Cement
Cement is a mixture of fine grey powder of calcium silicate and calcium aluminate. This note contains information on cement and its use.
Summary
Cement is a mixture of fine grey powder of calcium silicate and calcium aluminate. This note contains information on cement and its use.
Things to Remember
- Cement is a mixture of fine grey powder of calcium silicate and calcium aluminate.
- Paste formed by mixture of powdered form of the limestone and clay along with water to obtain a paste is called slurry.
- The red colored pea sized balls of calcium silicate and calcium aluminate formed by heating slurry is called cement clinkers.
- Cement is used for making houses, buildings, roads, etc.
MCQs
No MCQs found.
Subjective Questions
Q1:
What do you mean by transpose of a matrix? Illustrate with example.
Type: Short Difficulty: Easy
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Answer: <p>Transpose of matrix: A matrix obtained by interchanging rows and columns of the given matrix is called transpose of the matrix.</p> <p>A = (aamp;bamp;cdamp;eamp;f)</p> <p>A<sup>T</sup> or A<sup>I</sup> =(aamp;dbamp;ecamp;f)</p> <p></p>
Q2:
If (417−3) (2−113)= (x−111y), find the value of x and y.
Type: Short Difficulty: Easy
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Answer: <p>(4amp;17amp;−3) (2amp;−11amp;3)= (xamp;−111amp;y)</p> <p>or, (4times;2+1times;1amp;4times;−1+1times;37times;2+1times;−3amp;7times;−1+3times;−3) = (xamp;−111amp;y)</p> <p>or, (8+1amp;−4+314−3amp;−7−9) = (xamp;−111amp;y)</p> <p>or, (9amp;−111amp;−16) = (xamp;−111amp;y)</p> <p>∴ x = 9 and y = -16 <sub>Ans</sub></p>
Q3:
Give A = [−21−18], find:
(i) The transpose of A, (ii) The value of determinant A.
Type: Short Difficulty: Easy
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Answer: <p>A = [−2amp;1−1amp;8]</p> <p>Transpose of A (A<sup>T</sup>) or A<sup>I</sup>= [−2amp;−11amp;8]</p> <p>|A| = |−2amp;1−1amp;8| = -16 -(-1)= -16 + 1 = - 15 Ans</p>
Q4:
If M = [327−3] and N = [2−3−50] calculate the value of 2M + 3N.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>M = [3amp;27amp;−3] and N = [2amp;−3−5amp;0]</p> <p>2M = 2[3amp;27amp;−3] =[6amp;414amp;−6]</p> <p>3N = 3[2amp;−3−5amp;0] =[6amp;−9−15amp;0]</p> <p>2M + 3N =[6amp;414amp;−6] +[6amp;−9−15amp;0] = [6+6amp;4−914−15amp;−6+0] =[12amp;−5−1amp;−6] <sub>Ans</sub></p>
Q5:
If A = (2350) and B = (−342−5) then find the value of 2A - 3B.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>A = (2amp;35amp;0) and B = (−3amp;42amp;−5)</p> <p>2A - 3B = 2 (2amp;35amp;0) - 3 (−3amp;42amp;−5)</p> <p>=(4amp;610amp;0) -(−9amp;126amp;−15)</p> <p>=(4+9amp;6−1210−6amp;0+15)</p> <p>=(13amp;−64amp;15) <sub>Ans</sub></p>
Q6:
If A = (1−3−24) and B = (423−5), write the transpose of A + B.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>A = (1amp;−3−2amp;4) and B = (4amp;23amp;−5)</p> <p>A + B =(1amp;−3−2amp;4) +(4amp;23amp;−5)</p> <p>=(1+4amp;−3+2−2+3amp;4−5)</p> <p>=(5amp;−11amp;−1)</p> <p>(A + B)<sup>T</sup> =(5amp;1−1amp;−1) <sub>Ans</sub></p>
Q7:
A2×3 amd B3×3 are two matrices. Are AB and BA defined? Write with reasons.
Type: Short Difficulty: Easy
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Answer: <p>A<sub>2×3</sub> and B<sub>3×3</sub></p> <p>AB is defined because the number of column of A is equal to the number of rows of B.</p> <p>A<sub>2×3</sub> and B<sub>3×3</sub>= AB<sub>2×3</sub></p> <p>BA is not defined because the number of column of B is not equal to the number of rows of A.</p> <p>B<sub>3×3</sub> and A<sub>2×3</sub> , BA does not exist.</p>
Q8:
Prove that matrix A = (−32−96) has no inverse.
Type: Short Difficulty: Easy
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Answer: <p>A = (−3amp;2−9amp;6)</p> <p>|A| = |−3amp;2−9amp;6| = -18 + 18 = 0</p> <p>Hence, A<sup>-1</sup> does not exist.</p> <p>∴ Given matrix has no inverse matrix. <sub>Proved</sub></p>
Q9:
Construct a matrix of order 2×2 whose (ij)th element is given by aij = 2→i - 3→j.
Type: Short Difficulty: Easy
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Answer: <p>a<sub>11</sub> = 2(1) - 3(1) = 2 - 3 = -1</p> <p>a<sub>12</sub>= 2(1) - 3(2) = 2 - 6 = -4</p> <p>a<sub>21</sub>= 2(2) - 3(1) = 4 - 3 = 1</p> <p>a<sub>22</sub>= 2(2) - 3(2) = 4 - 6 = -2</p> <p>∴ Required matrix = (−1amp;−41amp;−2)<sub>2×2 Ans</sub></p> <p></p> <p></p> <p></p>
Q10:
Find the matrix x, y = (3224) and 2x + y = (10−32).
Type: Short Difficulty: Easy
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Answer: <p>2x + y =(1amp;0−3amp;2)</p> <p>or, 2x + (3amp;22amp;4) = (1amp;0−3amp;2)</p> <p>or, 2x =(1amp;0−3amp;2) -(3amp;22amp;4)</p> <p>or, 2x =(1−3amp;0−2−3−2amp;2−4)</p> <p>or, 2x =(−2amp;−2−5amp;−2)</p> <p>or, x = (−22amp;−22−52amp;−22)</p> <p>∴ x =(−1amp;−1−52amp;−1) <sub>Ans</sub></p>
Q11:
Construct a 3×3 matrix A whose elements are given by aij = 3i + 2j.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>a<sub>ij</sub> = 3i + 2j</p> <p>a<sub>11</sub> = 3× 1 + 2 × 1 = 3 + 2 = 5</p> <p>a<sub>12</sub>= 3× 1 + 2 × 2= 3 + 4 = 8</p> <p>a<sub>13</sub>= 3× 1 + 2 × 3 = 3 + 6 = 9</p> <p>a<sub>21</sub> = 3× 2 + 2 × 1 = 6 + 2 = 8</p> <p>a<sub>22</sub>= 3× 2 + 2 × 2 = 6 + 4 = 10</p> <p>a<sub>23</sub>= 3× 2 + 2 × 3 = 6 + 6 = 12</p> <p>a<sub>31</sub> = 3× 3 + 2 × 1 = 9 + 2 = 11</p> <p>a<sub>32</sub>= 3× 3 + 2 × 2 = 9 + 4 = 13</p> <p>a<sub>33</sub>= 3× 3 + 2 × 3 = 9 + 6 = 15</p> <p>∴ The required matrix is A = [5amp;7amp;98amp;10amp;1211amp;13amp;15] <sub>Ans</sub></p>
Q12:
If A + B = [7025] and A - B = [3003].
Type: Short Difficulty: Easy
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Answer: <p>If:</p> <p>A + B = [7amp;02amp;5] ..................(1)</p> <p>A - B = [3amp;00amp;3]....................(2)</p> <p>Adding (1) and (2), we get:</p> <p>A + B + A - B = [7amp;02amp;5] + [3amp;00amp;3]</p> <p>or, 2A = [10amp;02amp;8]</p> <p>or, 12 × 2A = 12 [10amp;02amp;8]</p> <p>∴ A = [5amp;01amp;4]</p> <p>Now,</p> <p>From (1)</p> <p>A + B =[7amp;02amp;5]</p> <p>or,[5amp;01amp;4] + B=[7amp;02amp;5]</p> <p>or, B =[7amp;02amp;5] -[5amp;01amp;4]</p> <p>or, B =[7−5amp;0−02−1amp;5−4]</p> <p>∴ B =[2amp;01amp;1]</p>
Q13:
If A = [3−520] and B = [−140−3], find the determinant of 2AT - 3BT.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>A = [3amp;−52amp;0] and B = [−1amp;40amp;−3]</p> <p>A<sup>T</sup> =[3amp;2−5amp;0] and B =[−1amp;04amp;−3]</p> <p>Now,</p> <p>2A<sup>T</sup> - 3B<sup>T</sup> = 2 [3amp;2−5amp;0] - 3[−1amp;04amp;−3]</p> <p>=[6amp;4−10amp;0] -[−3amp;012amp;−9]</p> <p>=[6+3amp;4−0−10−12amp;0+9]</p> <p>=[9amp;4−22amp;9]</p> <p>Again,</p> <p>|2AT−3BT| =|9amp;4−22amp;9| = 81 + 88 = 169 <sub>Ans</sub></p>
Q14:
Define inverse of a matrix with example.
Type: Short Difficulty: Easy
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Answer: <p>Two matrices A and B are said to be inverse to each other, if AB = I = BA then B is the inverse of A. i.e. A<sup>-1</sup> = B. Similarly, A is called the inverse of B i.e. B<sup>-1</sup> = A.</p>
Q15:
If A = (1x01) , B = (32−54) and determinant of A2 - BT = 8, find the value of x.
