Chapter β€’ All Mathematical truths are relative and conditional. β€” C.P. STEINMETZ β€’ 4.1 Introduction In the previous chapter, we have studied about matrices and algebra of matrices. We have also learnt that a system of algebraic equations can be expressed in the form of matrices. This means, a system of linear equations like a1 x + b1 y = c1 a2 x + b2 y = c2 ⎑ab ⎀xc⎑⎀ ⎑⎀ 11 1can be represented as = . Now, this ⎒ βŽ₯⎒βŽ₯ ⎒βŽ₯ ⎣aby c222 ⎦⎣⎦ ⎣⎦ system of equations has a unique solution or not, is determined by the number a1 b2 – a2 b1. (Recall that if a1 b1 P.S. Laplace β‰  or, ab – abβ‰  0, then the system of linear122 1a2 b2 (1749-1827) equations has a unique solution). The number ab – ab122 1 ab⎑11 ⎀which determines uniqueness of solution is associated with the matrix A =⎒βŽ₯ab⎣22 ⎦ and is called the determinant of A or det A. Determinants have wide applications in Engineering, Science, Economics, Social Science, etc. In this chapter, we shall study determinants up to order three only with real entries. Also, we will study various properties of determinants, minors, cofactors and applications of determinants in finding the area of a triangle, adjoint and inverse of a square matrix, consistency and inconsistency of system of linear equations and solution of linear equations in two or three variables using inverse of a matrix. 4.2 Determinant To every square matrix A = [aij] of order n, we can associate a number (real or complex) called determinant of the square matrix A, where aij = (i, j)th element of A. This may be thought of as a function which associates each square matrix with a unique number (real or complex). If M is the set of square matrices, K is the set of numbers (real or complex) and f: M β†’ K is defined by f(A) = k, where A ∈ M and k∈ K, then f(A) is called the determinant of A. It is also denoted by |A| or det A or Ξ”. ab⎑ab⎀If A = ⎒βŽ₯ , then determinant of A is written as |A| = = det (A)cd cd⎣⎦ Remarks (i) For matrix A, |A| is read as determinant of A and not modulus of A. (ii) Only square matrices have determinants. 4.2.1 Determinant of a matrix of order one Let A = [a] be the matrix of order 1, then determinant of A is defined to be equal to a 4.2.2 Determinant of a matrix of order two a a⎑ 11 12 ⎀Let A = ⎒βŽ₯ be a matrix of order 2 Γ— 2,a a⎣ 21 22 ⎦then the determinant of A is defined as: = aa– aadet (A) =|A| = Ξ” = 11222112 24 Example 1 Evaluate .–1 2 24 Solution We have = 2(2) – 4(–1) = 4 + 4 = 8. –1 2xx+ 1Example 2 Evaluate x–1 x Solution We have x x+1 = x(x) – (x+ 1) (x– 1) = x2 – (x2 – 1) = x2 – x2 + 1 = 1 x–1 x 4.2.3 Determinant of a matrix of order 3 Γ— 3 Determinant of a matrix of order three can be determined by expressing it in terms of second order determinants. This is known as expansion of a determinant along a row (or a column). There are six ways of expanding a determinant of order 3 corresponding to each of three rows (R1, R2 and R3) and three columns (C1, C2 and C3) giving the same value as shown below. Consider the determinant of square matrix A = [aij]3 Γ— 3 aaa11 12 13 aaai.e., |A | = 2122 23 aaa3132 33 Expansion along first Row (R1) by (–1)(1 + 1) [(–1)sum of suffixes in a11] and with the Step 1 Multiply first element a11 of R1second order determinant obtained by deleting the elements of first row (R1) and first column (C1) of | A | as a11 lies in R1 and C1, aa22 23(–1)1 + 1 a11i.e., aa32 33 by (–1)1 + 2 [(–1)sum of suffixes in a12] and the secondStep 2 Multiply 2nd element a12 of R1order determinant obtained by deleting elements of first row (R1) and 2nd column (C2) of | A | as a12 lies in R1 and C2, aa21 23(–1)1 + 2 a12i.e., aa31 33 by (–1)1 + 3 [(–1)sum of suffixes in aStep 3 Multiply third element a13 of R113] and the second order determinant obtained by deleting elements of first row (R1) and third column (C3) of | A | as a13 lies in R1 and C3, aa21 22 (–1)1 + 3 a13i.e., aa31 32 Step 4 Now the expansion of determinant of A, that is, | A | written as sum of all three terms obtained in steps 1, 2 and 3 above is given by aa22 23 det A=|A| = (–1)1 + 1 a11 aa32 33 aa21 22++ (–1) 13 a13 aa31 32 ++(–1) 12 a12 aa21 23 aa31 33 or |A| = a (aa – aa) – a (aa – aa)1122 3332 231221 3331 23(aa – aa)+ a1321323122= aaa – aaa – aaa + aaa + aaa1122331132 231221331231 23132132 – aaa... (1)1331 22 Expansion along second row (R2) aaa1112 13 aaa| A | = 2122 23 aaa3132 33 Expanding along R2,we get aa12 13+| A | = (–1) 21 a21 aa32 33 2 + 2+ (–1) a22 aa11 13 aa31 33 aa11 12 ++(–1) 23 a23 aa31 32 =– – ) + a22 – )a21 (a12 a33a32 a13 (a11 a33a31 a13– a (aa – aa)2311323112| A| =– a21 aa + aaa + aaa – aaa – aaa12332132 132211332231 13231132 + a23 a31 a12 = aaa – aaa – aaa + aaa + aaa112233112332122133122331132132 – aaa... (2)1331 22 Expansion along first Column (C1) a aa11 12 13 a aa| A | = 21 22 23 a aa31 32 33 By expanding along C1, we get aa22 23 a (–1) 11| A | = 11 + aa32 33 2 + 1+ a21 ( 1) βˆ’ aa12 13 aa32 33 aa12 13 +a (–1) 31 + 31 aa22 23 = a (aa – aa) – a (aa – aa) + a (aa – aa)112233233221123313323112231322| A| = aaa – aaa – aaa + aaa + aaa11 2233112332211233211332311223 – a31 a13 a22 – – + a12 + a13 = a11 a22 a33a11 a23 a32a12 a21 a33a23 a31a21 a32 – ... (3)a13 a31 a22 Clearly, values of |A| in (1), (2) and (3) are equal. It is left as an exercise to the reader to verify that the values of |A| by expanding along R3, C2 and C3 are equal to the value of |A| obtained in (1), (2) or (3). Hence, expanding a determinant along any row or column gives same value. Remarks (i) For easier calculations, we shall expand the determinant along that row or column which contains maximum number of zeros. (ii) While expanding, instead of multiplying by (–1)i + j, we can multiply by +1 or –1 according as (i + j) is even or odd. ⎑ 22 ⎀⎑ 11 ⎀(iii) Let A = ⎒βŽ₯ and B = ⎒βŽ₯ . Then, it is easy to verify that A = 2B. Also40 20⎣⎦⎣⎦|A| = 0 – 8 = – 8 and |B| = 0 – 2 = – 2. Observe that, |A| = 4(– 2) = 22|B| or |A| = 2n |B|, where n = 2 is the order of square matrices A and B. In general, if A = kB where A and B are square matrices of order n, then | A| = kn | B |, where n = 1, 2, 3 12 4 Example 3 Evaluate the determinant Ξ” = –1 3 0 . 41 0 Solution Note that in the third column, two entries are zero. So expanding along third column (C3), we get –1 3 1 2 1 2–0 + 0Ξ” = 4 41 41 –1 3 = 4 (–1 – 12) – 0 + 0 = – 52 0 sin Ξ± – cos Ξ± Example 4 Evaluate Ξ” = –sin Ξ± 0 sin Ξ² . cos Ξ± – sin Ξ² 0 Solution Expanding along R1, we get Ξ” = 0 sin – sin sin – sin 0 – sin – cos –sin 0 cos 0 cos Ξ² Ξ± Ξ² Ξ±Ξ± Ξ±Ξ² Ξ± Ξ± = 0 – sin Ξ± (0 – sin Ξ² cos Ξ±) – cos Ξ± (sin Ξ± sin Ξ² – 0) = sin Ξ± sin Ξ² cos Ξ± – cos Ξ± sin Ξ± sin Ξ² = 0 0 –sin Ξ² Example 5 Find values of x for which 3 x 1 x = 3 4 2 1 . Solution We have i.e. i.e. 3 x 1 x = 3 2 4 1 3 – x2 = 3 – 8 x2 =8 Hence x = Β± 22 Evaluate the determinants in Exercises 1 and 2. 24 1. –5 –1 cos ΞΈ – sin ΞΈ x2– x + 1 x –1 2. (i) (ii)sin ΞΈ cos ΞΈ x + 1 x + 1 ⎑12⎀3. If A = ⎒βŽ₯ , then show that | 2A | = 4 | A | 42⎣⎦ ⎑ 101 ⎀ ⎒βŽ₯4. If A = 01 2 , then show that | 3 A | = 27 | A | ⎒βŽ₯ ⎒ 004 βŽ₯⎣⎦ 5. Evaluate the determinants 2–1 –2 3–1 –2 3– 4 5 (i) 0 0 –1 (ii) 1 1 –2 3–5 0 2 3 1 012 0 2–1 –23 0 –10 –3 (iii) (iv) 3–5 0 ⎑ 11 –2 ⎀ 6. If A = ⎒ 21 –3 βŽ₯ , find | A | ⎒βŽ₯ ⎒ 54 –9 βŽ₯⎣⎦ 7. Find values of x, if 24 2x4 23 x3 = =(ii)(i) 51 6 x 45 2x5 x2 628. If , then xis equal to= 18 x 18 6 (A)6 (B)Β± 6 (C)–6 (D)0 4.3 Properties of Determinants In the previous section, we have learnt how to expand the determinants. In this section, we will study some properties of determinants which simplifies its evaluation by obtaining maximum number of zeros in a row or a column. These properties are true for determinants of any order. However, we shall restrict ourselves upto determinants of order 3 only. Property 1 The value of the determinant remains unchanged if its rows and columns are interchanged. aaa 123 bb b Verification Let Ξ” = 123 cc c 123 Expanding along first row, we get bb 23Ξ” = a1 cc 23 βˆ’ a2 bb 13 cc 13 = a(bc– bc) – a(b12 3322+ a3 bb12 cc 12 c– bc) + a(bc– bc)133131221By interchanging the rows and columns of Ξ”, we get the determinant abc 111 abc Ξ”1 = 222 abc 333 Expanding Ξ”1 along first column, we get = a1 (b2 – ) – a2 (b1 c3 – b3 ) + a3 (b1 – )Ξ”1 c3c2 b3c1c2b2 c1Hence Ξ” = Ξ”1 Remark It follows from above property that if A is a square matrix, then det (A) = det (Aβ€²), where Aβ€² = transpose of A. 2–3 5 Example 6 Verify Property 1 for Ξ” = 60 4 15 –7 Solution Expanding the determinant along first row, we have 04 64 602 – (–3) + 5Ξ” = 5–7 1 –7 15 = 2 (0 – 20) + 3 (– 42 – 4) + 5 (30 – 0) = – 40 – 138 + 150 = – 28 By interchanging rows and columns, we get 26 1 Ξ”1 = –3 0 5 (Expanding along first column) 54 –7 = 2 0 5 4–7 – (–3) 6 1 4–7 + 5 6 0 1 5 = 2 (0 – 20) + 3 (– 42 – 4) + 5 (30 – 0) = – 40 – 138 + 150 = – 28 Clearly Ξ” = Ξ”1 Hence, Property 1 is verified. Property 2 If any two rows (or columns) of a determinant are interchanged, then signof determinant changes. aaa123 bb bVerification Let Ξ” = 123 cc c123 Expanding along first row, we get Ξ” = a1 (bc – bc) – a(bc – bc) + a (bc – bc)23322 133131221Interchanging first and third rows, the new determinant obtained is given by cc c123 bb bΞ”1 = 123 aaa123 Expanding along third row, we get = a (cb – bc) – a (cb – cb) + a (bc – bc)Ξ”1 123232133132112=– [a1 (bc – bc) – a (bc – bc) + a (bc – bc)]23322133131221Clearly Ξ”1 =– Ξ” Similarly, we can verify the result by interchanging any two columns. Example 7 Verify Property 2 for Ξ” = 2–3 6 0 1 5 5 4 –7 . 