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Class 10 Class 12

A typist charges Rs. 145 for typing 10 English and 3 Hindi pages, while charges for typing 3 English and 10 Hindi pages are Rs. 180. Using matrices, find the charges of typing one English and one Hindi page separately. However, typist charged only Rs. 2 per page from a poor student Shyam for 5 Hindi pages. How much less was charged from this poor boy? Which values are reflected in this problem?

Let charges for typing one English page be Rs. x.

Let charges for typing one Hindi page be Rs.y.

Thus from the given statements, we have,

10x + 3y = 145

3x + 10y = 180

Thus the above system can be written as,

$open square brackets table row 10 3 row 3 10 end table close square brackets open square brackets table row straight x row straight y end table close square brackets space space equals space open square brackets table row 145 row 180 end table close square brackets rightwards double arrow space AX space equals space straight B comma space where comma space straight A space equals space open square brackets table row 10 3 row 3 10 end table close square brackets comma space straight X space space equals space open square brackets table row straight x row straight y end table close square brackets space and space open square brackets table row 145 row 180 end table close square brackets Multiply space straight A to the power of negative 1 end exponent space on space both space the space sides comma space we space have comma straight A to the power of negative 1 end exponent space straight x space AX space equals space straight A to the power of negative 1 end exponent straight B rightwards double arrow space IX space space equals space straight A to the power of negative 1 end exponent straight B rightwards double arrow space straight X space equals space straight A to the power of negative 1 end exponent straight B Thus comma space we space need space to space find space the space inverse space of space the space matrix space straight A. We space know space that comma space if space space straight P space equals space open square brackets table row straight a straight b row straight c straight d end table close square brackets space then space straight p to the power of negative 1 end exponent space equals space fraction numerator 1 over denominator ad minus bc end fraction space open square brackets table row straight d cell negative straight b end cell row cell negative straight c end cell straight a end table close square brackets Thus comma space straight A to the power of negative 1 end exponent space equals space fraction numerator 1 over denominator 10 space straight x space 10 minus 3 space straight x space 3 end fraction open square brackets table row 10 cell negative 3 end cell row cell negative 3 end cell 10 end table close square brackets equals fraction numerator 1 over denominator 100 minus 9 end fraction open square brackets table row 10 cell negative 3 end cell row cell negative 3 end cell 10 end table close square brackets equals 1 over 91 open square brackets table row 10 cell negative 3 end cell row cell negative 3 end cell 10 end table close square brackets therefore space space straight X space equals space 1 over 91 open square brackets table row 10 cell negative 3 end cell row cell negative 3 end cell 10 end table close square brackets space space open square brackets table row 145 row 180 end table close square brackets equals 1 over 91 open square brackets table row cell 10 space straight x space 145 space minus end cell cell 3 straight x space 180 end cell row cell negative 3 space straight x space 145 space plus end cell cell 10 straight x 180 end cell end table close square brackets space equals 1 over 91 open square brackets table row 910 row 1365 end table close square brackets equals open square brackets table row 10 row 15 end table close square brackets rightwards double arrow space open square brackets table row straight x row straight y end table close square brackets space equals space open square brackets table row 10 row 15 end table close square brackets therefore comma straight x space equals 10 space and space straight y space equals space 15 Amount space taken space from space shyam space equals 2 space straight x space 5 space equals Rs. space 10 Actual space rate space equals space 15 space straight x space 5 space equals Rs. space 75 Difference space amount space equals space 75 minus 10 space equals space 65 Rs. space 65 space was space less space charged space from space the space poor space boy space Shyam.$
Humanity is reflected in this problem.

2328 Views

If space straight A space equals space open square brackets table row 1 0 2 row 0 2 1 row 2 0 3 end table close square brackets space and space straight A cubed minus 6 straight A squared plus 7 straight A space plus kI subscript 3 space equals space 0 space find space straight k.