Type: Short Difficulty: Easy
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Answer: <p>A = (1amp;x0amp;1) , B =(3amp;2−5amp;4)</p> <p>A<sup>2</sup> - B<sup>T</sup></p> <p>=(1amp;x0amp;1) .(1amp;x0amp;1) -(3amp;−52amp;4)</p> <p>=(1+0amp;x+x0+0amp;0+1) -(3amp;−52amp;4)</p> <p>=(1amp;2x0amp;1) -(3amp;−52amp;4)</p> <p>=(−2amp;2x+5−2amp;−3)</p> <p>Now,</p> <p>|−2amp;2x+5−2amp;−3| = 8</p> <p>or, 6 + 4x + 10 = 8</p> <p>or, 4x = 8 - 16</p> <p>or, x = −84</p> <p>∴ x = -2 <sub>Ans</sub></p>
Q16:
If (2x+3y−23z−y) = (5−812−40−34) (72), find the value of x, y and z.
Type: Short Difficulty: Easy
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Answer: <p>(2x+3y−23z−y) = (5amp;−812amp;−40−3amp;4)(72)</p> <p>or,(2x+3y−23z−y) = (35−1684−80−21+8) = (194−13)</p> <p>Taking the corresponding elements of the equal matrix</p> <p>2x + 3 = 19</p> <p>or, 2x = 19 - 3</p> <p>or, x = 162</p> <p>∴ x = 8</p> <p>y - 2 = 4</p> <p>or, y = 4 + 2</p> <p>∴ y = 6</p> <p>3z - 1 = - 13</p> <p>or, 3z = - 13 + 1</p> <p>or, z = −123</p> <p>∴ z = -4</p> <p>∴ x = 8, y = 6 and z = -4 <sub>Ans</sub></p>
Q17:
Define the transpose of a matrix.If A = (135−247), find AT.
Type: Short Difficulty: Easy
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Answer: <p>A matrix obtained by interchanging rows and columns of the given matrix is called transpose of the matrix.</p> <p>A = (1amp;3amp;5−2amp;4amp;7)</p> <p>A<sup>T</sup> =A = (1amp;−23amp;45amp;7) <sub>Ans</sub></p>
Q18:
If (xy) = (3546) (12), find the value of x and y.
Type: Short Difficulty: Easy
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Answer: <p>(xy) = (3amp;54amp;6)(12)</p> <p>or,(xy) =(3+104+12)</p> <p>or,(xy) =(1316)</p> <p>∴ x = 13 and y = 16 <sub>Ans</sub></p>
Q19:
Find the value of (xy). If (−100−2) (xy) = (−24).
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>(−1amp;00amp;−2)(xy) =(−24)</p> <p>or,(−x+00−2) =(−24)</p> <p>or,(−x−2y) =(−24)</p> <p>Taking corresponding elements of the equal matrix</p> <p>-x = -2</p> <p>∴ x = 2</p> <p>-2y = 4</p> <p>or, y = 4−2</p> <p>∴ y = -2</p> <p>∴ x = 2 & y = -2 <sub>Ans</sub></p>
Q20:
Find the value of x and y: (113y) (x1) = (41).
Type: Short Difficulty: Easy
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Answer: <p>(1amp;13amp;y) (x1) =(41)</p> <p>or, (x+13x+y) =(41)</p> <p>Taking corresponding elements of the equal matrix:</p> <p>x + 1 = 4</p> <p>or, x = 4 -1</p> <p>∴ x = 3</p> <p>3x + y = 1</p> <p>or, 3 × 3 + y = 1</p> <p>or, 9 + y = 1</p> <p>or, y = 1 - 9</p> <p>∴ y = -8</p> <p>∴ x = 2 and y = -8 <sub>Ans</sub></p>
Q21:
If A = (34−25) and B = (21−13), find the determinant of 3A - 4B.
Type: Short Difficulty: Easy
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Answer: <p>Given matrix A = (3amp;4−2amp;5) and B =(2amp;1−1amp;3)</p> <p>3A - 4B = 3(3amp;4−2amp;5) - 4(3amp;4−2amp;5)</p> <p>=(9amp;12−6amp;15) +(−8amp;−44amp;−12)</p> <p>=(9−8amp;12−4−6+4amp;15−12)</p> <p>=(1amp;8−2amp;3)</p> <p>|3A−4B|</p> <p>=|1amp;8−2amp;3|</p> <p>= 1× 3 - 8 (-2)</p> <p>= 3 + 16</p> <p>= 19 <sub>Ans</sub></p>
Q22:
If P = (1−1−11) and Q = (2314); find the determinant of PQ.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>P = (1amp;−1−1amp;1) and Q =(2amp;31amp;4)</p> <p>PQ = (1times;2+1times;−1amp;1times;3+4times;−1−1times;2+1times;1amp;−1times;3+1times;4 )</p> <p>=(2−1amp;3−4−2+1amp;−3+4)</p> <p>=(1amp;−1−1amp;1)</p> <p>|PQ| =|1amp;−1−1amp;1|</p> <p>= -1× 1 - (-1)× (-1)</p> <p>= 1 -1</p> <p>= 0 <sub>Ans</sub></p>
Q23:
If P = (132−1) and Q = (123−4), find the determinant of 2P - 3Q.
Type: Short Difficulty: Easy
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Answer: <p>2P - 3Q = 2 (1amp;32amp;−1) - 3(1amp;23amp;−4)</p> <p>=(2amp;64amp;−2) -(3amp;69amp;−12)</p> <p>=(2−3amp;6−64−9amp;−2+12)</p> <p>=(−1amp;0−5amp;10)</p> <p>|2P−3Q| =|−1amp;0−5amp;10| = -1× 10 - (-5)× 0 = -10 + 0 = -10 <sub>Ans</sub></p>
Q24:
If P = (20−13) and Q = (24−62), find the determinant of 5P - 12Q + 2I.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>P = (2amp;0−1amp;3), Q = (2amp;4−6amp;2) and I = (1amp;00amp;1)</p> <p>5P - 12Q + 2I = 5 (2amp;0−1amp;3) - 12(2amp;4−6amp;2) +2 (1amp;00amp;1) = (10amp;0−5amp;15) - (1amp;2−3amp;1) + (2amp;00amp;2) = (12amp;0−5amp;17) - (1amp;2−3amp;1) = (11amp;−2−2amp;16)</p> <p>|5P−(12)Q+2I| =|11amp;−2−2amp;16| = 16× 11 - (-2)× (-2) = 176 - 4 = 172 <sub>Ans</sub></p> <p></p>
Q25:
If A = (35−60) and B = (18−6123) find the determinant of 3A - 13B.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>A = (3amp;5−6amp;0) and B =(18amp;−612amp;3)</p> <p>3A - 13B</p> <p>= 3(3amp;5−6amp;0) - 13(18amp;−612amp;3)</p> <p>=(9amp;15−18amp;0) -(6amp;−24amp;1)</p> <p>= (9−6amp;15+2−18−4amp;0−1)</p> <p>=(3amp;17−22amp;−1)</p> <p>|3A−(13)B| =(3amp;17−22amp;−1) = -3 +374 = 371 <sub>Ans</sub></p>
Q26:
Which matrix multiplies to the matrix(1134) to get a matrix(4562)?
Type: Short Difficulty: Easy
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Answer: <p>Let: M =(aamp;cbamp;d)</p> <p>(aamp;cbamp;d)(1amp;13amp;4) =(4amp;56amp;2)</p> <p>or,(atimes;1+ctimes;3amp;atimes;1+ctimes;4btimes;1+dtimes;3amp;btimes;1+dtimes;4) =(4amp;56amp;2)</p> <p>or,(a+3camp;a+4cb+3damp;b+4d) =(4amp;56amp;2)</p> <p>Equating both sides,</p> <p>a + 3c = 4</p> <p>or, a = 4 - 3c .........................(1)</p> <p>a + 4c = 5................................(2)</p> <p>Putting the value of a in equation (2)</p> <p>4 - 3c + 4c = 5</p> <p>or, 4 + c = 5</p> <p>or, c = 5 - 4</p> <p>∴ c = 1</p> <p>Putting the value of c in equation (1)</p> <p>∴ a = 4 - 3c = 4 - 3× 1 = 4 - 3 = 1</p> <p>b + 3d = 6</p> <p>b = 6 - 3d..........................(3)</p> <p>b + 4d = 2.........................(4)</p> <p>Putting the value of b in equation (4)</p> <p>6 - 3d + 4d = 2</p> <p>or, 6 + d = 2</p> <p>or, d = 2 - 6</p> <p>∴ d = -4</p> <p>Puting the value of d in equation (3)</p> <p>∴ b = 6 - 3 × (-4) = 6 + 12 = 18</p> <p>∴ (aamp;bcamp;d) = (1amp;118amp;−4) <sub>Ans</sub></p>
Q27:
If A = (−140−3) and B = (3−520) find the determinant of 2AT - 3BT.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>A = (−1amp;40amp;−3) and B =(3amp;−52amp;0)</p> <p>A<sup>T</sup> =(−1amp;04amp;−3) and B<sup>T</sup> =(3amp;2−5amp;0)</p> <p>2A<sup>T</sup> - 3B<sup>T</sup></p> <p>= 2(−1amp;04amp;−3) - 3(3amp;2−5amp;0)</p> <p>=(−2amp;08amp;−6) - (9amp;615amp;0)</p> <p>=(−2−9amp;0−68−15amp;−6−0)</p> <p>=(−11amp;−6−7amp;−6)</p> <p>|2AT−3BT| =|−11amp;−6−7amp;−6| = -11 ×(-6) - (-7) ×(-6) = 66 - 42 = 24 <sub>Ans</sub></p> <p></p>
Q28:
If P = (−301−2) and Q = (1−234), find the determinant of 5P - 2Q - 3I.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>P = (−3amp;01amp;−2) and Q = (1amp;−23amp;4) and I =(1amp;00amp;1)</p> <p>5P - 2Q - 3I</p> <p>= 5 (−3amp;01amp;−2) - 2(1amp;−23amp;4) -3(1amp;00amp;1)</p> <p>=(−15amp;05amp;−10) -(2amp;−46amp;8) -(3amp;00amp;3)</p> <p>=(−15−2−3amp;0−(−4)−05−6−0amp;−10−8−3)</p> <p>=(−20amp;4−1amp;−21)</p> <p>|5P−2Q−3I| =|−20amp;4−1amp;−21| = -20 × (-21) - 4 × (-1) = 420 + 4 = 424 <sub>Ans</sub></p>
Q29:
If M = [3−212] and N = [134−1], find the determinant of 2M + 3N.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>M = [3amp;−21amp;2] and N = [1amp;34amp;−1]</p> <p>2M + 3N</p> <p>= 2[3amp;−21amp;2] + 3[1amp;34amp;−1]</p> <p>=[6amp;−42amp;4] +[3amp;912amp;−3]</p> <p>=[6+3amp;−4+92+12amp;4−3]</p> <p>=[9amp;514amp;1]</p> <p>|2M+3N| =[9amp;514amp;1] = 9× 1 - 5× 14 = 9 - 70 = -61 <sub>Ans</sub></p>
Q30:
If P = (1−23−1) and Q = (3−421), find the determinant of 3P - 2Q.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>P = (1amp;−23amp;−1) and Q =(3amp;−42amp;1)</p> <p>3P - 2Q</p> <p>= 3 (1amp;−23amp;−1) - 2 (3amp;−42amp;1)</p> <p>= (3amp;−6 9amp;−3) -(6amp;−84amp;2)</p> <p>=(3−6amp;−6+89−4amp;−3−2)</p> <p>=(−3amp;25amp;−5)</p> <p>|3P−2Q| =(−3amp;25amp;−5) = -3× (-5) - 5× 2 = 15 - 10 = 5 <sub>Ans</sub></p>
Q31:
If A = (46x3), B = (−4578) and the determinant of A - B - 5I is 14, calculate the value of x.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>A = (4amp;6xamp;3) and B = (−4amp;57amp;8) and I =(1amp;00amp;1)</p> <p>Now,</p> <p>A - B - 5I</p> <p>=(4amp;6xamp;3) -(−4amp;57amp;8) - 5(1amp;00amp;1)</p> <p>=(4−(−4)amp;6−5x−7amp;3−8) -(5amp;00amp;5)</p> <p>=(8amp;1x−7amp;−5) -(5amp;00amp;5)</p> <p>=(8−5amp;1−0x−7−0amp;−5−5)</p> <p>=(3amp;1x−7amp;−10)</p> <p>We have,</p> <p>|A−B−2I| = 14</p> <p>or,|3amp;1x−7amp;−10| = 14</p> <p>or, 3× -10 - 1× (x-7) = 14</p> <p>or, -30 - x + 7 = 14</p> <p>or, -x = 14 - 7 + 30</p> <p>or, -x = 37</p> <p>∴ x = -37 <sub>Ans</sub></p> <p></p>
Q32:
If the inverse of matrix A = (1−20x) and B = (14y2), determine the values of x and y.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>A = (1amp;−20amp;x) and B =(1amp;4yamp;2)</p> <p>If A and B are inverse matrix, then</p> <p>AB = I, where I =(1amp;00amp;1)</p> <p>or,(1amp;−20amp;x)(1amp;4yamp;2) =(1amp;00amp;1)</p> <p>or,(1times;1+ytimes;(−2)amp;1times;4+2times;(−2)0times;1+ytimes;xamp;0times;4+2times;x) =(1amp;00amp;1)</p> <p>or,(−2y+1amp;0xyamp;2x) =(1amp;00amp;1)</p> <p>Taking corresponding elements:</p> <p>-2y + 1 = 1</p> <p>or, -2y = 1 - 1</p> <p>or, y = 02</p> <p>∴ y = 0</p> <p>Similarly,</p> <p>2x = 1</p> <p>∴ x= 12</p> <p>∴ x = 12 and y = 0 <sub>Ans</sub></p>
Q33:
If A = (7432) then find A-1.
Type: Short Difficulty: Easy
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Answer: <p>Given,</p> <p>A = (7amp;43amp;2)</p> <p>|A| = |7amp;43amp;2| = 7× 2 - 4× 3 = 14 - 12 = 2≠ 0.</p> <p>It is possible to find A<sup>-1</sup>.</p> <p>A<sup>-1</sup> = 1|A| Adj (A)</p> <p>A = 1|A| (2amp;−4−3amp;7) = 12(2amp;−4−3amp;7) = ((22)amp;(−42)(−32)amp;(72)) =(1amp;−2(−32amp;(72))</p> <p>∴ A<sup>-1</sup> =(1amp;−2(−32amp;(72)) <sub>Ans</sub></p>
Q34:
If the inverse of the matrix (2m759) is the matrix (9n−54). Calculate the value of m and n.
Type: Short Difficulty: Easy
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Answer: <p>Let:</p> <p>A =(2mamp;75amp;9) and A<sup>-1</sup>=(9amp;n−5amp;4)</p> <p>We know that,</p> <p>AA<sup>-1</sup> = I, where I =(1amp;00amp;1)</p> <p>or,(2mamp;75amp;9) (9amp;n−5amp;4) =(1amp;00amp;1)</p> <p>or,(18m−35amp;2mn+2845−45amp;5n+36) =(1amp;00amp;1)</p> <p>or,(18m−35amp;2mn+280amp;5n+36) =(1amp;00amp;1)</p> <p>Taking the corresponding elements both sides</p> <p>18m - 35 = 1</p> <p>or, 18 m = 1 + 35</p> <p>or, m = 3618</p> <p>∴ m = 2</p> <p>5n + 36 = 1</p> <p>or, 5n = 1 - 36</p> <p>or, n = −355</p> <p>∴ n = -7</p> <p>∴ m = 2 and n = -7 <sub>Ans</sub></p>
Q35:
If |x−1x−2xx−3| = 0, find the value of x.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>|x−1amp;x−2xamp;x−3| = 0</p> <p>or, (x - 1) (x - 3) - x(x - 2) = 0</p> <p>or, x<sup>2</sup> - 4x + 3 - x<sup>2</sup> + 2x = 0</p> <p>or, - 2x + 3 = 0</p> <p>or, 2x = 3</p> <p>∴ x = 32 <sub>Ans</sub></p>
Q36:
If A = [3−5−42] show that A2 - 5A = 14I, where I is an identity matrix.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>A = [3amp;−5−4amp;2]</p> <p>L.H.S. = A<sup>2</sup> - 5A = A× A - 5× A</p> <p>= [3amp;−5−4amp;2]× [3amp;−5−4amp;2] - 5 [3amp;−5−4amp;2]</p> <p>= [3times;3+(−5)times;(−4)amp;3times;(−5)+(−5)times;2−4times;3+2times;(−4)amp;−4times;(−5)+2times;2] + [−5times;3amp;−5times;(−5)−5times;(−4)amp;−5times;2]</p> <p>= [9+20amp;−15−10−12−8amp;20+4] + [−15amp;2520amp;−10]</p> <p>= [29amp;−25−20amp;24] + [−15amp;2520amp;−10]</p> <p>= [29−15amp;−25+25−20+20amp;24−10]</p> <p>= [14amp;00amp;14]</p> <p>= 14 [1amp;00amp;1]</p> <p>= 14I <sub>Ans</sub></p>
Q37:
If the inverse of the matrix (x2x−9−y3) is the matrix (35yx). Find the value of x and y.
Type: Short Difficulty: Easy
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Answer: <p>We know that,</p> <p>(xamp;2x−9−yamp;3)(3amp;5yamp;x) = I where I =(1amp;00amp;1)</p> <p>or,(3x+y(2x−9)amp;5x+x(2x−9)−3y+3yamp;−5y+3x) =(1amp;00amp;1)</p> <p>or,(3x+2xy−9yamp;5x+2x2−9x0amp;−5y+3x) =(1amp;00amp;1)</p> <p>or,(3x+2xy−9yamp;2x2−4x0amp;3x−5y) =(1amp;00amp;1)</p> <p>Taking the corresponding element of the equal matrix</p> <p>2x<sup>2</sup> - 4x = 0</p> <p>2x(x - 2) = 0</p> <p>Either: 2x = 0 ∴x = 0</p> <p>Or, x - 2 = 0 ∴x = 2</p> <p>3x - 5y = 1</p> <p>or, 3x - 1 = 5y</p> <p>or, y = 3x−15</p> <p>If x = 0 then y = \frac {3 × 0 - 1}{5} = -15</p> <p>If x = 2 then y =\frac {3 × 2 - 1}{5} = 1</p> <p>∴ x = 0 or 2</p> <p>y = -15 or 1 <sub>Ans</sub></p>
Q38:
if I is the unit matrix of order 2×2 and 3A - 4I = 5(1−20−1) then find the matrix A.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>3A - 4I = 5(1amp;−20amp;−1)</p> <p>or, 3A - 4(1amp;00amp;1) =(5amp;−100amp;−5)</p> <p>or, 3A -(4amp;00amp;4) =(5amp;−100amp;−5)</p> <p>or, 3A =(5amp;−100amp;−5) +(4amp;00amp;4)</p> <p>or, 3A = (5+4amp;−10+00+0amp;−5+4) </p> <p>or, 3A =(9amp;−100amp;−1)</p> <p>or, A = ((93)amp;(−103)(03)amp;(−13))</p> <p>∴ A =(3amp;(−103)0amp;(−13)) <sub>Ans</sub></p> <p></p>
Q39:
If A = (3512), find the determinant of A2 + 5A-1 - 14I, where I is a 2×2 matrix.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>A =(3amp;51amp;2) and I =(1amp;00amp;1)</p> <p>A<sup>2</sup> + 5A<sup>-1</sup> - 14I</p> <p>=(3amp;51amp;2)⋅(3amp;51amp;2) + 5(3amp;51amp;2)<sup>-1</sup> -14(1amp;00amp;1)</p> <p>=(9+5amp;15+103+2amp;5+4) + 56−5(2amp;−5−1amp;3) -(14amp;00amp;14)</p> <p>=(14amp;255amp;9) +(10amp;−25−5amp;15) -(14amp;00amp;14)</p> <p>=(24amp;00amp;24) -(14amp;00amp;14)</p> <p>=(10amp;00amp;10)</p> <p>|A2+5A−1−14A| =|10amp;00amp;10| = 100 - 0 = 100 <sub>Ans</sub></p>
Q40:
If A = (42−11) prove that A2 - 5A + 6I = 0 where I is an 2×2 unit matrix.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>A = (4amp;2−1amp;1), I = (1amp;00amp;1)</p> <p>L.H.S.</p> <p>= A<sup>2</sup> - 5A + 6I</p> <p>= (4amp;2−1amp;1)⋅ (4amp;2−1amp;1) - 5 (4amp;2−1amp;1) + 6 (1amp;00amp;1)</p> <p>= (16−2amp;8+2−4−1amp;−2+1) - (20amp;10−5amp;5) + (6amp;00amp;6)</p> <p>= (14amp;10−5amp;−1) - (20amp;10−5amp;5) + (6amp;00amp;6)</p> <p>= (14−20+6amp;10−10+05+5+0amp;−1−5+6)</p> <p>= (0amp;00amp;0)</p> <p>= 0</p> <p>= R.H.S <sub>Proved</sub></p> <p></p>
Q41:
if A-1 = (4−131−3), find the matrix A.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>A<sup>-1</sup> =(4amp;−131amp;−3)</p> <p>Adj1 (A) =(−3amp;−113amp;4)</p> <p>|A| = -12 + 13 = 1</p> <p>A = 1|A| Adj.(A) = 11(−3amp;13−1amp;4) =(−3amp;13−1amp;4) <sub>Ans</sub></p>
Q42:
If M = (4005), find a matrix P such that MP = (1224).
Type: Short Difficulty: Easy
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Answer: <p>Let:</p> <p>P = (aamp;bcamp;d)</p> <p>MP =(1amp;22amp;4)</p> <p>or,(4amp;00amp;5)⋅(aamp;bcamp;d) =(1amp;22amp;4)</p> <p>or,(4a+0amp;4b+00+5camp;0+5d) =(1amp;22amp;4)</p> <p>or,(4aamp;4b5camp;5d) =(1amp;22amp;4)</p> <p>Taking corresponding elements of the equal matrix:</p> <p>4a = 1</p> <p>∴ a = 14</p> <p>4b = 2</p> <p>∴ b = 12</p> <p>5c = 2</p> <p>∴ c = 25</p> <p>5d = 4</p> <p>∴ d = 45</p> <p>∴ P =(14amp;1225amp;45) <sub>Ans</sub></p>
Q43:
Define inverse of a matrix. Find the inverse A-1 to matrix A if A = (2134).
Type: Short Difficulty: Easy
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Answer: <p>Two matrices A and B are said to be inverse to each other, if AB = I = BA then B is the inverse of A. i.e. A<sup>-1</sup> = B. Similarly, A is called the inverse of B i.e. B<sup>-1</sup> = A.</p> <p>A =(2amp;13amp;4)</p> <p>A<sup>-1</sup>= 1|A| Adj.(A)</p> <p>|A| =|2amp;13amp;4| = 8 - 3 = 5</p> <p>A<sup>-1</sup> = 15(4amp;−1−3amp;2) =(45amp;−15−35amp;25) <sub>Ans</sub></p>
Q44:
If A = (1153) find the A-1.
Type: Short Difficulty: Easy
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Answer: <p>Here,</p> <p>A =(1amp;15amp;3)</p> <p>|A| =|1amp;15amp;3| = 3 - 5 = -2</p> <p>A<sup>-1</sup>=1|A| Adj.(A)</p> <p>= 1−2|3amp;−1−5amp;1|</p> <p>=(−32amp;1252amp;−12) <sub>Ans</sub></p>
Q45:
Solve by matrix method:
4x - 3y = 11
3x + 7y = -1
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>4x - 3y = 11..........................................(1)</p> <p>3x + 7y = -1..........................................(2)</p> <p>Given equations in matrix form:</p> <p>(4amp;−33amp;7)(xy) =(11−1)</p> <p>We know,</p> <p>AX = B</p> <p>where. A =(4amp;−33amp;7), X =(xy) and B =(11−1)</p> <p>|A| =|4amp;−33amp;7| = 7× 4 - 3× -3 = 28 + 9 = 37≠ 0</p> <p>Hence, it has unique solution.</p> <p>X = A<sup>-1</sup>B where A<sup>-1</sup> =1|A| Adj.(A) = 137(7amp;3−3amp;4)</p> <p>X = A<sup>-1</sup>B = 137(7amp;3−3amp;4)(11−1)</p> <p>= 137(77amp;−3−33amp;−4)</p> <p>= 137(7437)(7437−3737)</p> <p>= (2−1)</p> <p>∴ x = 2 and y = -1 <sub>Ans</sub></p> <p></p>
Q46:
Solve by matrix method:
2x + 3y - 18 = 0
3x - 2y - 1 = 0
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>2x + 3y = 18......................(1)</p> <p>3x - 2y = 1 .........................(2)</p> <p>Given equation in matrix form:</p> <p>(2amp;33amp;−2)(xy) = (181), AX = B</p> <p>where, A =(2amp;33amp;−2), X =(xy) and B =(181)</p> <p>|A| =|2amp;33amp;−2| = 2× (-2) - 3× 3 = -4 - 9 = -13≠ 0</p> <p>It has a unique solution.</p> <p>A<sup>-1</sup>=1|A| Adj.A =(−2amp;−3−3amp;2)</p> <p>X = A<sup>-1</sup>B</p> <p>or, X = 1−13(−2amp;−3−3amp;2)(181)</p> <p>or, X = 1−13(−2times;18+1times;−3−3times;18+2times;1)</p> <p>or, X = 1−13(−36−3−54+2)</p> <p>or, X =(−39−13−52−13)</p> <p>or, X =(34)</p> <p>∴ x = 3 and y = -4 <sub>Ans</sub></p>
Q47:
Solve by matrix method:
3x - 2y = 5
x + y = 5
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>3x - 2y = 5 ......................(1)</p> <p>x + y = 5 ...........................(2)</p> <p>Given equation in matrix form:</p> <p>(3amp;−21amp;1)(xy) =(55)</p> <p>We know, AX = B</p> <p>where, A =(3amp;−21amp;1), X =(xy) and B =(55)</p> <p>|A| =|3amp;−21amp;1| = 3× 1 -1× -2 = 5≠ 0</p> <p>It has a unique solution.</p> <p>A<sup>-1</sup>=1|A| Adj. A = 15(1amp;2−1amp;3) =(15amp;25−15amp;35)</p> <p>X = A<sup>-1</sup>B =(15amp;25−15amp;35)(55) =(1+2−1+3) =(32)</p> <p>∴ x = 3 and y = 2 <sub>Ans</sub></p>
Q48:
Solve by matrix method:
3x + 5y = 21
2x + 3y = 13
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>3x + 5y = 21................................(1)</p> <p>2x + 3y = 13................................(2)</p> <p>Given equation in matrix form:</p> <p>(3amp;52amp;3)(xy) =(2113)</p> <p>We know, AX = B</p> <p>where, A =(3amp;52amp;3), X =(xy) and B =(2113)</p> <p>|A| =|3amp;52amp;3| = 3× 3 - 5× 2 = 9 -10 = -1≠ 0</p> <p>It has a unique solution.</p> <p>X = A<sup>-1</sup>B</p> <p>A<sup>-1</sup>=1|A| Adj. A = 1−1(3amp;−5−2amp;3) =(−3amp;52amp;−3)</p> <p>X = A<sup>-1</sup>B</p> <p>=(−3amp;52amp;−3)(2113)</p> <p>=(−63+6542−39)</p> <p>= (23)</p> <p>∴ x = 2 and y = 3 <sub>Ans</sub></p> <p></p> <p></p>
Q49:
Solve by matrix method:
2x + 3y = 5
5x - 2y = 3
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>2x + 3y = 5.........................(1)</p> <p>5x - 2y = 3..........................(2)</p> <p>Given equation in matrix form:</p> <p>(2amp;35amp;−2) (xy) =(53) or, AX = B</p> <p>where, A =(2amp;35amp;−2), X =(xy) and B =(53)</p> <p>|A| =|2amp;35amp;−2| = -4 - 15 = -19≠ 0</p> <p>It has a unique solution.</p> <p>X = A<sup>-1</sup>B</p> <p>A<sup>-1</sup>=1|A| Adj. A = 1−19(−2amp;−3−5amp;2)</p> <p>X = A<sup>-1</sup>B</p> <p>or, X =1−19(−2amp;−3−5amp;2)(53)</p> <p>or, X =1−19(−10−9−25+6)</p> <p>or, X =(−19−19−19−19)</p> <p>∴ X =(11)</p> <p>∴ x = 1 and y = 1 <sub>Ans</sub></p>
Q50:
Solve by matrix method:
3x- 5y = 3
4x + 3y = 4
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>3x - 5y = 3..........................(1)</p> <p>4x + 3y = 4.........................(2)</p> <p>Given equation in matrix form;</p> <p>(3amp;−54amp;3)(xy) =(34) or, AX = B</p> <p>where, A = (3amp;−54amp;3), X =(xy) and B =(34)</p> <p>|A| =|3amp;−54amp;3| = 9 + 20 = 29≠ 0</p> <p>It has a unique solution.</p> <p>X = A<sup>-1</sup>B</p> <p>X =1|A|(3amp;5−4amp;3)(34)</p> <p>or, X = 129 (3times;3+5times;4−4times;3+3times;4)</p> <p>or, X = 129 (9+20−12+12)</p> <p>or, X = 129 (290)</p> <p>or, X =(2929029)</p> <p>∴ X =(10)</p> <p>∴ x = 1 and y = 0 <sub>Ans</sub></p>
Q51:
Solve by matrix method:
2x + y = 3
3x + 2y = 2
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>2x + y = 3..................(1)</p> <p>3x + 2y = 2...............(2)</p> <p>Given equation in matrix form:</p> <p>(2amp;13amp;2)(xy) =(32), AX = B</p> <p>where, A =(2amp;13amp;2), X =(xy) and B =(32)</p> <p>|A| =|2amp;13amp;2| = 2 × 2 - 3 × 1 = 4 - 3 = 1 ≠ 0</p> <p>It has a unique solution.</p> <p>A<sup>-1</sup>=1|A| Adj. A = 11 (2amp;−1−3amp;2)</p> <p>=(2amp;−1−3amp;2)</p> <p>X = A<sup>-1</sup>B</p> <p>or, X =(2amp;−1−3amp;2)(32)</p> <p>or, X =(2times;3−1times;2−3times;3+2times;2)</p> <p>or, X =(6−2−9+4)</p> <p>∴ X = (4−5)</p> <p>∴ x = 4 and y = -5 <sub>Ans</sub></p>
Q52:
Given: A = [ab0c], B = [2003] and AB = A + B, find the values of a, b and c.
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>A = [aamp;b0amp;c] and B =[2amp;00amp;3]</p> <p>AB =[aamp;b0amp;c][2amp;00amp;3] =[atimes;2+btimes;0amp;atimes;0+btimes;30times;2+ctimes;0amp;0times;0+ctimes;3] =[2aamp;3b0amp;3c]</p> <p>A + B =[aamp;b0amp;c] +[2amp;00amp;3] =[a+2amp;b0amp;c+3]</p> <p>From Question,</p> <p>[2aamp;3b0amp;3c] =[a+2amp;b0amp;c+3]</p> <p>Taking the corresponding elements of the equal matrix,</p> <p>a + 2 = 2a</p> <p>or, 2a - a = -2</p> <p>∴ a = -2</p> <p>3b = b</p> <p>or, 3b - b = 0</p> <p>or, 2b = 0</p> <p>∴ b = 0</p> <p>3c = c + 3</p> <p>or, 3c - c = 3</p> <p>or, 2c = 3</p> <p>∴ c = 32</p> <p>∴ a = 2, b = 0 and c = 32 <sub>Ans</sub></p>
Q53:
Solve by matrix method:
2x + 3y + 4 = 0
-5x + 4y + 13 = 0
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>2x + 3y = -4....................................(1)</p> <p>-5x + 4y = -13...............................(2)</p> <p>Given equation in matrix form:</p> <p>(2amp;3−5amp;4)(xy) =(−4−13), AX = B</p> <p>where, A =(2amp;3−5amp;4), X =(xy) and B =(−4−13)</p> <p>|A| =|2amp;3−5amp;4| = 2 × 4 - (-5) × 3 = 8 + 15 = 23 ≠ 0</p> <p>It has a unique solution.</p> <p>A<sup>-1</sup>= 1|A| Adj. A = 123 (4amp;−35amp;2) =(423amp;−323523amp;223)</p> <p>We know that,</p> <p>AX = B</p> <p>X = A<sup>-1</sup>B</p> <p>or, X =(423amp;−323523amp;223)(−4−13)</p> <p>or, X =(−1623+3923−2023−2623)</p> <p>or, X = (−16+3923−20−2623)</p> <p>or, X = (2323−4623)</p> <p>∴ X =(1−2 )</p> <p>∴ x = 1 and y = -2<sub>Ans</sub></p> <p></p>
Q54:
Solve the following given equations by matrix method:
4x + 3y = 5
y - 3x = -7
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>4x + 3y = 5.........................................(1)</p> <p>y - 3x = -7...........................................(2)</p> <p>Given equation in matrix form:</p> <p>(4amp;3−3amp;1)(xy) =(5−7), AX =B</p> <p>where, A =(4amp;3−3amp;1), X =(xy) and B =(5−7)</p> <p>|A| =|4amp;3−3amp;1| = 4 × 1 - 3 × -3 = 4 + 9 = 13 ≠ 0</p> <p>It has unique solution.</p> <p>A<sup>-1</sup>=1|A| Adj.A = 113(1amp;−33amp;4) =(113amp;−313313amp;413)</p> <p>We know that,</p> <p>AX = B</p> <p>X = A<sup>-1</sup>B</p> <p>or, X =(113amp;−313313amp;413)(5−7)</p> <p>or, X =(513+21131513−2813)</p> <p>or, X =(5+211315−2813)</p> <p>or, X =(2613−1313)</p> <p>∴ X =(2−1)</p> <p>∴ x = 2 and y = -1 <sub>Ans</sub></p>
Q55:
Solve the given equation by matrix method:
2x - y = 1
2y + x = 3
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>2x - y = 1...............................(1)</p> <p>x + 2y = 3..............................(2)</p> <p>Given equation in matrix form:</p> <p>(2amp;−11amp;2)(xy) =(13), AX = B</p> <p>where, A =(2amp;−11amp;2), X = (xy) and B =(13)</p> <p>|A| =|2amp;−11amp;2| = 2 × 2 - 1 × -1 = 4 + 1 = 5 ≠ 0</p> <p>It has a unique solution.</p> <p>A<sup>-1</sup>= 1|A| Adj.A = 15(2amp;1−1amp;2) =(−25amp;15−15amp;25)</p> <p>We know that,</p> <p>AX = B</p> <p>X = A<sup>-1</sup>B</p> <p>or, X =(−25amp;15−15amp;25)(13)</p> <p>or, X =(25+35−15+65)</p> <p>or, X =(2+35−1+65)</p> <p>or, X =(5555)</p> <p>∴ X =(11)</p> <p>∴ x = 1 and y = 1 <sub>Ans</sub></p>
Q56:
Solve by matrix method:
x + 2y = 8
2x + 3y = 11
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>x + 2y = 8........................................(1)</p> <p>2x + 3y = 11..................................(2)</p> <p>Given equation in matrix form:</p> <p>(1amp;22amp;3)(xy) =(811) , AX = B</p> <p>where, A =(1amp;22amp;3), X =(xy) and B =(811)</p> <p>|A| =|1amp;22amp;3| = 1 × 3 - 2 × 2 = 3 - 4 = -1 ≠ 0</p> <p>It has a unique solution.</p> <p>A<sup>-1</sup>= 1|A| Adj.A = 1−1(3amp;−2−2amp;1) = (−3amp;22amp;−1)</p> <p>We know that,</p> <p>AX = B</p> <p>X = A<sup>-1</sup>B</p> <p>or, X =(−3amp;22amp;−1)(811)</p> <p>or, X =(−24+2216−11)</p> <p>∴ X =(−25)</p> <p>∴ x = -2 and y = 5 <sub>Ans</sub></p>
Q57:
Solve by matrix method:
2x + 3y = 5
2y - 3x = - 1
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>2x + 3y = 5...........................(1)</p> <p>-3x + 2y = -1.......................(2)</p> <p>Given equation in matrix form:</p> <p>(2amp;3−3amp;2)(xy) =(5−1), AX = B</p> <p>where, A =(2amp;3−3amp;2), X =(xy) and B =(5−1)</p> <p>|A| =|2amp;3−3amp;2| = 2 × 2 - (-3) × 3 = 4 + 9 = 13 ≠ 0</p> <p>It has a unique solution.</p> <p>A<sup>-1</sup>= 1|A| Adj.A = 113(2amp;−33amp;2)</p> <p>We know that,</p> <p>AX = B</p> <p>X = A<sup>-1</sup>B</p> <p>or, X =113(2amp;−33amp;2)(5−1)</p> <p>or, X = 113(10+315−2)</p> <p>or, X = 113(1313)</p> <p>or, X =(13131313)</p> <p>∴ X = (11)</p> <p>∴ x = 1 and y = 1 <sub>Ans</sub></p>
Q58:
Solve by matrix method:
3x + y = 51
4x - 3y = 3
Type: Long Difficulty: Easy
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Answer: <p>Given equation:</p> <p>3x + y = 51...........................(1)</p> <p>4x - 3y = 3............................(2)</p> <p>Given equation in matrix form:</p> <p>(3amp;14amp;−3)(xy) =(513), AX = B</p> <p>where, A =(3amp;14amp;−3), X = (xy) and B =(513)</p> <p>|A| =|3amp;14amp;−3| = 3 × (-3) - 4 × 1 = -9 - 4 = -13 ≠ 0</p> <p>It has a unique solution.</p> <p>A<sup>-1</sup>= 1|A| Adj.A = 1−13(−3amp;−1−4amp;3) =(313amp;113413amp;−313)</p> <p>We know that,</p> <p>AX = B</p> <p>X = A<sup>-1</sup>B</p> <p>or, X =(313amp;113413amp;−313)(513)</p> <p>or, X =(15313amp;31320413amp;−913)</p> <p>or, X =(153+313204−913)</p> <p>or, X =(1561319513)</p> <p>∴ X =(125)</p> <p>∴ x = 12 and y = 5 <sub>Ans</sub></p> <p></p>
Q59:
Solve by matrix method:
2x - 5y = 1
7x + 3y = 24
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>2x - 5y = 1.................................(1)</p> <p>7x + 3y = 24.............................(2)</p> <p>Given equation in matrix form:</p> <p>(2amp;−57amp;3)(xy) =(124), AX = B</p> <p>where, A =(2amp;−57amp;3), X =(xy) and B =(124)</p> <p>|A| =|2amp;−57amp;3| = 2 × 3 - 7 × (-5) = 6 + 35 = 41 ≠ 0 </p> <p>It has a unique solution.</p> <p>A<sup>-1</sup>= 1|A| Adj.A = 141(3amp;5−7amp;2) =(341amp;541−741amp;241)</p> <p>We know that,</p> <p>AX = B</p> <p>X = A<sup>-1</sup>B</p> <p>or, X =(341amp;541−741amp;241)(124)</p> <p>or, X =\begin {pmatrix} \frac {3 × 1}{41} & \frac {5 × 24}{41}\\ \frac {-7 × 1}{41} & \frac {2 × 24}{41}\\ \end {pmatrix}</p> <p>or, X =(341amp;12041−741amp;4841)</p> <p>or, X =(3+12041−7+4841)</p> <p>or, X =(123414141)</p> <p>∴ X =|31|</p> <p>∴ x = 3 and y = 1 <sub>Ans</sub></p> <p></p>
Q60:
Solve by matrix method;
x - 8y = 39
2x - 3y - 13 = 0
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>x - 8y = 39..........................................(1)</p> <p>2x - 3y = 13.......................................(2)</p> <p>Given equation in matrix form:</p> <p>\begin {pmatrix} 1 & -8\\ 2 & -3\\ \end {pmatrix}\begin {pmatrix} x\\ y\\ \end {pmatrix} =\begin {pmatrix} 39\\ 13\\ \end {pmatrix}, AX = B</p> <p>where, A =\begin {pmatrix} 1 & -8\\ 2 & -3\\ \end {pmatrix}, X =\begin {pmatrix} x\\ y\\ \end {pmatrix} and B =\begin {pmatrix} 39\\ 13\\ \end {pmatrix}</p> <p>\begin {vmatrix} A\\ \end {vmatrix} =\begin {vmatrix} 1 & -8\\ 2 & -3\\ \end {vmatrix} = 1 × (-3) - 2 × (-8) = -3 + 16 = 13 ≠ 0</p> <p>It has a unique solution.</p> <p>A<sup>-1</sup>= \frac {1}{\begin {vmatrix} A\\ \end {vmatrix}} Adj.A = \frac {1}{13}\begin {pmatrix} -3 & -8\\ -2 & 1\\ \end {pmatrix}</p> <p>We know that,</p> <p>AX =B</p> <p>X = A<sup>-1</sup>B</p> <p>or, X =\frac {1}{13}\begin {pmatrix} -3 & -8\\ -2 & 1\\ \end {pmatrix}\begin {pmatrix} 39\\ 13\\ \end {pmatrix}</p> <p>or, X = \frac {1}{13}\begin {pmatrix} -117 + 104\\ -78 + 13\\ \end {pmatrix}</p> <p>or, X = \frac {1}{13}\begin {pmatrix} -13\\ -65\\ \end {pmatrix}</p> <p>or, X =\begin {pmatrix} \frac {-13}{13}\\ \frac {-65}{13}\\ \end {pmatrix}</p> <p>∴ X = \begin {pmatrix} -1\\ -5\\ \end {pmatrix}</p> <p>∴ x = -1 and y = -5 <sub>Ans</sub></p>
Q61:
Solve by matrix method:
\frac 2x + \frac 3y = 2
\frac 4x - \frac 9y = -1
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>\frac 2x + \frac 3y = 2..............................................(1)</p> <p>\frac 4x - \frac 9y = -1.............................................(2)</p> <p>Given equation in matrix form:</p> <p>\begin {pmatrix} 2 & 3\\ 4 & -9\\ \end {pmatrix}\begin {pmatrix} \frac 1x\\ \frac1y\\ \end {pmatrix} =\begin {pmatrix} 2\\ -1\\ \end {pmatrix}, AX = B</p> <p>where, A =\begin {pmatrix} 2 & 3\\ 4 & -9\\ \end {pmatrix}, X =\begin {pmatrix} \frac 1x\\ \frac1y\\ \end {pmatrix} and B = \begin {pmatrix} 2\\ -1\\ \end {pmatrix}</p> <p>\begin {vmatrix} A\\ \end {vmatrix} =\begin {vmatrix} 2 & 3\\ 4 & -9\\ \end {vmatrix} = 2 × (-9) - 4 × 3 = -18 - 12 = -30 ≠ 0</p> <p>It has a unique solution.</p> <p>A<sup>-1</sup>= \frac {1}{\begin {vmatrix} A\\ \end {vmatrix}} Adj.A = \frac {1}{-30}\begin {pmatrix} -9 & -3\\ -4 & 2\\ \end {pmatrix} =\begin {pmatrix} \frac {9}{30} & \frac {3}{30}\\ \frac {4}{30} & \frac {-2}{30}\\ \end {pmatrix}</p> <p>We know that,</p> <p>AX = B</p> <p>X = A<sup>-1</sup>B</p> <p>or, X =\begin {pmatrix} \frac {9}{30} & \frac {3}{30}\\ \frac {4}{30} & \frac {-2}{30}\\ \end {pmatrix}\begin {pmatrix} 2\\ -1\\ \end {pmatrix}</p> <p>or, X =\begin {pmatrix} \frac {9×2}{30} - \frac {3×1}{30}\\ \frac {4×2}{30} + \frac {2×1}{30}\\ \end {pmatrix}</p> <p>or, X =\begin {pmatrix} \frac {18-3}{30}\\ \frac {8+2}{30}\\ \end {pmatrix}</p> <p>or, X = \begin {pmatrix} \frac {15}{30}\\ \frac {10}{30}\\ \end {pmatrix}</p> <p>∴ X = \begin {pmatrix} \frac 12\\ \frac 13\\ \end {pmatrix}</p> <p>i.e.\begin {pmatrix} \frac 1x\\ \frac1y\\ \end {pmatrix} =\begin {pmatrix} \frac 12\\ \frac 13\\ \end {pmatrix}</p> <p>Taking the corresponding part of the equal matrix:</p> <p>\frac 1x = \frac 12</p> <p>∴ x = 2</p> <p>\frac 1y = \frac 13</p> <p>∴ y = 3</p> <p>∴ x = 2 and y = 3 <sub>Ans</sub></p>
Q62:
Solve by matrix method:
\frac {3x + 5y}{4} = \frac {7x + 3y}{5} = 4
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>\frac {3x + 5y}{4} =4, \frac {7x + 3y}{5} = 4</p> <p>3x + 5y = 16..............................(1)</p> <p>7x + 3y = 20..............................(2)</p> <p>Given equation in matrix form:</p> <p>\begin {pmatrix} 3 & 5\\ 7 & 3\\ \end {pmatrix}\begin {pmatrix} x\\ y\\ \end {pmatrix} =\begin {pmatrix} 16\\ 20\\ \end {pmatrix}, AX =B</p> <p>where, A =\begin {pmatrix} 3 & 5\\ 7 & 3\\ \end {pmatrix}, X =\begin {pmatrix} x\\ y\\ \end {pmatrix} and B =\begin {pmatrix} 16\\ 20\\ \end {pmatrix}</p> <p>\begin {vmatrix} A\\ \end {vmatrix} =\begin {vmatrix} 3 & 5\\ 7 & 3\\ \end {vmatrix} = 3 × 3 - 7 × 5 = 9 - 35 = -26 ≠ 0</p> <p>It has a unique solution.</p> <p>A<sup>-1</sup>= \frac {1}{\begin {vmatrix} A\\ \end {vmatrix}} Adj.A = \frac {1}{-26}\begin {pmatrix} 3 & -5\\ -7 & 3\\ \end {pmatrix}</p> <p>We know that:</p> <p>AX = B</p> <p>X = A<sup>-1</sup>B</p> <p>or, X =\frac {1}{-26}\begin {pmatrix} 3 & -5\\ -7 & 3\\ \end {pmatrix}\begin {pmatrix} 16\\ 20\\ \end {pmatrix}</p> <p>or, X =\frac {1}{-26}\begin {pmatrix} 48 - 100\\ -112 + 60\\ \end {pmatrix}</p> <p>or, X =\frac {1}{-26}\begin {pmatrix} -52\\ -52\\ \end {pmatrix}</p> <p>or, X =\begin {pmatrix} \frac {-52}{-26}\\ \frac {-52}{-26}\\ \end {pmatrix}</p> <p>∴ X =\begin {pmatrix} 2\\ 2\\ \end {pmatrix}</p> <p>∴ x = 2 and y = 2 <sub>Ans</sub></p>
Q63:
Solve the equation by matrix method: x = \frac 23y and 4x - 3y = 1.
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>Given equations are:</p> <p>x = \frac 23y i.e. 3x - 2y = 0................(1)</p> <p>4x - 3y = 1............................(2)</p> <p>Given equation in matrix form;</p> <p>\begin {pmatrix} 3 & -2\\ 4 & -3\\ \end {pmatrix}\begin {pmatrix} x\\ y\\ \end {pmatrix} =\begin {pmatrix} 0\\ 1\\ \end {pmatrix}, AX = B</p> <p>where, A =\begin {pmatrix} 3 & -2\\ 4 & -3\\ \end {pmatrix}, X =\begin {pmatrix} x\\ y\\ \end {pmatrix} and B =\begin {pmatrix} 0\\ 1\\ \end {pmatrix}</p> <p>\begin {vmatrix} A\\ \end {vmatrix} =\begin {vmatrix} 3 & -2\\ 4 & -3\\ \end {vmatrix} = 3 × -3 - 4 × -2 = -9 + 8 = -1 ≠ 0</p> <p>It has a unique solution.</p> <p>A<sup>-1</sup>= \frac {1}{\begin {vmatrix} A\\ \end {vmatrix}} Adj.A = \frac {1}{-1}\begin {pmatrix} -3 & -2\\ -4 & 3\\ \end {pmatrix} = \begin {pmatrix} 3 & 2\\ 4 & -3\\ \end {pmatrix}</p> <p>We know that,</p> <p>AX = B</p> <p>X = A<sup>-1</sup>B</p> <p>or, X =\begin {pmatrix} 3 & 2\\ 4 & -3\\ \end {pmatrix}\begin {pmatrix} 0\\ 1\\ \end {pmatrix}</p> <p>or, X =\begin {pmatrix} 3 × 0 + 2 × 1\\ 4 × 0 - 3 × 1\\ \end {pmatrix}</p> <p>or, X =\begin {pmatrix} 0 + 2\\ 0 - 3\\ \end {pmatrix}</p> <p>∴ X =\begin {pmatrix} 2\\ -3\\ \end {pmatrix}</p> <p>∴ x = 2 and y = -3 <sub>Ans</sub></p>
Q64:
Solve by matrix method:
\frac 4x + 3y = 11
3xy - 8 - 5x = 0
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>\frac 4x + 3y = 11.................................(1)</p> <p>3xy - 8 -5x = 0</p> <p>or, \frac {3xy}{x} - \frac 8x - \frac {5x}{x} = 0</p> <p>or, -\frac 8x + 3y = 5...................................(2)</p> <p>Given equation in matrix form:</p> <p>\begin {pmatrix} 4 & 3\\ -8 & 3\\ \end {pmatrix}\begin {pmatrix} \frac 1x\\ y\\ \end {pmatrix} =\begin {pmatrix} 11\\ 5\\ \end {pmatrix}, AX = B</p> <p>where, A =\begin {pmatrix} 4 & 3\\ -8 & 3\\ \end {pmatrix}, X =\begin {pmatrix} \frac 1x\\ y\\ \end {pmatrix} and B =\begin {pmatrix} 11\\ 5\\ \end {pmatrix}</p> <p>\begin {vmatrix} A\\ \end {vmatrix} =\begin {vmatrix} 4 & 3\\ -8 & 3\\ \end {vmatrix} = 4 × 3 - (-8) × 3 = 12 + 24 = 36 ≠ 0</p> <p>It has a unique solution.</p> <p>A<sup>-1</sup>= \frac {1}{\begin {vmatrix} A\\ \end {vmatrix}} Adj.A = \frac {1}{36}\begin {pmatrix} 3 & -3\\ 8 & 4\\ \end {pmatrix} =\begin {pmatrix} \frac {3}{36} & \frac {-3}{36}\\ \frac {8}{36} & \frac {4}{36}\\ \end {pmatrix} =\begin {pmatrix} \frac {1}{12} & \frac {-1}{12}\\ \frac {2}{9} & \frac {1}{9}\\ \end {pmatrix}</p> <p>We know that,</p> <p>AX = B</p> <p>X = A<sup>-1</sup>B</p> <p>or, X =\begin {pmatrix} \frac {1}{12} & \frac {-1}{12}\\ \frac {2}{9} & \frac {1}{9}\\ \end {pmatrix}\begin {pmatrix} 11\\ 5\\ \end {pmatrix}</p> <p>or, X =\begin {pmatrix} \frac {11}{12} & \frac {-5}{12}\\ \frac {22}{9} & \frac {5}{9}\\ \end {pmatrix}</p> <p>or, X = \begin {pmatrix} \frac {11 - 5}{12}\\ \frac {22 + 5}{9}\\ \end {pmatrix}</p> <p>or, X = \begin {pmatrix} \frac {6}{12}\\ \frac {27}{9}\\ \end {pmatrix}</p> <p>∴X = \begin {pmatrix} \frac {1}{2}\\ 3\\ \end {pmatrix}</p> <p>i.e.\begin {pmatrix} \frac 1x\\ y\\ \end {pmatrix} =\begin {pmatrix} \frac {1}{2}\\ 3\\ \end {pmatrix}</p> <p>Taking the corresponding part of the equal matrix:</p> <p>\frac 1x = \frac 12</p> <p>∴ x = 2</p> <p>∴ y = 3</p> <p>∴ x = 2 and y = 3 <sub>Ans</sub></p> <p></p>
Q65:
Solve by matrix method:
\frac 4x + 3y = 11
3xy - 8 - 5x = 0
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>\frac 4x + 3y = 11.................................(1)</p> <p>3xy - 8 -5x = 0</p> <p>or, \frac {3xy}{x} - \frac 8x - \frac {5x}{x} = 0</p> <p>or, -\frac 8x + 3y = 5...................................(2)</p> <p>Given equation in matrix form:</p> <p>\begin {pmatrix} 4 & 3\\ -8 & 3\\ \end {pmatrix}\begin {pmatrix} \frac 1x\\ y\\ \end {pmatrix} =\begin {pmatrix} 11\\ 5\\ \end {pmatrix}, AX = B</p> <p>where, A =\begin {pmatrix} 4 & 3\\ -8 & 3\\ \end {pmatrix}, X =\begin {pmatrix} \frac 1x\\ y\\ \end {pmatrix} and B =\begin {pmatrix} 11\\ 5\\ \end {pmatrix}</p> <p>\begin {vmatrix} A\\ \end {vmatrix} =\begin {vmatrix} 4 & 3\\ -8 & 3\\ \end {vmatrix} = 4 × 3 - (-8) × 3 = 12 + 24 = 36 ≠ 0</p> <p>It has a unique solution.</p> <p>A<sup>-1</sup>= \frac {1}{\begin {vmatrix} A\\ \end {vmatrix}} Adj.A = \frac {1}{36}\begin {pmatrix} 3 & -3\\ 8 & 4\\ \end {pmatrix} =\begin {pmatrix} \frac {3}{36} & \frac {-3}{36}\\ \frac {8}{36} & \frac {4}{36}\\ \end {pmatrix} =\begin {pmatrix} \frac {1}{12} & \frac {-1}{12}\\ \frac {2}{9} & \frac {1}{9}\\ \end {pmatrix}</p> <p>We know that,</p> <p>AX = B</p> <p>X = A<sup>-1</sup>B</p> <p>or, X =\begin {pmatrix} \frac {1}{12} & \frac {-1}{12}\\ \frac {2}{9} & \frac {1}{9}\\ \end {pmatrix}\begin {pmatrix} 11\\ 5\\ \end {pmatrix}</p> <p>or, X =\begin {pmatrix} \frac {11}{12} & \frac {-5}{12}\\ \frac {22}{9} & \frac {5}{9}\\ \end {pmatrix}</p> <p>or, X = \begin {pmatrix} \frac {11 - 5}{12}\\ \frac {22 + 5}{9}\\ \end {pmatrix}</p> <p>or, X = \begin {pmatrix} \frac {6}{12}\\ \frac {27}{9}\\ \end {pmatrix}</p> <p>∴X = \begin {pmatrix} \frac {1}{2}\\ 3\\ \end {pmatrix}</p> <p>i.e.\begin {pmatrix} \frac 1x\\ y\\ \end {pmatrix} =\begin {pmatrix} \frac {1}{2}\\ 3\\ \end {pmatrix}</p> <p>Taking the corresponding part of the equal matrix:</p> <p>\frac 1x = \frac 12</p> <p>∴ x = 2</p> <p>∴ y = 3</p> <p>∴ x = 2 and y = 3 <sub>Ans</sub></p> <p></p>
Q66:
Solve by matrix method:
4x + \frac y5 = 7
3x + \frac y4 = 5
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>4x + \frac y5 = 7</p> <p>or, \frac {20x + y}{5} = 7</p> <p>or, 20x + y = 35.........................................(1)</p> <p>3x + \frac y4 = 5</p> <p>or, \frac {12x + y}{4} = 5</p> <p>or, 12x + y = 20..........................................(2)</p> <p>Given equation in matrix form:</p> <p>\begin {pmatrix} 20 & 1\\ 12 & 1\\ \end {pmatrix}\begin {pmatrix} x\\ y\\ \end {pmatrix} =\begin {pmatrix} 35\\ 20\\ \end {pmatrix}, AX =B</p> <p>where, A =\begin {pmatrix} 20 & 1\\ 12 & 1\\ \end {pmatrix}, X =\begin {pmatrix} x\\ y\\ \end {pmatrix} and B =\begin {pmatrix} 35\\ 20\\ \end {pmatrix}</p> <p>\begin {vmatrix} A\\ \end {vmatrix} =\begin {vmatrix} 20 & 1\\ 12 & 1\\ \end {vmatrix} = 20 × 1 - 12 × 1 = 20 - 12 = 8 ≠ 0</p> <p>It has a unique solution.</p> <p>A<sup>-1</sup>= \frac {1}{\begin {vmatrix} A\\ \end {vmatrix}} Adj.A = \frac {1}{8}\begin {pmatrix} 1 & -1\\ -12 & 20\\ \end {pmatrix} =\begin {pmatrix} \frac {1}{8} & \frac {-1}{8}\\ \frac {-12}{8} & \frac {20}{8}\\ \end {pmatrix} = \begin {pmatrix} \frac {1}{8} & \frac {-1}{8}\\ \frac {-3}{2} & \frac {5}{2}\\ \end {pmatrix}</p> <p>We know that,</p> <p>AX = B</p> <p>or, X = A<sup>-1</sup>B</p> <p>or, X =\begin {pmatrix} \frac {1}{8} & \frac {-1}{8}\\ \frac {-3}{2} & \frac {5}{2}\\ \end {pmatrix}\begin {pmatrix} 35\\ 20\\ \end {pmatrix}</p> <p>or, X =\begin {pmatrix} \frac {35}{8} - \frac {20}{8}\\ \frac {-105}{2} + \frac {100}{2}\\ \end {pmatrix}</p> <p>or, X =\begin {pmatrix} \frac {35 - 20}{8}\\ \frac {-105 + 100}{2}\\ \end {pmatrix}</p> <p>∴ X =\begin {pmatrix} \frac {15}{8}\\ \frac {-5}{2}\\ \end {pmatrix}</p> <p>∴ x = \frac {15}{2} and y = \frac {-5}{2} <sub>Ans</sub></p> <p></p>
Q67:
Solve by the matrix method:
\frac 3x + \frac 4y = 2
\frac 9x - \frac 2y = \frac 52
Type: Long Difficulty: Easy
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Answer: <p>Here,</p> <p>\frac 3x + \frac 4y = 2..........................................(1)</p> <p>\frac 9x - \frac 2y = \frac 52....................(2)</p> <p>Given equation in the form of matrix:</p> <p>\begin {pmatrix} 3 & 4\\ 9 & -2\\ \end {pmatrix}\begin {pmatrix} \frac 1x\\ \frac1y\\ \end {pmatrix} =\begin {pmatrix} 2\\ \frac52\\ \end {pmatrix}, AX = B</p> <p>where, A =\begin {pmatrix} 3 & 4\\ 9 & -2\\ \end {pmatrix}, X =\begin {pmatrix} \frac 1x\\ \frac1y\\ \end {pmatrix} and B =\begin {pmatrix} 2\\ \frac52\\ \end {pmatrix}</p> <p>\begin {vmatrix} A\\ \end {vmatrix} =\begin {vmatrix} 3 & 4\\ 9 & -2\\ \end {vmatrix} = 3 × (-2) - 9 × 4 = -6 - 36 = -42 ≠ 0</p> <p>It has a unique solution.</p> <p>A<sup>-1</sup>= \frac {1}{\begin {vmatrix} A\\ \end {vmatrix}} Adj.A = \frac {1}{-42}\begin {pmatrix} -2 & -4\\ -9 & 3\\ \end {pmatrix} = \begin {pmatrix} \frac {-2}{-42} & \frac {-4}{-42}\\ \frac {-9}{-42} & \frac {3}{-42}\\ \end {pmatrix} =\begin {pmatrix} \frac {1}{21} & \frac {2}{21}\\ \frac {9}{42} & \frac {3}{42}\\ \end {pmatrix}</p> <p>We know that:</p> <p>AX = B</p> <p>or, X = A<sup>-1</sup>B</p> <p>or, X =\begin {pmatrix} \frac {1}{21} & \frac {2}{21}\\ \frac {9}{42} & \frac {3}{42}\\ \end {pmatrix}\begin {pmatrix} 2\\ \frac52\\ \end {pmatrix}</p> <p>or, X =\begin {pmatrix} \frac {2}{21} + \frac {5}{21}\\ \frac {9}{21} - \frac {15}{84}\\ \end {pmatrix}</p> <p>or, X =\begin {pmatrix} \frac {2 + 5}{21}\\ \frac {36 - 15}{84}\\ \end {pmatrix}</p> <p>or, X =\begin {pmatrix} \frac {7}{21}\\ \frac {21}{84}\\ \end {pmatrix}</p> <p>∴ X =\begin {pmatrix} \frac {1}{3}\\ \frac {1}{4}\\ \end {pmatrix}</p> <p>i.e. \begin {pmatrix} \frac 1x\\ \frac1y\\ \end {pmatrix} = \begin {pmatrix} \frac {1}{3}\\ \frac {1}{4}\\ \end {pmatrix}</p> <p>Taking the corresponding parts of the equal matrix:</p> <p>\frac 1x = \frac 13</p> <p>∴ x = 3</p> <p>\frac 1y = \frac 14</p> <p>∴ y = 4</p> <p>∴ x = 3 and y = 4 <sub>Ans</sub></p> <p></p>
Q68:
Solve by matrix method:
In a company the men get Rs. 25 a day and the women get Rs. 20 a day. 50 people are employed and the total wages are Rs. 1150 a day. Find the member of men and women employed in the company?
Type: Long Difficulty: Easy
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Answer: <p>Suppose the number of men = x</p> <p>Suppose the number of women = y</p> <p>Now,</p> <p>1 men get Rs. 25 a day.</p> <p>x men get Rs. 25x a day.</p> <p>1 women get Rs. 20 a day.</p> <p>y women get Rs. 20y a day.</p> <p>According to the question,</p> <p>x + y = 50.....................................(1)</p> <p>25x + 20y = 1150...................(2)</p> <p>Equation (1) and (2) in matrix form:</p> <p>\begin {bmatrix} 1 & 1\\ 25 & 20\\ \end {bmatrix}\begin {bmatrix} x\\ y\\ \end {bmatrix} =\begin {bmatrix} 50\\ 1150\\ \end {bmatrix}</p> <p>where:</p> <p>A =\begin {bmatrix} 1 & 1\\ 25 & 20\\ \end {bmatrix} is a coefficient matrix.</p> <p>X =\begin {bmatrix} x\\ y\\ \end {bmatrix} is a variable matrix.</p> <p>B =\begin {bmatrix} 50\\ 1150\\ \end {bmatrix} is a constant matrix.</p> <p>\begin {vmatrix} A\\ \end {vmatrix} =\begin {vmatrix} 1 & 1\\ 25 & 20\\ \end {vmatrix} = 1 × 20 - 25 × 1 = 20 - 25 = -5 ≠ 0</p> <p>It has a unique solution.</p> <p>A<sup>-1</sup>= \frac {1}{\begin {vmatrix} A\\ \end {vmatrix}} Adj.A = \frac {1}{-5}\begin {pmatrix} 20 & -1\\ -25 & 1\\ \end {pmatrix} =\begin {pmatrix} \frac {20}{-5} & \frac {-1}{-5}\\ \frac {-25}{-5} & \frac {1}{-5}\\ \end {pmatrix} = \begin {pmatrix} -4 & \frac {1}{5}\\ 5& \frac {1}{-5}\\ \end {pmatrix}</p> <p>We know that,</p> <p>AX = B</p> <p>or, X = A<sup>-1</sup>B</p> <p>or, X =\begin {pmatrix} -4 & \frac {1}{5}\\ 5& \frac {1}{-5}\\ \end {pmatrix}\begin {bmatrix} 50\\ 1150\\ \end {bmatrix}</p> <p>or, X =\begin {bmatrix} -200 + 230\\ 250 - 230\\ \end {bmatrix}</p> <p>∴ X =\begin {bmatrix} 30\\ 20\\ \end {bmatrix}</p> <p>∴ x = 30 and y = 20 <sub>Ans</sub></p> <p></p>
Q69:
Solve by matrix method;
1 kg potatoes and 1 kg onions together cost Rs. 50.5 kg potatoes and 3 kg onions together cost Rs. 190. Find the cost of per kg potatoes and onions?
Type: Long Difficulty: Easy
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Answer: <p>Suppose the cost of potatoes for 1 kg be Rs. x and the cost of onions for 1 kg be Rs. y.</p> <p>Here,</p> <p>The cost of 1 kg potatoes be Rs. x.</p> <p>The cost of 5 kg potatoes be Rs. 5x.</p> <p>The cost of 1 kg onions be Rs. y.</p> <p>The cost of 1 kg onions be Rs. 3y.</p> <p>According to question:</p> <p>x + y = 50...................................(1)</p> <p>5x + 3y = 190..........................(2)</p> <p>Equation (1) and (2) can be written in matrix form.</p> <p>\begin {bmatrix} 1 & 1\\ 5 & 3\\ \end {bmatrix}\begin {bmatrix} x\\ y\\ \end {bmatrix} =\begin {bmatrix} 50\\ 190\\ \end {bmatrix}</p> <p>where:</p> <p>A =\begin {bmatrix} 1 & 1\\ 5 & 3\\ \end {bmatrix} is a coefficient matrix.</p> <p>X =\begin {bmatrix} x\\ y\\ \end {bmatrix} is a variable matrix.</p> <p>B =\begin {bmatrix} 50\\ 190\\ \end {bmatrix} is a constant matrix.</p> <p>\begin {vmatrix} A\\ \end {vmatrix} =\begin {vmatrix} 1 & 1\\ 5 & 3\\ \end {vmatrix} = 1 × 3 - 5 × 1 = 3 - 5 = -2 ≠ 0</p> <p>So, the system has unique solution.</p> <p>A<sup>-1</sup>= \frac {1}{\begin {vmatrix} A\\ \end {vmatrix}} Adj.A = \frac {1}{-2}\begin {pmatrix} 3 & -1\\ -5 & 1\\ \end {pmatrix} = \begin {pmatrix} \frac {-3}{2} & \frac {1}{2}\\ \frac {5}{2} & \frac {-1}{2}\\ \end {pmatrix}</p> <p>We know that,</p> <p>AX = B</p> <p>X = A<sup>-1</sup>B</p> <p>or, X =\begin {pmatrix} \frac {-3}{2} & \frac {1}{2}\\ \frac {5}{2} & \frac {-1}{2}\\ \end {pmatrix}\begin {bmatrix} 50\\ 190\\ \end {bmatrix}</p> <p>or, X =\begin {bmatrix} -75 + 95\\ 125 - 95\\ \end {bmatrix}</p> <p>∴ X =\begin {bmatrix} 20\\ 30\\ \end {bmatrix}</p> <p>∴x = 20 and y = 30</p> <p>Hence, the cost per kg potatoes is Rs. 20 and the cost per kg onions is Rs. 30. <sub>Ans</sub></p>
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Cement
Cement is a mixture of fine grey powder of calcium silicate and calcium aluminate.
Slurry:
While manufacturing cement, the limestone and clay are first crushed in crushes and then grinned into powder.That powder mixed with water to obtain a paste is called slurry.
Cement clinkers:
While heating slurry in rotating kilin in 1600 degree Celsius, then it forms the red colored pea sized balls of calcium silicate and calcium aluminate called cement clinkers.
Note: 2 to 3% gypsum is added to cement to increase its setting time.
Uses
- It is used for making houses, buildings, roads, etc.
- A thick paste of cement, sand (1:3 ratio) and water is called mortar. It is used for plastering walls and joining bricks, stones, etc.
- A mixture of cement, gravel, coarse, sand and water is called concrete which is used for roofing and flooring.
- Reinforced Concrete Cement (RCC) is a mixture of cement, gravel, sand and water which is present in the framework of iron rods. It is used for making pillars, roofs of buildings, bridges, etc.
Lesson
Materials Used In Daily Life
Subject
Science
Grade
Grade 10
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