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XI Basic Mathematics Solution

Matrices and Determinants Exercise: 5.2 Class 11 Basic Mathematics Solution [NEB UPDATED]

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XI Basic Mathematics Solution Exercise 5.2

Exercise 5.2

1. Evaluate the determinant of the following matrices.

a.(1234)

Solution:

Let A = (1234)

Or, |A|= |1234| = 1*4 – 2 * 3 = –2. 

b.(1−11−1)

Solution:

Let A = (1−11−1)

Or, |A|= |1−11−1| = –1 + 1= 0. 

c.(123−104123)

Solution:

Let A = (123−104123), Then

|A|= |123−104123|.

Or, |A| = 1|0423| – 2 |−1413| + 3 |−1012|

= 1(0 – 8) – 2(3 – 4) + 3(–2 – 0) = 0 

d.(−103214−2−31)

Solution:

Let A = (−103214−2−31), Then

Or, |A|= |−103214−2−31|.

Or, |A| = –1|14−3−1| – 0 |24−2−1| + 3 |21−2−3|

= –1(–1 + 12) – 0 + 3(–6 + 2) = –23. 

e. (3−22410231)

Solution:

Let A = (3−22410231), Then

Or, |A|= |3−22410231|.

Or, |A| = 3|1031| – (–2)|4021| + 2 |4123|

= 3(1 – 0) + 2(4 – 0) + 2(12 – 2) = 31.

f.(2−20−124−456)

Solution:

Let A = (2−20−124−456), Then

Or, |A|= |2−20−124−456|.

Or, |A| = 2|2456| + 2|−14−46| + 0 |−12−45|

= 2(12 – 20) + 2(–6 + 16) + 0 = 4.

= 4.

 

2. Solve for x.

a.|x23−1012−20|=0

Solution:

|x23−1012−20|=0

Or, |x23−1012−20|=0

Or, x |01−20|−2|−1120|+3|−102−2| = 0

Or, x(0 + 2) – 2(0 – 2) + 3(2 – 0) = 0

Or, 2x + 4 + 6 = 0

Or, 2x + 10 = 0

Or, x = –5. 

b.|x3333x223|=0

Solution:

|x3333x223|=0

Or, |x3333x223|=0

Or, x |3x33|−3|3x23|+3|3323| = 0

Or, x(9 – 3x) – 3(9 – 2x) + 3(9 – 6) = 0

Or, 9x – 3x2 – 27 + 6x + 9 = 0

Or, –3x2 + 15x – 18 = 0

Or, x2 – 5x + 6 = 0

Or, (x – 2)(x – 3) = 0

Either, x – 2 = 0.

So, x = 2

Or, x – 3 = 0

So, x = 3

So, x = 2 or 3.

 

3. Without expanding the determinants, show that the values of each of the following determinants is zero.

(i)|61924718327|

Solution:

Or, |61924718327|=|6192473∗63∗13∗9|=3|619247619| (taking 3 common from R3).

= 3.0 = 0   (So, R1 = R3). 

(ii)|1ab+c1bc+a1ca+b|

Soln:

Or, |1ab+c1bc+a1ca+b|=|1ab+c+a1bc+a+b1ca+b+c|(C3→ C2 + C3)

= (a + b + c) .|1a11b11c1|  (Taking a + b + c common from C3).

= (a + b + c). 0 = 0   (C1 = C3). 

(iii)|(a−b)(b−c)(c−a)(b−c)(c−a)(a−b)(c−a)(a−b)(b−c)|

Solution:

Or, |(a−b)(b−c)(c−a)(b−c)(c−a)(a−b)(c−a)(a−b)(b−c)|=|a−b+b−cb−cc−ab−c+c−ac−aa−bc−a+a−ba−bb−c|(C1→ C1 + C2)

=|a−cb−cc−ab−ac−aa−bc−ba−bb−c|= −|c−ab−cc−aa−bc−aa−bb−ca−bb−c| (Taking –1 common from C1).

