NCERT Solutions for Class 12 Maths<\/a> Chapter 4 Determinants Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-4-miscellaneous-exercise\/<\/p>\nNCERT Solutions for Class 12 Maths Chapter 4 Determinants Miscellaneous Exercise<\/h2>\n
<\/p>\n
Question 1.
\nProve that the determinant \\(\\left|\\begin{array}{ccc}
\nx & \\sin \\theta & \\cos \\theta \\\\
\n-\\sin \\theta & -x & 1 \\\\
\n\\cos \\theta & 1 & x
\n\\end{array}\\right|\\) is independent of \u03b8.
\nSolution:
\n<\/p>\n
Question 2.
\nWithout expanding the determinant, prove that \\(\\left|\\begin{array}{lll}
\na & a^{2} & b c \\\\
\nb & b^{2} & c a \\\\
\nc & c^{2} & a b
\n\\end{array}\\right|=\\left|\\begin{array}{lll}
\n1 & a^{2} & a^{3} \\\\
\n1 & b^{2} & b^{3} \\\\
\n1 & c^{2} & c^{3}
\n\\end{array}\\right|\\)
\nSolution:
\n<\/p>\n
<\/p>\n
Question 3.
\nEvaluate \\(\\left|\\begin{array}{ccc}
\n\\cos \\alpha \\cos \\beta & \\cos \\alpha \\sin \\beta & -\\sin \\alpha \\\\
\n-\\sin \\beta & \\cos \\beta & 0 \\\\
\n\\sin \\alpha \\cos \\beta & \\sin \\alpha \\sin \\beta & \\cos \\alpha
\n\\end{array}\\right|\\)
\nSolution:
\n<\/p>\n
Question 4.
\nIf a, b and c are real numbers, and \u2206 = \\(\\left|\\begin{array}{lll}
\nb+c & c+a & a+b \\\\
\nc+a & a+b & b+c \\\\
\na+b & b+c & c+a
\n\\end{array}\\right|\\) = 0 Show that either a + b + c = 0 or a = b = c.
\nSolution:
\n
\nExpanding along C,
\n= 2(a + b + c) ((c – b)(b – c) – (a- b)(a – c))
\n= 2(a + b + c) (bc – c\u00b2 – b\u00b2 + bc – a\u00b2 + ac + ab – bc)
\n= – (a + b + c) (2a\u00b2 + 2b\u00b2 + 2c\u00b2 – 2ab – 2bc – 2ac)
\n= – (a + b + c) (a\u00b2 + b\u00b2 – 2ab + b\u00b2 + c\u00b2 – 2bc + c\u00b2 + a\u00b2 – 2ac)
\n= – (a + b + c) ((a – b)\u00b2 + (b – c)\u00b2 + (c – a)\u00b2)
\n\u2234 \\(\\left|\\begin{array}{lll}
\nb+c & c+a & a+b \\\\
\nc+a & a+b & b+c \\\\
\na+b & b+c & c+a
\n\\end{array}\\right|\\) \u21d2 – (a + b + c)((a – b)\u00b2 + (b – c)\u00b2 + (c – a)\u00b2) = 0
\n\u21d2 (a + b + c) = 0 or (a – b)\u00b2 + (b – c)\u00b2 + (c – a)\u00b2 = 0
\n\u21d2 a + b + c = 0 or a – b = 0, b – c = 0 and c – a = 0
\n\u21d2 a + b + c = 0 or a = b = c<\/p>\n
<\/p>\n
Question 5.
\nSolve the equation \\(\\left|\\begin{array}{ccc}
\nx+a & x & x \\\\
\nx & x+a & x \\\\
\nx & x & x+a
\n\\end{array}\\right|\\) = 0, a \u2260 0
\nSolution:
\n<\/p>\n
Question 6.
\nProve that \\(\\left|\\begin{array}{ccc}
\na^{2} & b c & a c+c^{2} \\\\
\na^{2}+a b & b^{2} & a c \\\\
\na b & b^{2}+b c & c^{2}
\n\\end{array}\\right|\\) = 4a\u00b2b\u00b2c\u00b2
\nSolution:
\n<\/p>\n
Question 7.
\nIf A-1<\/sup> = \\(=\\left[\\begin{array}{ccc}
\n3 & -1 & 1 \\\\
\n-15 & 6 & -5 \\\\
\n5 & -2 & 2
\n\\end{array}\\right]\\) and B = \\(\\left[\\begin{array}{ccc}
\n1 & 2 & -2 \\\\
\n-1 & 3 & 0 \\\\
\n0 & -2 & 1
\n\\end{array}\\right]\\), find (AB)-1<\/sup>
\nSolution:
\n<\/p>\n<\/p>\n
Question 8.