2–3 5 6 04Solution Ξ” = = – 28 (See Example 6) 1 5–7Interchanging rows R2 and R3 i.e., R2 ↔ R3, we have 2–3 5 1 5–7 Ξ”1 = 604 Expanding the determinant Ξ”1 along first row, we have 5–7 1 –7 15 – (–3) + 5Ξ”1 = 2 04 64 60 = 2 (20 – 0) + 3 (4 + 42) + 5 (0 – 30) = 40 + 138 – 150 = 28 Clearly Ξ”1 =– Ξ” Hence, Property 2 is verified. Property 3 If any two rows (or columns) of a determinant are identical (all corresponding elements are same), then value of determinant is zero. Proof If we interchange the identical rows (or columns) of the determinant Ξ”, then Ξ” does not change. However, by Property 2, it follows that Ξ” has changed its sign Therefore Ξ” =– Ξ” or Ξ” =0 Let us verify the above property by an example. 32 3 Example 8 Evaluate Ξ” = 22 3 32 3 Solution Expanding along first row, we get Ξ” = 3 (6 – 6) – 2 (6 – 9) + 3 (4 – 6) = 0 – 2 (–3) + 3 (–2) = 6 – 6 = 0 Here R1 and R3 are identical. Property 4 If each element of a row (or a column) of a determinant is multiplied by a constant k, then its value gets multiplied by k. abc111 abcVerification Let Ξ” = 222 abc333 and Ξ”1 be the determinant obtained by multiplying the elements of the first row by k. Then ka kb kc 111 abcΞ”1 = 222 abc333 Expanding along first row, we get = k a1 – ) – k b1 (a2 – ) + k c1 (a2 b3 – )Ξ”1(b2 c3b3 c2c3c2 a3b2 a3= k [a (bc – bc) – b (ac – ca) + c (ab – ba)]123321232312323= k Ξ” 1ka 1kb 1kc 1a 1b 1c Hence 2a 2b 2c = k 2a 2b 2c 3a 3b 3c 3a 3b 3c Remarks (i) By this property, we can take out any common factor from any one row or any one column of a given determinant. (ii) If corresponding elements of any two rows (or columns) of a determinant are proportional (in the same ratio), then its value is zero. For example aa a1 23 bb bΞ” = = 0 (rows R1 and R2 are proportional)123 kaka ka123 102 18 36 1 34 17 3 6 Example 9 Evaluate 102 18 36 6(17)6(3)6(6) 173 6 Solution Note that 1 3 4 = 1 3 4 6= 1 3 4 0= 17 3 6 17 3 6 17 3 6 (Using Properties 3 and 4) Property 5 If some or all elements of a row or column of a determinant are expressed as sum of two (or more) terms, then the determinant can be expressed as sum of two (or more) determinants. 1 1a +Ξ» 2 2a +Ξ» 3 3a +Ξ» 1a a 2 3a 1Ξ» 2Ξ» 3Ξ» For example, 1b 2b 3b = 1b b 2 3b + 1b 2b 3b 1c 2c 3c 1c c 2 3c 1c 2c 3c a +Ξ» a +Ξ» a +Ξ»1 12 233 bb bVerification L.H.S. = 12 3 cc c123 Expanding the determinants along the first row, we get Ξ” =(a+ Ξ»1) (bc– cb) – (a+ Ξ»2) (bc– bc)1232321331+ (a+ Ξ») (bc– bc)331221= a(bc– cb) – a(bc– bc) + a(bc– bc)123232133131221+ Ξ» (bc– cb) – Ξ» (bc– bc) + Ξ» (bc– bc)123232133131221(by rearranging terms)aa a1 23 bb b = 123 cc c123 +λλ Ξ»123 bb b = R.H.S.123 cc c123 Similarly, we may verify Property 5 for other rows or columns. abc a+ 2xb + 2yc + 2z= 0Example 10 Show that xyz a b c abc a b c Solution We have 2 2 2a xb yc z+ + + = abc + 2x 2y 2z x y z xyz x y z (by Property 5) = 0 + 0 = 0 (Using Property 3 and Property 4) Property 6 If, to each element of any row or column of a determinant, the equimultiples of corresponding elements of other row (or column) are added, then value of determinant remains the same, i.e., the value of determinant remain same if we apply the operation Ri β†’ Ri+ kRjor Ciβ†’ Ci+ kCj. Verification aa a 123 bbb Let Ξ” = 123 cc c123 and Ξ”1 = a+ kc a + kc a + kc112 233 bb b ,12 3 cc c12 3 where Ξ”1 is obtained by the operation R1 β†’ R1 + kR3 . Here, we have multiplied the elements of the third row (R3) by a constant kand added them to the corresponding elements of the first row (R1). Symbolically, we write this operation as R1 β†’ R1 + kR3. Now, again aaa 1 23 bbb Ξ”1 = 1 23 cc c 1 23 +kckc kc1 23 b b b (Using Property 5)1 23 c c c1 23 = Ξ” + 0 (since R1 and R3 are proportional) Hence Ξ” = Ξ”1 Remarks (i) If Ξ”1 is the determinant obtained by applying Riβ†’ kRior Ciβ†’ kCito the determinant Ξ”, then Ξ”1 = kΞ”. (ii) If more than one operation like Riβ†’ Ri+ kRjis done in one step, care should be taken to see that a row that is affected in one operation should not be used in another operation. A similar remark applies to column operations. a ab + ++abc 32 3 + 2b 4a+ 3b+ 2caa = a. 3a6a+ 3b10 a+ 6b+ 3c Example 11 Prove that Solution Applying operations R2 β†’ R2 – 2R1 and R3 β†’ R3 – 3R1 to the given determinant Ξ”, we have aa ba b c + ++ 0 a 2ab +Ξ” = 0 3a 7a+ 3b Now applying R3 β†’ R3 – 3R2 , we get aa ba b c + ++ 0 a 2ab +Ξ” = 00 a Expanding along C1, we obtain a 2 +ab Ξ” = a + 0 + 0 0 a = a(a2 – 0) = a(a2) = a3 Example 12 Without expanding, prove that xyyzzx + ++ zxy =0 111 Ξ” = Solution Applying R1 β†’ R1 + R2 to Ξ”, we get x yzx yzxyz ++++ ++ zxyΞ” = 111 Since the elements of R1 and R3 are proportional, Ξ” = 0. Example 13 Evaluate 1 abc 1 bca 1 cab Ξ” = Solution Applying R2 β†’ R2 – R1 and R3 β†’ R3 – R1, we get 1 a bc ab )0 bacβˆ’ ( βˆ’Ξ” = ac )0 cabβˆ’ ( βˆ’ Taking factors (b– a) and (c– a) common from R2 and R3, respectively, we get 1 abc Ξ” =(baca )( βˆ’βˆ’ ) 01– c 01– b =(b– a) (c– a) [(– b+ c)] (Expanding along first column) =(a– b) (b– c) (c– a) bc+ aa b ca + b =4abc cc + Example 14 Prove that ab bc + aa b ca + b cc + Solution Let Ξ” = ab Applying Rβ†’ R – R – R to Ξ”, we get1 1230 –2 c –2 b bc + ab cca + b Ξ” = Expanding