straight A space equals open square brackets table row 1 0 2 row 0 2 1 row 2 0 3 end table close square brackets straight A squared space equals space AA space equals space open square brackets table row 1 0 2 row 0 2 1 row 2 0 3 end table close square brackets open square brackets table row 1 0 2 row 0 2 1 row 2 0 3 end table close square brackets equals open square brackets table row 5 0 8 row 2 4 5 row 8 0 13 end table close square brackets straight A cubed space equals space straight A squared. straight A space equals space open square brackets table row 5 0 8 row 2 4 5 row 8 0 13 end table close square brackets open square brackets table row 1 0 2 row 0 2 1 row 2 0 3 end table close square brackets space equals open square brackets table row 21 0 34 row 12 8 23 row 34 0 55 end table close square brackets therefore comma space straight A cubed minus 6 straight A squared plus 7 straight A plus KI subscript 3 space equals space 0 rightwards double arrow space open square brackets table row 21 0 34 row 12 8 23 row 34 0 55 end table close square brackets minus 6 open square brackets table row 5 0 8 row 2 4 5 row 8 0 13 end table close square brackets plus 7 open square brackets table row 1 0 2 row 0 2 1 row 2 0 3 end table close square brackets plus straight k open square brackets table row 1 0 0 row 0 1 0 row 0 0 1 end table close square brackets space equals 0 rightwards double arrow space straight k equals 2

860 Views

If A is a 3 x 3 matrix |3A| = k|A|, then write the value of k.

|3A| = k |A|

|3A| = 27|A|

k = 27

920 Views

$If space straight A space equals space open parentheses table row cell cos space straight alpha end cell cell sin space straight alpha end cell row cell negative sin space straight alpha end cell cell cos space straight alpha end cell end table close parentheses comma space find space straight alpha space satisfying space 0 space less than space straight alpha space less than space straight pi over 2 space when space straight A space plus space straight A to the power of straight T space space equals space square root of 2 straight I subscript 2 semicolon space Where space straight A to the power of straight T space is space transpose space of space straight A.$

straight A space equals space open square brackets table row cell cos space straight alpha end cell cell sin space straight alpha end cell row cell negative sin space straight alpha end cell cell cos space straight alpha end cell end table close square brackets space space 0 space less than space straight alpha space less than space straight pi over 2 straight A space plus space straight A to the power of straight T space equals space square root of 2 straight I subscript 2 space open square brackets table row cell cos space straight alpha end cell cell sin space straight alpha end cell row cell negative sin space straight alpha end cell cell cos space straight alpha end cell end table close square brackets space plus space open square brackets table row cell cos space straight alpha end cell cell negative sin space straight alpha end cell row cell sin space straight alpha end cell cell cos space straight alpha end cell end table close square brackets space space equals space square root of 2 space end root open square brackets table row 1 0 row 0 1 end table close square brackets open square brackets table row cell 2 cos space straight alpha end cell 0 row 2 cell 2 cos space straight alpha end cell end table close square brackets space space equals space open square brackets table row cell square root of 2 end cell 0 row 0 cell square root of 2 end cell end table close square brackets 2 space cos space straight alpha space equals space square root of 2 cos space straight alpha space space equals space fraction numerator square root of 2 over denominator 2 end fraction space equals space fraction numerator 1 over denominator square root of 2 end fraction straight alpha space equals space straight pi over 4

4254 Views

Show that the four points A(4,5,1), B(0,-1,-1), C(3,9,4) and D(-4,4,4)
are coplanar.

Given four points A(4,5,1), B(0,-1,-1), C(3,9,4) and D(-4,4,4).
therefore,

$AB with rightwards arrow on top space equals space left parenthesis negative 4 i with hat on top minus 6 j with hat on top minus 2 k with hat on top right parenthesis AC with rightwards arrow on top space equals space left parenthesis negative i with hat on top plus 4 j with hat on top plus 3 k with hat on top right parenthesis AD with rightwards arrow on top space equals space left parenthesis negative 8 i with hat on top plus j with hat on top plus 3 k with hat on top right parenthesis therefore space space vertical line AB with rightwards arrow on top space space AC with rightwards arrow on top space space AD with rightwards arrow on top vertical line space equals space open vertical bar table row cell negative 4 end cell cell negative 6 end cell cell negative 2 end cell row cell negative 1 end cell 4 3 row cell negative 8 end cell 1 3 end table close vertical bar equals space minus 4 left parenthesis 12 space plus space 3 right parenthesis space plus space 6 left parenthesis negative 3 space plus space 24 right parenthesis space minus space 2 left parenthesis 1 space plus space 32 right parenthesis equals space minus 60 space plus space 126 space minus space 66 equals 0 Hence comma space Four space points space straight A comma space straight B comma space straight C comma straight D space are space coplanar.$

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