\nLet A = \\(\\left[\\begin{array}{ccc}
\n1 & -2 & 1 \\\\
\n-2 & 3 & 1 \\\\
\n1 & 1 & 5
\n\\end{array}\\right]\\). Verify that i. [adj A]-1<\/sup> = adj(A-1<\/sup>) ii. (A-1<\/sup>)-1<\/sup> = A
\nSolution:
\n
\nFrom (1) and (2) we get (adj A)-1<\/sup> = adj(A-1<\/sup>).
\n<\/p>\nQuestion 9.
\nEvaluate \\(\\left|\\begin{array}{ccc}
\nx & y & x+y \\\\
\ny & x+y & x \\\\
\nx+y & x & y
\n\\end{array}\\right|\\)
\nSolution:
\n<\/p>\n
<\/p>\n
Question 10.
\nEvaluate \\(\\left|\\begin{array}{ccc}
\n1 & x & y \\\\
\n1 & x+y & y \\\\
\n1 & x & x+y
\n\\end{array}\\right|\\)
\nSolution:
\n<\/p>\n
Question 11.
\n\\(\\left|\\begin{array}{lll}
\n\\alpha & \\alpha^{2} & \\beta+\\gamma \\\\
\n\\beta & \\beta^{2} & \\gamma+\\alpha \\\\
\n\\gamma & \\gamma^{2} & \\alpha+\\beta
\n\\end{array}\\right|\\) = (\u03b1 – \u00df) (\u00df – \u03b3) (\u03b3 – \u03b1) (\u03b1 + \u00df + \u03b3)
\nSolution:
\n
\nTaking out (\u00df – \u03b1) from R2<\/sub> and (\u03b3 – \u03b1) from
\nR3<\/sub>
\nExpanding along C3<\/sub>,
\n= (\u03b1 + \u00df + \u03b3)(\u03b3 – \u03b1)(\u03b3 – \u03b1)(\u03b3 + \u03b1 – \u00df – \u03b1)
\n= (\u03b1 + \u00df + \u03b3 )(\u00df – \u03b1)(\u03b3 – \u03b1)(\u03b3 – \u00df)
\n= (\u00df – \u03b3)(\u03b3 – \u03b1)(\u03b1 – \u00df)(\u03b1 + \u00df + \u03b3)<\/p>\nQuestion 12.
\n\\(\\left|\\begin{array}{lll}
\nx & x^{2} & 1+p x^{3} \\\\
\ny & y^{2} & 1+p y^{3} \\\\
\nz_{1} & z^{2} & 1+p z^{3}
\n\\end{array}\\right|\\) = (1 + pxyz) (x – y) (y – z) (z – x)
\nSolution:
\n
\nExpanding along C1<\/sub>,
\n= (1 + pxyz) (y – x) (z – x) (z + x – y – x)
\n= (1 + pxyz) (y – x) (z – x) (z – y)
\n= (1 + pxyz) (x – y) (y – z) (z – x)<\/p>\nQuestion 13.
\n\\(\\left|\\begin{array}{ccc}
\n3 a & -a+b & -a+c \\\\
\n-b+a & 3 b & -b+c \\\\
\n-c+a & -c+b & 3 c
\n\\end{array}\\right|\\) = 3(a + b + c) (ab + bc + ca)
\nSolution:
\n
\nExpanding along C1<\/sub>,
\n(a + b + c)\\(\\left|\\begin{array}{ll}
\n2 b+a & -b+a \\\\
\n-c+a & 2 c+a
\n\\end{array}\\right|\\)
\n= (a + b + c) [(2b + a) (2c + a) – (a – c) (a – b)]
\n= (a + b + c) [4bc + 2ab + 2ca + a\u00b2 – (a\u00b2 – ab – ac + bc)]
\n= (a + b + c) [3bc + 3ab + 3ac] = 3 (a + b + c) (ab + bc + ca)<\/p>\n<\/p>\n
Question 14.
\n\\(\\left|\\begin{array}{ccc}
\n1 & 1+p & 1+p+q \\\\
\n2 & 3+2 p & 4+3 p+2 q \\\\
\n3 & 6+3 p & 10+6 p+3 q
\n\\end{array}\\right|\\) = 1.
\nSolution:
\n<\/p>\n
Question 15.
\n\\(\\left|\\begin{array}{lll}
\n\\sin \\alpha & \\cos \\alpha & \\cos (\\alpha+\\delta) \\\\
\n\\sin \\beta & \\cos \\beta & \\cos (\\beta+\\delta) \\\\
\n\\sin \\gamma & \\cos \\gamma & \\cos (\\gamma+\\delta)
\n\\end{array}\\right|\\) = 0
\nSolution:
\n<\/p>\n
Question 16.
