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Question
Find x, y, z and t, if
`3[[x y],[z t]]=[[x 6],[-1 2t]]+[[4 x+y],[z+t 3]]`
Sum
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Solution
`3 [[x y],[z t]]=[[x 6],[-1 2t]]+[[4 x+y],[z+t 3]]`
`⇒ [[3x 3y],[3z 3t]]` = `[[x+ 4 6+x+y],[-1+z+t 2t+3]]`
∴ 3x=x+4
⇒3x−x=4
⇒2x=4
⇒x=2
Also,
3y=6+x+y
⇒3y−y=6+x
⇒2y=6+x ...(1)
Putting the value of x in eq. (1), we get
2y=6+2
⇒2y=8
⇒y=4
Now,
3t=2t+3
⇒3t−2t=3
⇒t=3
3z=−1+z+t
⇒3z−z=−1+t
⇒2z=−1+t ...(2)
Putting the value of t in eq. (2), we get
2z=−1+3
⇒2z=2
⇒ z=1
∴ x=2, y=4, z=1 and t=3
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