= 0 (C1 = C3). 

(iv)|1bca(b+c)1cab(c+a)1abc(a+b)|

Solution:

Or, |1bca(b+c)1cab(c+a)1abc(a+b)|=|1bcab+ac1cabc+ba1abca+cb|

= |1bcab+ac+bc1cabc+ab+ca1abac+bc+ab|(C1C1+C2)

 = (ab + bc + ca) |1bc11ca11ab1| (Taking ab + bc common from C3).

= (ab + bc + ca) . 0 = 0   (C1 = C3). 

(v)|1bcbc(b+c)1caca(c+a)1abab(a+b)|

Solution:

Or, |1bcbc(b+c)1caca(c+a)1abab(a+b)|=1abc|aabcabc(b+c)babcabc(c+a)cabcabc(a+b)|R1→aR1R2→bR2R3→ac

= 1abc.abc.abc|a1b+cb1c+ac1a+b|   (taking abc common from C2 and C3).

 = abc |a1b+c+ab1c+a+bc1a+b+c| (C3→C3 + C1)

= abc(a + b + c) |a11b11c11|  (taking a + b + c common from C3).

= abc (a + b + c).0 = 0  (C2 = C3).

(vi)|1ww2ww21w21w|

Solution:

Or, |1ww2ww21w21w|=|1+w+w2ww2w+w2+1w21w2+1+w1w| (C1→ C1 + C2 + C3)

= |0ww20w2101w| = 0  (Each element of C1 = 0).

 

4) Evaluate:

(i)|516171567111|

Solution:

|516171567111|

Or, |516171567111|=|51−161−171−1567111| (R1→R1 – R3)

= |506070567111|

= 10 |567567111| (Taking 10 common from R1).

= 10 * 0 = 0.   (R1 = R2). 

(ii)|161923151822131720|

Solution:

|161923151822131720|

Or, |161923151822131720|=|16−1519−1823−22151822131720| (R1→R1 – R3)

= |111151822131720|=|1−111−115−181822−1813−171720−17|C1→C1−C2C3→C3−C2

= |010−3184−4173|

= –1 |−34−43|

= –1(–9 + 16) = –7.

 

5) Without expanding the determinants, prove that

(i)|1784−34−5102|= |84214−57−310| 

Solution:

Or, |1784−34−5102|=|14−57−310842| (Rows and columns are interchanged).

= −|14−58427−310| (Two adjacent rows are interchanged).

= |84214−57−310| (Two adjacent rows are interchanged).

(ii)|357921785|+|937258175|

Solution:

Or, |357921785| and Δ′=|937258175| (Rows and columns are interchanged).

Or, Δ = |357921785| = |397528715| (Rows and columns are interchanged).

= −|937258175| (Two adjacent columns are interchanged.)

= −Δ′So, Δ+Δ′ = 0. 

(iii)|1xx21yy21zz2|=|1xyz1yzx1zxy|

Solution:

Or, |1xx21yy21zz2|=1xyz|xyzxx2xyzyy2xyzzz2|

= 1xyz.xyz|yz1xzx1yxy1z|  (x,y,z are taken common from R1,R2 and R3).

= |1yzx1zxy1xyz| (Two adjacent column are interchanged).

= |1xyz1yzx1zxy| (Two adjacent columns are interchanged). 

(iv)|1bcb+c1cac+a1aba+b|=|1aa21bb21cc2|

Solution:

L.H.S. =|1bcb+c1cac+a1aba+b|=|1bc(a+b+c)−a1ca(b+c+a)−b1ab(a+b+c)−c|

= |1bca+b+c1cab+c+a1aba+b+c|−|1bca1cab1abc|

= (a + b + c)|1bc11ca11ab1|−1abc|aabca2babcb2cabcc2|

= (a + b + c). 0 – 1abc.abc |a1a2b1b2c1c2| = |1aa21bb21cc2| = R.H.S. 