\nSolve the system of the following equations \\(\\frac { 2 }{ x }\\) + \\(\\frac { 3 }{ y }\\) + \\(\\frac { 10 }{ z }\\) = 4
\n\\(\\frac { 4 }{ x }\\) – \\(\\frac { 6 }{ y }\\) + \\(\\frac { 5 }{ z }\\) = 1
\n\\(\\frac { 6 }{ x }\\) + \\(\\frac { 9 }{ y }\\) – \\(\\frac { 20 }{ z }\\) = 2
\nSolution:
\nLet \\(\\frac { 1 }{ x }\\) = a, \\(\\frac { 1 }{ y }\\) = b and \\(\\frac { 1 }{ z }\\) = c
\n\u2234 The given equations become
\n2a + 3b + 10c = 4
\n4a – 6b + 5c = 1
\n6a + 9b – 20c = 2
\nThe equations can be written in the form AX = B
\nWhere A = \\(\\left[\\begin{array}{ccc}
\n2 & 3 & 10 \\\\
\n4 & -6 & 5 \\\\
\n6 & 9 & -20
\n\\end{array}\\right]\/latex]; X = [latex]\\left[\\begin{array}{l}
\na \\\\
\nb \\\\
\nc
\n\\end{array}\\right]\\); B = \\(\\left[\\begin{array}{l}
\n4 \\\\
\n1 \\\\
\n2
\n\\end{array}\\right]\\)
\n
\n\u2234 a = \\(\\frac { 1 }{ 2 }\\); \\(\\frac { 1 }{ 3 }\\) = b and \\(\\frac { 1 }{ 15 }\\) = c
\nHence x = \\(\\frac { 1 }{ a }\\) = 2; y = \\(\\frac { 1 }{ b }\\) = 3; z = \\(\\frac { 1 }{ c }\\) = 5<\/p>\n
<\/p>\n
Question 17.
\nIf a, b, c are in A.P. then determinant
\n\\(\\left|\\begin{array}{lll}
\nx+2 & x+3 & x+2 a \\\\
\nx+3 & x+4 & x+2 b \\\\
\nx+4 & x+5 & x+2 c
\n\\end{array}\\right|\\) is
\na. 0
\nb. 1
\nc. x
\nd. 2x
\nSolution:
\na. 0
\n<\/p>\n
Question 18.
\nIf x, y, z are nonzero real numbers, then the inverse of matrix A = \\(\\left[\\begin{array}{lll}
\nx & 0 & 0 \\\\
\n0 & y & 0 \\\\
\n0 & 0 & z
\n\\end{array}\\right]\\) is
\na. \\(\\left[\\begin{array}{ccc}
\nx^{-1} & 0 & 0 \\\\
\n0 & y^{-1} & 0 \\\\
\n0 & 0 & z^{-1}
\n\\end{array}\\right]\\)<\/p>\n
b. \\(x y z\\left[\\begin{array}{ccc}
\nx^{-1} & 0 & 0 \\\\
\n0 & y^{-1} & 0 \\\\
\n0 & 0 & z^{-1}
\n\\end{array}\\right]\\)<\/p>\n
c. \\(\\frac{1}{x y z}\\left[\\begin{array}{lll}
\nx & 0 & 0 \\\\
\n0 & y & 0 \\\\
\n0 & 0 & z
\n\\end{array}\\right]\\)<\/p>\n
d. \\(\\frac{1}{x y z}\\left[\\begin{array}{lll}
\n1 & 0 & 0 \\\\
\n0 & 1 & 0 \\\\
\n0 & 0 & 1
\n\\end{array}\\right]\\)
\nSolution:
\n<\/p>\n
<\/p>\n
Question 19.
\nLet A = \\(\\left[\\begin{array}{ccc}
\n1 & \\sin \\theta & 1 \\\\
\n-\\sin \\theta & 1 & \\sin \\theta \\\\
\n-1 & -\\sin \\theta & 1
\n\\end{array}\\right]\\), where 0 \u2264 \u03b8 \u2264 2\u03c0. Then
\na. Det (A) = 0
\nb. Det (A) \u2208 (2, \u221e)
\nc. Det (A) \u2208 (2, 4)
\nd. Det (A) \u2208 [2, 4]
\nSolution:
\nd. Det (A) \u2208 [2, 4]
\n<\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 12 Maths Chapter 4 Determinants Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-4-miscellaneous-exercise\/ NCERT Solutions for Class 12 Maths Chapter 4 Determinants Miscellaneous Exercise Question 1. Prove that the determinant is independent of \u03b8. Solution: Question 2. Without expanding the determinant, prove that Solution: …<\/p>\n
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Miscellaneous Exercise<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 4 Determinants Miscellaneous Exercise - MCQ Questions<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n