along R1, we obtain ca b bb bc + a+ – (–2 ) c + (–2 ) bΞ” =0 ca + b ca + b cc =2 c (a b + b2 – bc) – 2 b (b c – c2 – ac) =2 a b c + 2 cb2 – 2 bc2 – 2 b2c + 2 bc2 + 2 abc =4 abc xx21+x3 Example 15 If x, y, z are different and Ξ”= =0 , then zz21+z3 yy21+ y3 show that 1 + xyz = 0 Solution We have xx21+x3 yy21+ y3Ξ” = zz21+z3 23xx21 xx x yy21 +yy2 y3 = (Using Property 5) 23zz21 zz z 1 xx2 1 xx2 βˆ’ 2 =(1) 1 yy2 +xyz 1 yy2 (Using C3 ↔ C2 and then C1 ↔ C2) 1 zz2 1 zz2 1 xx2 1 yy2 (1 +xyz)= 1 zz2 1 xx2 = (1+xyz) (Using Rβ†’ R–R and Rβ†’ R–R)0 y βˆ’xy2 βˆ’x2 2213310 zxzβˆ’ 2 βˆ’x2 Taking out common factor (y – x) from R2 and (z – x) from R3, we get 1 xx2 Ξ” = (1+ xyz) ( y–x) ( z–x) 01 y +x 01 zx+ = (1 + xyz) (y – x) (z – x) (z – y) (on expanding along C1) Since Ξ” = 0 and x, y, z are all different, i.e., x – y β‰  0, y – z β‰  0, z – x β‰  0, we get 1 + xyz = 0 Example 16 Show that 1+a 11 βŽ› 111 ⎞11+b 1 abc 1+++ =abcbcca ab = +++ ⎜⎟⎝ abc ⎠1 11+c Solution Taking out factors a,b,c common from R1, R2 and R3, we get 1 11+1 a aa 11 1abc +1L.H.S. = bb b 1 11 +1 c cc Applying R1β†’ R1 + R2 + R3, we have 111 111 1111+++ 1+++ 1+++ abc abc abc 11 1+1Ξ” = abc bb b 1 11 +1 c cc βŽ› 111 ⎞abc1+ ++= ⎜⎟⎝ abc⎠ 11 1 11 1+1bb b 111 +1 ccc Now applying C2 β†’ C2 – C1, C3 β†’ C3 – C1, we get βŽ› 111 βŽžΞ” = abc⎜1+ + + ⎟⎝ abc ⎠ βŽ› 111 ⎞100 1 10b 1 01 c abc1+++ ⎑ 1 1 –0 ⎀= ⎜ ⎟⎣( ) ⎦⎝ abc⎠ βŽ› 111 ⎞ = abc⎜1+ ++⎟= abc+ bc+ ca+ ab= R.H.S. ⎝ abc ⎠Using the property of determinants and without expanding in Exercises 1 to 7, prove that: xaxa + +=0yb yb zc zc 1. + 27 65 3. 38 75 59 86 =0 bcqr yz +++ car p zx ++ +5. ab pqxy + ++ =2 2. 4. apx bqy cr z abbcca βˆ’βˆ’βˆ’ bccaab βˆ’βˆ’βˆ’=0 caabbc βˆ’βˆ’βˆ’ 1 (bc )bca+ 1 () ca cab+ =0 1 () ab abc+ βˆ’a2 ab ac 0 a βˆ’b 2 22ba βˆ’b2 bc =4abc βˆ’a 0 βˆ’c =06. 7. bc 0 ca cb βˆ’c2 By using properties of determinants, in Exercises 8 to 14, show that: 1 aa2 ()ab ()bβˆ’cc=βˆ’ βˆ’a() 8. (i) 1 bb2 1 cc2 111 abc ab βˆ’c βˆ’ab=βˆ’ bc a++ c()()()( )(ii) 333abc xx2 yz yy2 zx 9. = (x – y) (y – z) (z – x) (xy + yz + zx) zz2 xy x+42x 2x 2x x+4 2x =(5x+4)(4 βˆ’x)2 10. (i) 2x 2x x+ 4 y+ky y y y+k y =k2 (3y+k)(ii) y y y+k ab aβˆ’βˆ’c 2a 2 2 bca 2b βˆ’βˆ’ b = ++ (abc)311. (i) 2c 2ccβˆ’βˆ’ab xy++2zx y z yz 2xy++ 2(x++ z)3 = y(ii) z xzx 2++y 1 x x2 x21 x =βˆ’(1 x3 )2 12. xx2 1 1+βˆ’ a2 b22ab βˆ’2b 13. a22 2a2ab 1βˆ’+ b 2b βˆ’2a 1 a22βˆ’βˆ’ b a2 +1 ab ac 2 2 2 2 2=+ + b(1 a )3 2ab b +1 bc =1 a bc+++ 14. ca cbc2 +1 Choose the correct answer in Exercises 15 and 16. 15. Let A be a square matrix of order 3 Γ— 3, then | kA| is equal to (A) k|A| (B) k2|A| (C) k3|A| (D) 3k |A| 16. Which of the following is correct (A) Determinant is a square matrix. (B) Determinant is a number associated to a matrix. (C) Determinant is a number associated to a square matrix. (D) None of these 4.4 Area of a Triangle In earlier classes, we have studied that the area of a triangle whose vertices are 1(x1, y1), (x2, y2) and (x3, y3), is given by the expression [x1(y2–y3) + x2 (y3–y1) +2 x3 (y1–y2)]. Now this expression can be written in the form of a determinant as x1 y1 1 1 x2 y2 1Ξ” = ... (1)2 x3 y3 1 Remarks (i) Since area is a positive quantity, we always take the absolute value of the determinant in (1). (ii) If area is given, use both positive and negative values of the determinant forcalculation. (iii) The area of the triangle formed by three collinear points is zero. Example 17 Find the area of the triangle whose vertices are (3, 8), (– 4, 2) and (5, 1). Solution The area of triangle is given by 38 1 Ξ” = 1 2 42 51 – 1 1 1 ⎑32–1 –8 –4–5 +1 –4– 10 ⎀= ⎣()( )( )⎦2 =1 (372 –14 ) 61 += 2 2 Example 18 Find the equation of the line joining A(1, 3) and B (0, 0) using determinants and find k if D(k, 0) is a point such that area of triangle ABD is 3sq units. Solution Let P (x, y) be any point on AB. Then, area of triangle ABP is zero (Why?). So 001 1 13 1 =02 xy1 1This gives 2 (y– 3 x) =0 or y = 3x, which is the equation of required line AB. Also, since the area of the triangle ABD is 3 sq. units, we have 131 1 00 1 =Β±3 2 k01 βˆ’3kThis gives, =Β±3 , i.e., k = m 2. 2 1. Find area of the triangle with vertices at the point given in each of the following : (i) (1, 0), (6, 0), (4, 3) (ii) (2, 7), (1, 1), (10, 8) (iii) (–2, –3), (3, 2), (–1, –8) 2. Show that points A (a, b + c), B (b, c + a), C (c, a + b) are collinear. 3. Find values of k if area of triangle is 4 sq. units and vertices are (i) (k, 0), (4, 0), (0, 2) (ii) (–2, 0), (0, 4), (0, k) 4. (i) Find equation of line joining (1, 2) and (3, 6) using determinants. (ii) Find equation of line joining (3, 1) and (9, 3) using determinants. 5. If area of triangle is 35 sq units with vertices (2, – 6), (5, 4) and (k, 4). Then k is (A) 12 (B) –2 (C) –12, –2 (D) 12, –2 4.5 Minors and Cofactors In this section, we will learn to write the expansion of a determinant in compact form using minors and cofactors. Definition 1 Minor of an element aij of a determinant is the determinant obtained by deleting its ith row and jth column in which element aij lies. Minor of an element aij is denoted by Mij. Remark Minor of an element of a determinant of order n(n β‰₯ 2) is a determinant of order n – 1. 123 Example 19 Find the minor of element 6 in the determinant Ξ”= 456 789 Solution Since 6 lies in the second row and third column, its minor M23 is given by 12 M23 = = 8 – 14 = – 6 (obtained by deleting R2 and C3 in Ξ”).78Definition 2 Cofactor of an element aij , denoted by Aij is defined by A = (–1)i + j M, where M is minor of aijijijij. 1–2 Example 20 Find minors and cofactors of all the elements of the determinant 4 3 Solution Minor of the element a is Mijij Here a11 = 1. So M11 = Minor of a11= 3 = Minor of the element a12 = 4M12 = Minor of the element a21 = –2M21M22 = Minor of the element a22 = 1 Now, cofactor of aij is Aij. So = (–1)1 + 1 M11 = (–1)2 (3) = 3A11= (–1)1 + 2 M12 = (–1)3 (4) = – 4A12= (–1)2 + 1 M21 = (–1)3 (–2) = 2A21A22 = (–1)2 + 2 M22 = (–1)4 (1) = 1 Example 21 Find minors and cofactors of the elements a11, a21 in the determinant aaa1112 13 aaaΞ” = 2122 23 aaa3132 33 Solution By definition of minors and cofactors, we have aa22 23 = aa– aaMinor of a11 = M11 = 2233 2332aa32 33 Cofactor of a11 = A11 = (–1)1+1 M11 = –a22 a33 a23 a32 aa12 13Minor of a21 = M21 = = a12 a33 – a13 a32aa32 33 Cofactor of a21 = A21 = (–1)2+1 M21 = (–1) (a12 ) = – a12a33 – a13 a32a33 + a13 a32 Remark Expanding the determinant Ξ”, in Example 21, along R1, we have aa22 23 Ξ” = (–1)1+1 a11 aa32 33 = aA + aA + a+ (–1)1+2 a12 aa21 23 aa31 33 + (–1)1+3 a13 111112 1213 13ijijA, where A is cofactor of aaa21 22 aa31 32 = sum of product of elements of R1 with their corresponding cofactors Similarly, Ξ” can be calculated by other five ways of expansion that is along R2, R3, C1, C2 and C3. Hence Ξ” = sum of the product of elements of any row (or column) with their corresponding cofactors. Ξ” = a11A21 + a12 A22 + a13 A23 aa aa aa12 13 11 13 11 12 = (–1)1+1 + a12 (–1)1+2 + a13 (–1)1+3a11aa aa aa32 33 31 33 31 32 11a 12a 13 a = 11a 12a 13 a = 0 (since R1 and R2 are identical) 31a 32a 33 a Similarly, we can try for other rows and columns. Example 22 Find minors and cofactors of the elements of the determinant 2 3– 5 60 4 15 7– and verify that a11= 0A31 + a12 A32 + a13 A33Solution We have M11 = M12 = 6 1 4 –7 M13 = 6 1 0 5 –35 M21 = 5 –7 M22 = 2 1 5 –7 M23 = 2 1 3 5 – –35 M31 = 04 04 = 0 –20 = –20; A11 = (–1)1+1 (–20) = –205 –7= – 42 – 4 = – 46; = 30 – 0 = 30; = 21 – 25 = – 4; = –14 – 5 = –19; = 10 + 3 = 13; = –12 – 0 = –12; A12A13A21A22A23A31= (–1)1+2 (– 46) = 46 = (–1)1+3 (30) = 30 = (–1)2+1 (– 4) = 4 = (–1)2+2 (–19) = –19 = (–1)2+3 (13) = –13 = (–1)3+1 (–12) = –12 25 M32 = = 8 – 30 = –22; = (–1)3+2 (–22) = 22A32642 –3 and = = 0 + 18 = 18; = (–1)3+3 (18) = 18M33A3360 Now a11 = 2, a12 = –3, a13 = 5; A31 = –12, A32 = 22, A33 = 18 So a11 A31 + a12 A32 + a13 A33 = 2 (–12) + (–3) (22) + 5 (18) = –24 – 66 + 90 = 0 Write Minors and Cofactors of the elements of following determinants: 2 –4 ac1. (i) (ii)03 bd 100 10 4 010 35 –12. (i) (ii) 001 01 2 538 2013. Using Cofactors of elements of second row, evaluate Ξ” = . 123 1 xyz 1 yzx4. Using Cofactors of elements of third column, evaluate Ξ” = . 1 zxy aaa1112 13 aaa5. If Ξ” = and Aijis Cofactors of aij, then value of Ξ” is given by2122 23 aaa3132 33 (A) aA+ aA + aA(B) aA+ aA + aA113112 3213 33 111112 2113 31 (C) a+ a + a(D) a+ a + a21 A1122 A1223 A13 11 A1121 A2131 A31 4.6 Adjoint and Inverse of a Matrix In the previous chapter, we have studied inverse of a matrix. In this section, we shall discuss the condition for existence of inverse of a matrix. To find inverse of a matrix A, i.e., A–1 we shall first define adjoint of a matrix. 4.6.1 Adjoint of a matrix Definition 3 The adjoint of a square matrix A = [aij] is defined as the transpose ofn Γ— nthe matrix [A], where A is the cofactor of the element a. Adjoint of the matrix A ijn Γ— nijijis denoted by adj A. ⎑aaa ⎀1112 13⎒βŽ₯Let A = aaa2122 23 ⎒βŽ₯ ⎒aaa βŽ₯⎣ 3132 33 ⎦ AAA AAA⎑ 1112 13 ⎀⎑ 11 21 31 ⎀ ⎒ βŽ₯⎒βŽ₯Then adj A=Transposeof A21 A22 A23 =A12 A22 A32⎒ βŽ₯⎒βŽ₯ ⎒AAA βŽ₯⎒AAA βŽ₯⎣ 31 32 33 ⎦⎣ 13 23 33 ⎦ ⎑23⎀Example 23 Findadj Afor A = ⎒βŽ₯14⎣⎦ Solution We have A11 = 4, A12 = –1, A21 = –3, A22 = 2 A A 4–3⎑ 11 21 ⎀⎑ ⎀ =Hence adj A = ⎒ βŽ₯⎒βŽ₯A A –12⎣ 12 22 ⎦⎣ ⎦ Remark For a square matrix of order 2, given by ⎑aa ⎀11 12 A = ⎒βŽ₯aa⎣ 21 22 ⎦ The adj A can also be obtained by interchanging a11 and a22 and by changing signs of a12 and a21, i.e., We state the following theorem without proof. Theorem 1 If A be any given square matrix of order n, then A(adj A) =(adj A) A = AI, where I is the identity matrix of order n Verification aaa⎑ 11 12 13 ⎀⎑A11 A21 A31 ⎀ ⎒βŽ₯ ⎒βŽ₯Let A = ⎒a21 a22 a23 βŽ₯ , then adj A = ⎒A12 A22 A32 βŽ₯ ⎒⎣a31 a32 a33 ⎦βŽ₯ ⎒⎣A13 A23 A33 βŽ₯⎦ Since sum of product of elements of a row (or a column) with corresponding cofactors is equal to |A| and otherwise zero, we have ⎑ A 00 ⎀⎑100⎀ βŽ₯ ⎒βŽ₯A 010 =A AA (adj A) = ⎒⎒ 0 0 = IβŽ₯ ⎒βŽ₯ ⎒ 00 A βŽ₯⎒001βŽ₯⎦⎣ ⎦⎣ ASimilarly, we can show (adj A) A = I A IHence A (adj A) = (adj A) A = ADefinition 4 A square matrix A is said to be singular if = 0. 