If two matrices $A = [\begin{matrix} 0 & 2 1 & - 1 \end{matrix}]$ and $B = [\begin{matrix} x & y 0 & 0 \end{matrix}]$ are such that $A B = O .$ Then $(x + 1, y - 2)$ is

The correct option is C $(1, - 2)$
$A B = [\begin{matrix} 0 & 2 1 & - 1 \end{matrix}] [\begin{matrix} x & y 0 & 0 \end{matrix}] = [\begin{matrix} 0 & 0 x & y \end{matrix}] = [\begin{matrix} 0 & 0 0 & 0 \end{matrix}]$
$\Rightarrow x = 0, y = 0$
$∴ (x + 1, y - 2) \equiv (1, - 2)$