(v) |bcaa2cabb2abcc2|= |1a2a31b2b31c2c3|

Solution:

L.H.S. = |bcaa2cabb2abcc2|=1abc|abca2a3abcb2b3abcc2c3|

= 1abc.abc|1a2a31b2b31c2c3|

= |1a2a31b2b31c2c3|  = R.H.S.

 

6) Show that:

(i)|111abcbccaab|=(a – b)(c – a)(b – c) 
Solution:

L.H.S. = |111abcbccaab|

= |100ab−ac−abcca−bcab−bc|(C2→C2−C1C3→C3−C1)

= |b−ac−ac(a−b)b(a−c)| 

= |−(a−b)(c−a)c(a−b)−b(c−a)|

= (a – b)(c – a) |−11c−b| = (a – b)(c – a)(b – c) 

= R.H.S. 

(ii)|abca2b2c2b+cc+aa+b|(C2→C2−C1C3→C3−C1)=(a – b)(b – c)(c – a)(a + b + c)

Solution:

L.H.S. = |abca2b2c2b+cc+aa+b|(C2→C2−C1C3→C3−C1)

= |a−(a−b)(c−a)a2−(a−b)(a+b)(c−a)(c+a)b+c(a−b)−(c−a)|

= (a – b)(c – a) |a−11a2−(a+b)(c+a)b+c1−1|

= (a – b)(c – a) |a−10a2−(a+b)c−bb+c10| (C­3àC3 + C2).

= –(a – b)(c – a)(c – b) |a−1b+c1|

= (a – b)(b – c)(c – a)(a + b + c) = R.H.S. 

(iii)|111abca3b3c3|=(a – b)(b – c)(c – a)(a + b + c)

Solution:

L.H.S. = |111abca3b3c3|=|1−111−1a−bbc−ba3−b3b3c3−b3|(C1→C1−C2C3→C3−C1)

= |010a−bbc−ba3−b3b3c3−b3|

= −1|a−bc−ba3−b3c3−b3|

= –(a – b)(c – b) |11a2+ab+b2c2+cb+b2|

= (a – b)(b – c)(c2 + cb + b2 – a2 – ab – b2)

= (a – b)(b – c){(c + a)(c – a) + b(c – a)}

= (a – b)(b – c)(c – a)(a + b + c) = R.H.S. 

(iv)|xyzx2y2z2yzzxxy|=(b-c)(c-a)(a-b)(a+b+c)

Solution:

L.H.S. = |xyzx2y2z2yzzxxy|=|x−yyz−yx2−y2y2z2−y2yz−zxzxxy−zx|(C1→C1−C2C3→C3−C2)

= (x – y)(z – y) |1y1x+yy2z+y−zzx−x|

= (x – y)(z – y)|0y1x−zy2z+y1zx−x| (C1àC1 – C3)

= (x – y)(z – y)(z – x) |0y11y2z+y1zx−x|  (Taking x – z common from C1).

= (x – y)(z – y)(z – x) |0y10y2−zxz+y1zx−x+x|  (R2àR2 – R3)

= (x – y)(z – y)(z – x).1. |y1y2−zxz+y+x|

= (x – y)(z – y)(z – x){yz + y2 + xy – y2 + zx}

= (x – y)(z – y)(z – x)(yz + zx + xy) = R.H.S. 

(v) |b+cabc+acaa+bbc|=(a + b + c)(a – c)2

Solution:

L.H.S. = |b+cabc+acaa+bbc|=|2(a+b+c)a+b+ca+b+cc+acaa+bbc| (C1àC1 + C2 + C3).

= (a + b + c)|211c+acaa+bbc|(Taking a + b + c common from R1).

= (a + b + c)|0110caa−cbc| (C1àC1 – C2 – C3)

= (a + b + c)(a – c) |11ca| = (a + b + c)(a – c)(a – c)

= (a + b + c)(a – c)2= R.H.S. 

(vi)|a−b−c2a2a2bb−c−a2b2c2cc−a−b|= (a + b + c)3 

Solution:

L.H.S. = |a−b−c2a2a2bb−c−a2b2c2cc−a−b|

=|a+b+ca+b+ca+b+c2bb−c−a2b2c2cc−a−b| (R1àR1 + R2 + R3).

= (a + b + c) |1112bb−c−a2b2c2cc−a−b|

= (a + b + c)|0110b−c−a2bc+a+b2cc−a−b| (C1àC1 – C3)

= (a + b + c)2|11b−c−a2b|

= (a + b + c)2(2b – b + c + a) = (a + b + c)3 = R.H.S.

 

(vii) |1+a1a2a3a11+a2a3a1a21+a3|= 1 + a1 + a2 + a3

Solution:

L.H.S. = |1+a1a2a3a11+a2a3a1a21+a3|=|1+a1+a2+a3a2a31+a1+a2+a31+a2a31+a1+a2+a3a2a+a3| (C1àC1 + C2 + C3).

= (1 + a1 + a2 + a3) |1a2a311+a2a31a21+a3|

= (1 + a1 + a2 + a3)|00−111+a2a31a21+a3| (R1àR1– R3)

= –(1 + a1 + a2 + a3)|11+a21a2|

= –(1 + a1 + a2 + a3)(a2 – 1 – a2) = 1 + a1 + a2 + a3 = R.H.S.

 

(viii) |1+x1111+y1111+z|= xyz(1x+1y+1z+1)

Solution:

L.H.S. = |1+x1111+y1111+z|

=xyz.|1+xx1y1z1x1+yy1z1x1y1+zz|

= xyz.|1x+11y1z1x1y+11z1x1y1z+1|

= xyz.|1x+1+1y+1z1y1z1x+1y+1+1z1y+11z1x+1y+1z+11y1z+1| (C1àC1 + C2 + C3)

= xyz.(1x+1y+1z+1)|11y1z11y+11z11y1z+1|

= xyz.(1x+1y+1z+1)|00−111y+11z11y1z+1|  (R1àR1 – R3)

= –xyz (1x+1y+1z+1)|11y+111y|

= –xyz (1x+1y+1z+1){1y−1y−1}

= xyz(1x+1y+1z+1) = R.H.S.

 

(ix)|a2bcc2+aca2+abb2acabb2+bcc2|= 4a2b2c2

Solution:

L.H.S. = |a2bcc2+aca2+abb2acabb2+bcc2|

=abc.|acc+aa+bbabb+cc|

= abc.|a−c−c−acc+aa+b−b−abab−b−c−cb+cc|

=abc.|−2ccc+a0ba−2cb+cc|

=  abc.|−2ccc+a0ba0b−a| (C3 àC3 – C1)

= abc(–2c) |bab−a| = –2abc2(–ab – ab)

= – 2abc2(–2ab) = 4a2b2c2 = R.H.S. 

(x)xyzxyz|x2+1xyxzxyy2+1yzxzyzz2+1|= (1 + x2 + y2 + z2)

Solution:

L.H.S. = xyzxyz|x2+1xyxzxyy2+1yzxzyzz2+1|

=1xyz.|x(x2+1)x2yx2zxy2y(y2+1)y2zxz2yz2z(z2+1)|

= 1xyz.xyz|x2+1x2x2y2y2+1y2z2z2z2+1|

=|1+x2+y2+z21+x2+y2+z21+x2+y2+z2y2y2+1y2z2z2z2+1|

=  (1+x2+y2+z2)|111y2y2+1y2z2z2z2+1|=  (1+x2+y2+z2)|100y210z201|(C2→C2−C1C3→C3−C1)

= (1+x2+y2+z2)|1001|

=(1+x2+y2+z2)(1 – 0) = (1 + x2 + y2 + z2) = R.H.S.

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#XI Basic Mathematics Solution

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