12 For example, the determinant of matrix A = is zero48Hence A is a singular matrix. Definition 5 A square matrix A is said to be non-singular if A β‰  0 12 Let A = ⎒βŽ₯ . Then⎑12⎀ A = = 4 – 6 = – 2 β‰  0.34 34⎣⎦ Hence A is a nonsingular matrix We state the following theorems without proof. Theorem 2 If A and B are nonsingular matrices of the same order, then AB and BA are also nonsingular matrices of the same order. Theorem 3 The determinant of the product of matrices is equal to product of their respective determinants, that is, AB = A B , where A and B are square matrices of the same order A 00 A 0 00 Remark We know that (adj A) A = AI = 0 A Writing determinants of matrices on both sides, we have A 00 (adj A)A0 A 0 00 = A 100 3A010i.e. |(adj A)| |A| = (Why?) 001 i.e. |(adj A)| |A| = |A|3 (1) i.e. |(adj A)| = |A|2 In general, if A is a square matrix of order n, then | adj (A)| = |A|n – 1. Theorem 4 A square matrix A is invertible if and only if A is nonsingular matrix. Proof Let A be invertible matrix of order n and I be the identity matrix of order n. Then, there exists a square matrix B of order n such that AB = BA = I Now AB = I. So AB = Ior A B = 1 (since I =1, AB = A B ) This gives A β‰  0. Hence A is nonsingular. A β‰  0Conversely, let A be nonsingular. Then Now A (adj A) = (adj A) A = A I (Theorem 1) βŽ› 1 βŽžβŽ› 1 ⎞ or A adj A = adj AA =I⎜⎟⎜ ⎟⎝ |A| ⎠⎝ |A| ⎠ 1 or AB = BA = I, where B = adj A |A| 1 Thus A is invertible and A–1 = adj A |A| 133 Example 24 If A = 14 3 , then verify that A adj A = |A| I. Also find A–1. 134 Solution We have A= 1(16 – 9) –3 (4 – 3) + 3 (3 – 4) = 1 β‰  0 Now A = 7, A = –1, A = –1, A = –3, A = 1,A = 0, A = –3, A = 0, A33 = 1 11 12 13 21 22 23 31 32 7⎑ 3βˆ’ 3βˆ’ ⎀ Therefore adj A = 1⎒ βŽ’βˆ’ 1 0 βŽ₯ βŽ₯ 1βŽ’βŽ£βˆ’ 0 1 βŽ₯⎦ ⎑133⎀⎑ 7 βˆ’3 βˆ’3⎀ ⎒βŽ₯⎒ βŽ₯Now A (adj A) = 14 3 βˆ’11 0⎒βŽ₯⎒ βŽ₯ ⎒134βŽ₯βŽ’βˆ’10 1βŽ₯⎣⎦⎣ ⎦ 7 βˆ’βˆ’33 330 βˆ’+ + 303⎀⎑ βˆ’++ ⎒βŽ₯ =743 340 βˆ’++ 303βˆ’βˆ’ βˆ’++ ⎒βŽ₯ ⎒734 βˆ’+ + 30 βˆ’+ + 4βŽ₯βˆ’βˆ’ 3 30⎣⎦ ⎑100⎀⎑100⎀ ⎒βŽ₯⎒βŽ₯ =010 = (1) 010 = A . I⎒βŽ₯⎒βŽ₯ ⎒001βŽ₯⎒001βŽ₯⎣⎦⎣⎦⎑ 7 βˆ’3 βˆ’3⎀⎑ 7 βˆ’3 βˆ’3⎀ 11 ⎒βŽ₯A–1Also adjA = βŽ’βŽ’βˆ’11 0βŽ₯βŽ₯ = βŽ’βˆ’11 0βŽ₯A 1 ⎒ 10 1βŽ₯ βŽ’βŽ£βˆ’10 1βŽ₯βŽ£βˆ’βŽ¦βŽ¦ ⎑23 ⎀⎑1 βˆ’2⎀Example 25 If A = ⎒βŽ₯and B =⎒ βŽ₯, then verify that (AB)–1 = B–1A–1.1 βˆ’4 βˆ’13⎣⎦ ⎣⎦⎑23⎀⎑1 βˆ’2⎀ ⎑ βˆ’15 ⎀Solution We have AB = =⎒βŽ₯⎒βŽ₯⎒ βŽ₯1 βˆ’4 βˆ’13 5 βˆ’14 ⎣⎦⎣⎦⎣ ⎦ ABSince, = –11 β‰  0, (AB)–1 exists and is given by 11 βŽ‘βˆ’14 βˆ’5⎀ 1 ⎑14 5⎀ =(AB)–1 = adj (AB) =βˆ’βŽ’ βŽ₯⎒βŽ₯11 βŽ£βˆ’5 βˆ’1⎦ 11 ⎣51⎦AB Further, A = –11 β‰  0 and B = 1 β‰  0. Therefore, A–1 and B–1 both exist and are given by 43 3211A–1,B= 1112 11 1 1 132 43 1145 1 ⎑14 5⎀Therefore BA = ⎒βŽ₯1111 12 115 1 1151⎣⎦ Hence (AB)–1 = B–1 A–1 23 Example 26 Show that the matrix A = satisfies the equation A2 – 4A + I = O, 12where I is 2 Γ— 2 identity matrix and O is 2 Γ— 2 zero matrix. Using this equation, find A–1. 2 ⎑23⎀⎑23⎀⎑7 12 ⎀Solution We have A =A.A ==⎒βŽ₯⎒βŽ₯⎒ βŽ₯1212 47⎣⎦⎣⎦⎣ ⎦ 2 ⎑712 ⎀⎑8 12 ⎀⎑10⎀⎑00⎀Hence A βˆ’4A += I βˆ’ +==O⎒ βŽ₯⎒ βŽ₯⎒βŽ₯⎒βŽ₯47 480100⎣ ⎦⎣ ⎦⎣⎦⎣⎦ Now A2 – 4A + I = O Therefore AA – 4A = – I or A A (A–1) – 4 A A–1 = – I A–1 (Post multiplying by A–1 because |A| β‰  0) or A (AA–1) – 4I = – A–1 or AI – 4I = – A–1 ⎑40⎀⎑23⎀⎑2 βˆ’3⎀ A–1βˆ’= or = 4I – A = ⎒βŽ₯⎒βŽ₯⎒ βŽ₯04 12 βˆ’12⎣⎦⎣⎦⎣ ⎦ ⎑2 βˆ’3⎀Aβˆ’1Hence =⎒ βŽ₯βŽ£βˆ’12 ⎦ Find adjoint of each of the matrices in Exercises 1 and 2. 1 12 121. 2. 2 35 34 201 Verify A (adj A) = (adj A) A = |A| I in Exercises 3 and 4 1 12 23 30 23. 4.46 10 3 Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11. 123 22 15 0245. 6.7.43 32 005 100 213 112 8. 33 0 9. 4 10 10. 02 3 521 721 324 ⎑10 0 ⎀ ⎒ βŽ₯11. 0cos Ξ± sin α⎒βŽ₯ 0sin cos βŽ₯⎒ Ξ±βˆ’Ξ± ⎣ ⎦ 37 68 12. Let A = and B = . Verify that (AB)–1 = B–1 A–1.13. If A= , show that A2 – 5A + 7I = O. Hence find A–1.2579 31 12 32 14. For the matrix A = , find the numbers a and b such that A2 + aA + bI = O. 11 11 1 15. For the matrix A = 12 3 2 13 Show that A3– 6A2 + 5A + 11 I = O. Hence, find A–1. 2 11 16. If A= 12 1 1 12 Verify that A3 – 6A2 + 9A – 4I = O and hence find A–1 17. Let A be a nonsingular square matrix of order 3 Γ— 3. Then | adj A| is equal to (A) |A| (B) |A|2 (C) |A|3 (D) 3|A| 18. If A is an invertible matrix of order 2, then det (A–1) is equal to 1 (A) det (A) (B) (C) 1 (D) 0 det (A) 4.7 Applications of Determinants and Matrices In this section, we shall discuss application of determinants and matrices for solving the system of linear equations in two or three variables and for checking the consistency of the system of linear equations. Consistent system A system of equations is said to be consistent if its solution (one or more) exists. Inconsistent system A system of equations is said to be inconsistent if its solution does not exist. 