Given the matrices A- 1 -1- 4 -1 - and B- 1 -1- 2 -2 - The two matrices X and Y are such that X - BA-1 and Y - A-1B- then the matrix 3X-Y is-

We have, ⇒ nbsp; X=BA^ 1⋯ (i) nbsp; nbsp; nbsp; nbsp;⇒ Y=A^ 1B ⋯ (ii) nbsp; nbsp; nbsp; nbsp; Also nbsp; A^ 1=13[ 1 amp;1; n ...

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Byju's Answer

Standard IX

Mathematics

Equality of Matrices

Given the mat...

Question

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:

[\begin{matrix} 4 & - 1 4 & 2 \end{matrix}]

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[\begin{matrix} 4 & - 1 - 4 & 2 \end{matrix}]

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[\begin{matrix} 4 & 1 4 & - 2 \end{matrix}]

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[\begin{matrix} 4 & 1 4 & 2 \end{matrix}]

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Solution

The correct option is A $[\begin{matrix} 4 & - 1 4 & 2 \end{matrix}]$
We have,
$\Rightarrow X = B A^{- 1} \dots (i)$
$\Rightarrow Y = A^{- 1} B \dots (i i)$

Also $A^{- 1} = \frac{1}{3} [\begin{matrix} - 1 & 1 - 4 & 1 \end{matrix}]$

$\Rightarrow X = B A^{- 1} = \frac{1}{3} [\begin{matrix} 1 & - 1 2 & - 2 \end{matrix}] [\begin{matrix} - 1 & 1 - 4 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 2 & 0 \end{matrix}]$

$\Rightarrow Y = A^{- 1} B = \frac{1}{3} [\begin{matrix} - 1 & 1 - 4 & 1 \end{matrix}] [\begin{matrix} 1 & - 1 2 & - 2 \end{matrix}] = \frac{1}{3} [\begin{matrix} 1 & - 1 - 2 & 2 \end{matrix}]$

$Now, 3 (X + Y) = 3 X + 3 Y = [\begin{matrix} 3 & 0 6 & 0 \end{matrix}] + [\begin{matrix} 1 & - 1 - 2 & 2 \end{matrix}] = [\begin{matrix} 4 & - 1 4 & 2 \end{matrix}]$

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Similar questions

Q. Given the matrices A and B as

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 A = B

and

A Y = B

, then the matrix

3 (X + Y)

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

and

B

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 A = B

and

A Y = B

. Then

3 (X + Y)

is equal to

Q. 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)

Q. Use the product of two matrices

A

and

B

, where

A = ⎡ ⎢ ⎣ \begin{matrix} - 5 & 1 & 3 7 & 1 & - 5 1 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦

and

B = ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 2 3 & 2 & 1 2 & 1 & 3 \end{matrix} ⎤ ⎥ ⎦

to solve the following system of linear equations,

x + y + 2 z = 1

;

3 x + 2 y + z = 7

;

2 x + y + 3 z = 2

, for

x, y

and

z

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Given the matrices A=[1−14−1] and B=[1−12−2]. The two matrices X and Y are such that X=BA−1 and Y=A−1B, then the matrix 3(X+Y) is:

The correct option is A [4−142]We have, ⇒X=BA−1⋯(i) ⇒Y=A−1B⋯(ii) Also A−1=13[−11−41] ⇒X=BA−1=13[1−12−2][−11−41]=[1020] ⇒Y=A−1B=13[−11−41][1−12−2]=13[1−1−22] Now, 3(X+Y)=3X+3Y=[3060]+[1−1−22]=[4−142]

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: