{"id":31843,"date":"2018-10-08T11:30:03","date_gmt":"2018-10-08T11:30:03","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=31843"},"modified":"2020-11-25T14:33:14","modified_gmt":"2020-11-25T09:03:14","slug":"ml-aggarwal-class-10-solutions-for-icse-maths-chapter-9-chapter-test","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/ml-aggarwal-class-10-solutions-for-icse-maths-chapter-9-chapter-test\/","title":{"rendered":"ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Matrices Chapter Test"},"content":{"rendered":"
These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths<\/a>. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Matrices Chapter Test.<\/p>\n ML Aggarwal Solutions<\/a>ICSE Solutions<\/a>Selina ICSE Solutions<\/a><\/p>\n Question 1.<\/strong><\/span> Question 2.<\/strong><\/span> Question 3.<\/strong><\/span> Question 4.<\/strong><\/span> Question 5.<\/strong><\/span> Question 6.<\/strong><\/span> Question 7.<\/strong><\/span> Question 8.<\/strong><\/span> Question 9.<\/strong><\/span> Question 10.<\/strong><\/span> Question 11.<\/strong><\/span> Question 12.<\/strong><\/span> Question 13.<\/strong><\/span> Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Matrices Chapter Test are helpful to complete your math homework.<\/p>\n
\nFind the values of a and below<\/strong>
\n\\(\\begin{bmatrix} a+3 & { b }^{ 2 }+2 \\\\ 0 & -6 \\end{bmatrix}=\\begin{bmatrix} 2a+1 & 3b \\\\ 0 & { b }^{ 2 }-5b \\end{bmatrix}\\)<\/strong>
\nSolution:<\/strong><\/span>
\n\\(\\begin{bmatrix} a+3 & { b }^{ 2 }+2 \\\\ 0 & -6 \\end{bmatrix}=\\begin{bmatrix} 2a+1 & 3b \\\\ 0 & { b }^{ 2 }-5b \\end{bmatrix}\\)
\ncomparing the corresponding elements
\na + 3 = 2a + 1
\n=> 2a – a =3 – 1
\n=> a = 2
\nb\u00b2 + 2 = 3b
\n=>b\u00b2 – 3b + 2 = 0
\n=> b\u00b2 – b – 2b + 2 = 0
\n=> b (b – 1) – 2 (b – 1) = 0
\n=> (b – 1) (b – 2) = 0.
\nEither b – 1 = 0, then b = 1 or b – 2 = 0,
\nthen b = 2
\nHence a = 2, 5 = 2 or 1 Ans.<\/p>\n
\nFind a, b, c and d if \\(3\\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix}=\\begin{bmatrix} 4 & a+b \\\\ c+d & 3 \\end{bmatrix}+\\begin{bmatrix} a & 6 \\\\ -1 & 2d \\end{bmatrix}\\)<\/strong>
\nSolution:<\/strong><\/span>
\nGiven
\n\\(3\\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix}=\\begin{bmatrix} 4 & a+b \\\\ c+d & 3 \\end{bmatrix}+\\begin{bmatrix} a & 6 \\\\ -1 & 2d \\end{bmatrix}\\)
\n<\/p>\n
\nFind X if Y = \\(\\begin{bmatrix} 3 & 2 \\\\ 1 & 4 \\end{bmatrix} \\) and 2X + Y = \\(\\begin{bmatrix} 1 & 0 \\\\ -3 & 2 \\end{bmatrix} \\)<\/strong>
\nSolution:<\/strong><\/span>
\nGiven
\n2X + Y = \\(\\begin{bmatrix} 1 & 0 \\\\ -3 & 2 \\end{bmatrix} \\)
\n=> 2X = 2X + Y = \\(\\begin{bmatrix} 1 & 0 \\\\ -3 & 2 \\end{bmatrix} \\) – Y
\n<\/p>\n
\nDetermine the matrices A and B when<\/strong>
\nA + 2B = \\(\\begin{bmatrix} 1 & 2 \\\\ 6 & -3 \\end{bmatrix} \\) and 2A – B = \\(\\begin{bmatrix} 2 & -1 \\\\ 2 & -1 \\end{bmatrix} \\)<\/strong>
\nSolution:<\/strong><\/span>
\nA + 2B = \\(\\begin{bmatrix} 1 & 2 \\\\ 6 & -3 \\end{bmatrix} \\)…..(i)
\n2A – B = \\(\\begin{bmatrix} 2 & -1 \\\\ 2 & -1 \\end{bmatrix} \\)…….(ii)
\nMultiplying (i) by 1 and (ii) by 2
\n<\/p>\n
\n(i) Find the matrix B if A = \\(\\begin{bmatrix} 4 & 1 \\\\ 2 & 3 \\end{bmatrix} \\) and A\u00b2 = A + 2B<\/strong>
\n(ii) If A = \\(\\begin{bmatrix} 1 & 2 \\\\ -3 & 4 \\end{bmatrix} \\), B = \\(\\begin{bmatrix} 0 & 1 \\\\ -2 & 5 \\end{bmatrix} \\)<\/strong>
\nand C = \\(\\begin{bmatrix} -2 & 0 \\\\ -1 & 1 \\end{bmatrix} \\) find A(4B – 3C)<\/strong>
\nSolution:<\/strong><\/span>
\nA = \\(\\begin{bmatrix} 4 & 1 \\\\ 2 & 3 \\end{bmatrix} \\)
\nlet B = \\(\\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix} \\)
\n
\n
\n<\/p>\n
\nIf A = \\(\\begin{bmatrix} 1 & 4 \\\\ 1 & 0 \\end{bmatrix} \\), B = \\(\\begin{bmatrix} 2 & 1 \\\\ 3 & -1 \\end{bmatrix} \\) and\u00a0<\/strong>C = \\(\\begin{bmatrix} 2 & 3 \\\\ 0 & 5 \\end{bmatrix} \\) compute (AB)C = (CB)A ?<\/strong>
\nSolution:<\/strong><\/span>
\nGiven
\nA = \\(\\begin{bmatrix} 1 & 4 \\\\ 1 & 0 \\end{bmatrix} \\),
\nB = \\(\\begin{bmatrix} 2 & 1 \\\\ 3 & -1 \\end{bmatrix} \\) and
\nC = \\(\\begin{bmatrix} 2 & 3 \\\\ 0 & 5 \\end{bmatrix} \\)
\n
\n<\/p>\n
\nIf A = \\(\\begin{bmatrix} 3 & 2 \\\\ 0 & 5 \\end{bmatrix} \\) and B = \\(\\begin{bmatrix} 1 & 0 \\\\ 1 & 2 \\end{bmatrix} \\) find the each of the following and state it they are equal:<\/strong>
\n(i) (A + B)(A – B)<\/strong>
\n(ii)A\u00b2 – B\u00b2<\/strong>
\nSolution:<\/strong><\/span>
\nGiven
\nA = \\(\\begin{bmatrix} 3 & 2 \\\\ 0 & 5 \\end{bmatrix} \\) and
\nB = \\(\\begin{bmatrix} 1 & 0 \\\\ 1 & 2 \\end{bmatrix} \\)
\n
\n<\/p>\n
\nIf A = \\(\\begin{bmatrix} 3 & -5 \\\\ -4 & 2 \\end{bmatrix} \\) find A\u00b2 – 5A – 14I<\/strong>
\nWhere I is unit matrix of order 2 x 2<\/strong>
\nSolution:<\/strong><\/span>
\nGiven
\nA = \\(\\begin{bmatrix} 3 & -5 \\\\ -4 & 2 \\end{bmatrix} \\)
\n
\n<\/p>\n
\nIf A = \\(\\begin{bmatrix} 3 & 3 \\\\ p & q \\end{bmatrix} \\) and A\u00b2 = 0 find p and q<\/strong>
\nSolution:<\/strong><\/span>
\nGiven
\nA = \\(\\begin{bmatrix} 3 & 3 \\\\ p & q \\end{bmatrix} \\)
\n<\/p>\n
\nIf A = \\(\\begin{bmatrix} \\frac { 3 }{ 5 } & \\frac { 2 }{ 5 } \\\\ x & y \\end{bmatrix} \\) and A\u00b2 = I, find x,y<\/strong>
\nSolution:<\/strong><\/span>
\nGiven
\nA = \\(\\begin{bmatrix} \\frac { 3 }{ 5 } & \\frac { 2 }{ 5 } \\\\ x & y \\end{bmatrix} \\)
\n
\n<\/p>\n
\nIf \\(\\begin{bmatrix} -1 & 0 \\\\ 0 & 1 \\end{bmatrix}\\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix}=\\begin{bmatrix} 1 & 0 \\\\ 0 & -1 \\end{bmatrix} \\) find a,b,c and d<\/strong>
\nSolution:<\/strong><\/span>
\nGiven
\n\\(\\begin{bmatrix} -1 & 0 \\\\ 0 & 1 \\end{bmatrix}\\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix}=\\begin{bmatrix} 1 & 0 \\\\ 0 & -1 \\end{bmatrix} \\)
\n<\/p>\n
\nFind a and b if<\/strong>
\n\\(\\begin{bmatrix} a-b & b-4 \\\\ b+4 & a-2 \\end{bmatrix}\\begin{bmatrix} 2 & 0 \\\\ 0 & 2 \\end{bmatrix}=\\begin{bmatrix} -2 & -2 \\\\ 14 & 0 \\end{bmatrix} \\)<\/strong>
\nSolution:<\/strong><\/span>
\nGiven
\n\\(\\begin{bmatrix} a-b & b-4 \\\\ b+4 & a-2 \\end{bmatrix}\\begin{bmatrix} 2 & 0 \\\\ 0 & 2 \\end{bmatrix}=\\begin{bmatrix} -2 & -2 \\\\ 14 & 0 \\end{bmatrix} \\)
\n<\/p>\n
\nIf A = \\(\\begin{bmatrix} { sec60 }^{ o } & { cos90 }^{ o } \\\\ { -3tan45 }^{ o } & { sin90 }^{ o } \\end{bmatrix} \\) and B = \\(\\begin{bmatrix} 0 & { cos45 }^{ o } \\\\ -2 & { 3sin90 }^{ o } \\end{bmatrix} \\)<\/strong>
\nFind (i) 2A – 3B (ii) A\u00b2 (iii) BA<\/strong>
\nSolution:<\/strong><\/span>
\nGiven
\nA = \\(\\begin{bmatrix} { sec60 }^{ o } & { cos90 }^{ o } \\\\ { -3tan45 }^{ o } & { sin90 }^{ o } \\end{bmatrix} \\) and
\nB = \\(\\begin{bmatrix} 0 & { cos45 }^{ o } \\\\ -2 & { 3sin90 }^{ o } \\end{bmatrix} \\)
\n
\n<\/p>\n