4.7.1 Solution of system of linear equations using inverse of a matrix Let us express the system of linear equations as matrix equations and solve them using inverse of the coefficient matrix. Consider the system of equations a1 x + by + cz = d1 1 1 a2 x + b2 y + c2 z = d2 ax + by + cz = d3 3 3 3 abcx d⎑ ⎑⎀ ⎒ βŽ₯⎒βŽ₯ ⎒βŽ₯ 111 ⎀ ⎑⎀ 1 Let A = a2 b2 c2 ,X =y and B =d2⎒ βŽ₯⎒βŽ₯ ⎒βŽ₯ ⎒abc z ⎒βŽ₯ βŽ₯ ⎒βŽ₯d⎣333 ⎦ ⎣⎦ ⎣⎦ 3 Then, the system of equations can be written as, AX = B, i.e., abcx d⎑111 ⎀⎑⎀ ⎑1 ⎀ ⎒ βŽ₯⎒βŽ₯ ⎒βŽ₯a2 b2 c2 y = d2⎒ βŽ₯⎒βŽ₯ ⎒βŽ₯ ⎒abc βŽ₯⎒⎣z βŽ₯⎦ d⎣333 ⎦⎒⎣3 βŽ₯⎦ Case I If A is a nonsingular matrix, then its inverse exists. Now AX = B or A–1 (AX) = A–1 B (premultiplying by A–1) or (A–1A) X = A–1 B (by associative property) or I X =A–1 B or X =A–1 B This matrix equation provides unique solution for the given system of equations as inverse of a matrix is unique. This method of solving system of equations is known as Matrix Method. Case II If A is a singular matrix, then |A| = 0. In this case, we calculate (adj A) B. If (adj A) B β‰  O, (O being zero matrix), then solution does not exist and the system of equations is called inconsistent. If (adj A) B = O, then system may be either consistent or inconsistent according as the system have either infinitely many solutions or no solution. Example 27 Solve the system of equations 2x + 5y = 1 3x + 2y = 7 Solution The system of equations can be written in the form AX = B, where ⎑25⎀ ⎑⎀ x 1⎑⎀ A = ,X = and B =⎒βŽ₯⎒βŽ₯ ⎒βŽ₯ 32 y 7⎣⎦⎣⎦ ⎣⎦ Now, A= –11 β‰  0, Hence, A is nonsingular matrix and so has a unique solution. 12 5 A–1Note that = 1132 12 51Therefore X = A–1B = – 1132 7 ⎑ x⎀ 133 3 i.e. ⎒βŽ₯ = ⎣ y⎦1111 1 Hence x = 3, y = – 1 Example 28 Solve the following system of equations by matrix method. 3x – 2y + 3z =8 2x + y – z =1 4x – 3y + 2z =4 Solution The system of equations can be written in the form AX = B, where ⎑3 βˆ’23 ⎀ x ⎑⎀ 8⎑⎀ ⎒ βŽ₯⎒βŽ₯ ⎒βŽ₯ A= 21 βˆ’1 ,X = y and B = 1⎒ βŽ₯⎒βŽ₯ ⎒βŽ₯ ⎒4 βˆ’32 βŽ₯ ⎒βŽ₯ z 4⎒βŽ₯ ⎣ ⎦⎣⎦ ⎣⎦ We see that A = 3 (2 – 3) + 2(4 + 4) + 3 (– 6 – 4) = – 17 β‰  0 Hence, A is nonsingular and so its inverse exists. Now = –1, = – 8, = –10A11A12A13= –5, = – 6, = 1A21A22A23= –1, = 9, = 7A31A32A33βŽ‘βˆ’1 βˆ’5 βˆ’1⎀ 1 ⎒βŽ₯A–1Therefore = βˆ’βˆ’8 βˆ’69⎒βŽ₯17 ⎒ 101 7 βŽ₯βŽ£βˆ’ ⎦ βŽ‘βˆ’1 βˆ’5 βˆ’18⎀⎑ ⎀ –1 1 ⎒ βŽ₯⎒βŽ₯ So X = AB = βˆ’βˆ’8 βˆ’69 1⎒ βŽ₯⎒βŽ₯ 17 ⎒ 101 7 βŽ₯⎒ βŽ₯ 4βŽ£βˆ’ ⎦⎣⎦ ⎑ x⎀ βŽ‘βˆ’17 1⎀ ⎑⎀ ⎒βŽ₯ 1 ⎒ βŽ₯⎒βŽ₯ i.e. y = βˆ’βˆ’34 = 2⎒βŽ₯ ⎒βŽ₯⎒βŽ₯ 17⎒ z βŽ₯ ⎒ βŽ₯⎒βŽ₯ 51 3βŽ£βŽ¦βŽ£βˆ’βŽ¦ ⎣⎦ Hence x = 1, y = 2 and z = 3. Example 29 The sum of three numbers is 6. If we multiply third number by 3 and add second number to it, we get 11. By adding first and third numbers, we get double of the second number. Represent it algebraically and find the numbers using matrix method. Solution Let first, second and third numbers be denoted by x, y and z, respectively. Then, according to given conditions, we have x + y + z =6 y + 3z =11 x + z =2y or x – 2y + z = 0 This system can be written as A X = B, where 111 x 6 A =01 3, X = y and B = 11 1 –21 z0 Here A 11 6 –(0–3) 0–1 9 0. Now we find adj A A = 1 (1 + 6) = 7, A = – (0 – 3) = 3, A = – 1111213A = – (1 + 2) = – 3, A = 0, A = – (– 2 – 1) = 3212223= (3 – 1) = 2, = – (3 – 0) = – 3, A = (1 – 0) = 1A31A3233⎑7 –3 2⎀ Hence adj A = ⎒3 0–3 βŽ₯ ⎒βŽ₯ βŽ’β€“13 1 βŽ₯⎣⎦ ⎑7 –32 ⎀11 Thus A –1 = adj (A) = ⎒30 –3βŽ₯ ⎒βŽ₯A 9 βŽ’β€“13 1 βŽ₯⎣⎦ Since X =A–1 B ⎑7 –32 ⎀⎑6 ⎀ 1 ⎒ βŽ₯⎒βŽ₯X = 30 –3 11 ⎒ βŽ₯⎒βŽ₯9 βŽ’β€“13 1 βŽ₯⎒0 βŽ₯⎣ ⎦⎣⎦ x ⎑42 βˆ’33 +0⎀ 91 1 ⎒βŽ₯ 1y 18 + 0 +0 182or = ⎒βŽ₯ = = 9 9zβŽ’βˆ’+633 +0βŽ₯ 273⎣⎦Thus x = 1, y = 2, z = 3 Examine the consistency of the system of equations in Exercises 1 to 6. 1. x + 2y = 2 2. 2x – y = 5 3. x + 3y = 5 2x + 3y = 3 x + y = 4 2x + 6y = 8 4. x + y + z = 1 5. 3x–y – 2z = 2 6. 5x – y + 4z = 5 2x + 3y + 2z = 2 2y – z = –1 2x + 3y + 5z = 2 ax + ay +2az = 4 3x – 5y = 3 5x – 2y + 6z = –1 Solve system of linear equations, using matrix method, in Exercises 7 to 14. 7. 5x + 2y = 4 8. 2x – y = –2 9. 4x – 3y = 3 7x + 3y = 5 3x + 4y = 3 3x – 5y = 7 10. 5x + 2y = 3 11. 2x + y + z = 1 12. x – y + z = 4 33x + 2y = 5 x – 2y – z = 2x + y – 3z = 02 3y – 5z = 9 x + y + z = 2 13. 2x + 3y +3 z = 5 14. x – y + 2z = 7 x – 2y + z = – 4 3x + 4y – 5z = – 5 3x – y – 2z = 3 2x – y + 3z = 12 ⎑2–3 5⎀ 15. If A = ⎒3 2–4βŽ₯, find A–1. Using A–1 solve the system of equations⎒βŽ₯ ⎒1 1–2 βŽ₯⎣⎦ 2x – 3y + 5z =11 3x + 2y – 4z = – 5 x + y – 2z = – 3 16. The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70. Find cost of each item per kg by matrix method. Miscellaneous Examples Example 30 If a, b, c are positive and unequal, show that value of the determinant abc bca is negative.Ξ” = cabSolution Applying C1 β†’ C1 + C2 + C3 to the given determinant, we get abcbc++ abc a++ c abc b Ξ” = ++a 1 = (a + b + c) bc 1 bc 1 ca 1 ab 0 cb– a – c=(a + b + c) (ApplyingRβ†’ R–R,andRβ†’R–R)2 213310 ab– b – c = (a + b + c) [(c – b) (b – c) – (a – c) (a – b)] (Expanding along C1) = (a + b + c)(– a2 – b2 – c2 + ab + bc + ca) –1 = (a + b + c) (2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca)2 –1 = (a + b + c) [(a – b)2 + (b – c)2+ (c – a)2]2 which is negative (since a + b + c > 0 and (a – b)2 + (b – c)2+ (c – a)2> 0) Example 31 If a, b, c, are in A.P, find value of 2y + 45y + 78y + a 3y + 56y + 89y + b 4 y + 67 y + 9 10 y + c Solution Applying Rβ†’ R+ R– 2R to the given determinant, we obtain1 1 3 200 0 3y + 56y + 89y + b = 0 (Since 2b = a + c)4y + 67 y + 9 10 y +c Example 32 Show that ( yz+)2 xy zx xy ( x+ z)2 yz Ξ” = = 2xyz (x + y + z)3 xz yz ( x+ y)2 Solution Applying R1 , R2 , R3 to Ξ” and dividing by xyz, we getβ†’ xR1β†’ yR2 β†’ z R32 22xyz xy xz 1222xy yxz yz Ξ” = xyz 2 xz2 yz 2 zx y Taking common factors x, y, z from C1 C2 and C3, respectively, we get 22( yz+)2 xx xyz 22y ( x + z)2 yΞ” = xyz 22zz ( x + y)2 Applying C2 β†’ C2– C1, C3 β†’ C3– C1, we have 222( yz) x2– ( y + z) x2 βˆ’( y + z)+ y2 ( x + z)2 βˆ’ y20 z20 ( x + y)2– z2 Ξ” = Taking common factor (x + y + z) from C2 and C3, we have y + z x–y + z x–y + z()2 () () 2y ( x + z) –y 0Ξ” =(x + y + z)2 2z 0 ( x + y) –z Applying Rβ†’ R – (R + R), we have1 1232 yz – 2 z– 2 y y2 x βˆ’ y + z 0 z20 x + y –z Ξ” =(x + y + z)2 11Applying Cβ†’ (C + C) and C C C , we get221331yz 2 yz 00 y2 y2 xzΞ” =(x + y + z)2 z z2 z2 xyy Finally expanding along R1, we have Ξ” = (x + y + z)2 (2yz) [(x + z) (x + y) – yz] = (x + y + z)2 (2yz) (x2 + xy + xz) =(x + y + z)3 (2xyz) 1 1– 2 2– 0 1 Example 33 Use product 02 3– 9 2 3– to solve the system of equations 3 2– 4 6 1 2– x – y + 2z =1 2y – 3z =1 3x – 2y + 4z =2 ⎑1 –12 ⎀⎑ –20 1 ⎀ ⎒βŽ₯⎒ βŽ₯Solution Consider the product 0 2 –3 92 –3⎒βŽ₯⎒ βŽ₯ ⎒3 –24 βŽ₯⎒ 61 –2βŽ₯⎣⎦⎣ ⎦ βŽ‘βˆ’βˆ’ + 2 912 0 βˆ’+ 2 134 102 +βˆ’ ⎀ 0 ⎒βŽ₯ = 01+8 βˆ’18 0 +βˆ’ 43 066 010βˆ’+ =⎒βŽ₯ 618 24 044 3 +βˆ’ 8βŽ₯ 001βŽ’βˆ’βˆ’+ βˆ’+ 6⎣⎦1 –12 –1 –20 1 Hence 0 2 –3 92 –3 3 –24 61 –2 Now, given system of equations can be written, in matrix form, as follows ⎑1–1 2 x 1⎀⎑⎀ ⎑⎀ ⎒ βŽ₯⎒βŽ₯ ⎒βŽ₯0 2 –3 y =1⎒ βŽ₯⎒βŽ₯ ⎒βŽ₯ ⎒3–2 4 z 2βŽ₯⎒βŽ₯⎒⎣βŽ₯⎦⎣ ⎦⎣⎦ x ⎑1 βˆ’12 βˆ’11 –20⎀⎑⎀ 11 ⎒ βˆ’βŽ₯⎒βŽ₯ or y =0 23 1= 92 –31⎒ βŽ₯⎒βŽ₯ z⎒⎣3 βˆ’24 2 –22βŽ₯⎒βŽ₯ 61⎦⎣⎦ βŽ‘βˆ’++ 202⎀⎑⎀ 0 ⎒ +βˆ’βŽ₯⎒βŽ₯926 =5= ⎒ βŽ₯⎒βŽ₯ ⎒614 3⎣ +βˆ’βŽ₯⎒βŽ₯⎦ ⎣⎦ Hence x = 0, y = 5 and z = 3 Example 34 Prove that abx + c +dx p +qx ac p ax +b cx +d px +q =(1βˆ’x2) bdqΞ” = uvw uvw Solution Applying R1 β†’ R1 – x R2 to Ξ”, we get 22 2a (1βˆ’x ) c(1βˆ’x ) p (1βˆ’x ) ax +b cx +d px +qΞ” = uv w acp = (1βˆ’x2) ax +b cx +d px +q uvw Applying R2 β†’ R2 – xR1, we get ac p Ξ” = (1βˆ’ x2) bdq uvw Miscellaneous Exercises on Chapter 4 x sin ΞΈ cos –sin ΞΈ –x 1 cosΞΈ1 x 1. Prove that the determinant 2. Without expanding the determinant, prove that cosΞ± cos Ξ² cos Ξ± sin Ξ² – sin Ξ± 3. Evaluate –sin Ξ² cos Ξ² sin cos Ξ± Ξ² sin sin Ξ± Ξ² 4. If a, band care real numbers, and bccaab ++ + caabbc +++Ξ” = = 0, abbcca +++ 0 . cos Ξ± Show that either a+ b+ c= 0 or a= b= c. xa+ xx 5. Solve the equation 6. Prove that a2 2aab ab ⎑ 3 –1 ⎒ x xa + x x x xa + bc 2acc = 0, aβ‰  0 ΞΈ is independent of ΞΈ. 2aa bc 1 2 3a a 2bb ca 1 2 3b b . 2cc ab 1 2 3c c –15 6 –5 and B = –1 3 0 ,find AB 7. If A–1 = ⎒ βŽ₯⎒ βŽ₯( ⎒ 5 –22 βŽ₯⎒ 0 –21 βŽ₯⎣ ⎦⎣⎦ b2 ac = 4a2b2c2 22bbc c1 ⎀⎑ 12 –2⎀ βŽ₯⎒ βŽ₯ –) 1 1 –21 8. Let A = –23 1 . Verify that 1 15 (i) [adj A]–1 = adj (A–1) (ii) (A–1)–1 = A x yx +y yx +yx x +yx y 9. Evaluate 1 xy 10. Evaluate 1 x +yy 1 x x+y Using properties of determinants in Exercises 11 to 15, prove that: 2Ξ±Ξ± Ξ²+Ξ³ 2Ξ²Ξ² Ξ³+Ξ±11. = (Ξ² – Ξ³) (Ξ³ – Ξ±) (Ξ± – Ξ²) (Ξ± + Ξ² + Ξ³) 2Ξ³Ξ³ Ξ±+Ξ²xx21 px 3 12. = (1 + pxyz) (x – y) (y– z) (z– x), where p is any scalar. yy21 py 3 zz21 pz3 3a –a+b –a+c –b+a 3b –b+c –c +a –c+b 3c 13. 1 pq11+p ++ 23 +2 p 43 p 2q++ 14. 36 +3 p 10 ++6 p 3q 16. Solve the system of equations 2 310 4 xy z = 3(a + b + c) (ab + bc + ca) = 1 15. sin Ξ± cos Ξ± ( )cos Ξ±+Ξ΄ sinΞ² cos Ξ² ( ) cos 0Ξ²+Ξ΄= sin Ξ³ cos Ξ³ ( ) cos Ξ³+Ξ΄ 4 – 6 5 1 x y z 6 9 – 20 2 x y z Choose the correct answer in Exercise 17 to 19. 17. If a, b, c, are in A.P, then the determinant x + 2 x + 3 x + 2a x + 3 x + 4 x + 2b is x + 4 x + 5 x +2c (A)0 (B)1 (C) x (D) 2x ⎑x 00⎀ ⎒βŽ₯18. If x, y, z are nonzero real numbers, then the inverse of matrix A 0 y 0 is=⎒ βŽ₯ ⎒00 zβŽ₯⎣⎦ βˆ’1 βˆ’1⎑x 00 ⎀⎑x 00 ⎀ ⎒βŽ₯ ⎒βŽ₯βˆ’1 βˆ’1(A) ⎒ 0 y 0 βŽ₯ (B) xyz ⎒ 0 y 0 βŽ₯ βŽ’βˆ’1 βŽ₯ βŽ’βˆ’1 βŽ₯0 0 z 00 z⎣⎦ ⎣⎦ ⎑x 00⎀⎑100⎀1 ⎒βŽ₯ 1 ⎒βŽ₯(C) 0 y 0 (D) 010⎒βŽ₯ ⎒βŽ₯xyz xyz⎒00 zβŽ₯ ⎒001βŽ₯⎣⎦ ⎣⎦ ⎑ 1 sin ΞΈ 1 ⎀ ⎒ βŽ₯19. Let A = βˆ’sin ΞΈ 1 sin ΞΈ , where 0 ≀ ΞΈ ≀ 2Ο€. Then⎒ βŽ₯ βŽ’βˆ’1 βˆ’sin ΞΈ 1 βŽ₯⎣ ⎦ (A) Det (A) = 0 (B) Det (A) ∈ (2, ∞) (C) Det (A) ∈ (2, 4) (D) Det (A) ∈ [2, 4]

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