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The CORRECT CHOICE is (d) \(\begin{bmatrix}46&0\\0&46\end{bmatrix}\)

The BEST explanation: Given that, A=\(\begin{bmatrix}5&-8\\2&6\end{bmatrix}\)

∴adj A=\(\begin{bmatrix}6&8\\-2&5\end{bmatrix}\)

A(adj A)=\(\begin{bmatrix}5&-8\\2&6\end{bmatrix}\begin{bmatrix}6&8\\-2&5\end{bmatrix}\)

=\(\begin{bmatrix}5×6+(-8)×(-2)&5×8+5×(-8)\\2×6+6×(-2)&2×8+6×5\end{bmatrix}\)=\(\begin{bmatrix}46&0\\0&46\end{bmatrix}\).

Right option is (C) 8

To explain I would say: The MINOR of the ELEMENT 1 can be obtained by DELETING the FIRST row and the first column

∴M11=8.

Correct choice is (B) M21=5, A21=-5

The best explanation: The element 3 is in the second ROW (i=2) and FIRST column(j=1).

∴M21=5 (obtained by deleting R2 and C1 in Δ)

A21=(-1)^1+2 M21=-1×5 =-5.

The correct option is (a) 2

For explanation I would say: Given that the vertices are (3,2), (1,2), (5,k)

Therefore, the area of the triangle with vertices (3,2), (1,2), (5,k) is given by

Δ=\(\frac{1}{2}\) \(\begin{Vmatrix}3&2&1\\1&2&1\\5&k&1\end{Vmatrix}\)=0

Applying R1→R1-R2, we get

\(\frac{1}{2}\begin{Vmatrix}2&0&0\\1&2&1\\5&k&1\end{Vmatrix}\)=0

Expanding ALONG R1, we get

\(\frac{1}{2}\) {2(2-k)-0+0}=0

2-k=0

k=2.

Correct option is (b) 0

To EXPLAIN I WOULD say: In the given determinant FORM the ELEMENTS are in A.P

Also the common difference of this A.P is 7

Thus the value of the given determinant = 0

The CORRECT option is (a) sin^2⁡θ

To EXPLAIN I WOULD say: Given that, A=\(\begin{bmatrix}-cos⁡θ&-tan⁡θ\\cot⁡ θ&cos⁡θ \end{bmatrix}\)

|A|=\(\begin{vmatrix}-cos⁡θ&-tan⁡θ\\cot⁡θ&cos⁡θ \end{vmatrix}\)

|A|=-cos⁡θ (cos⁡θ )-cotθ(-tan⁡θ)

|A|=-cos^2⁡θ+1=sin^2⁡θ.

The CORRECT answer is (a) -sin(a – b) sin(b – c) sin(c – a)

EASIEST explanation: We have, \(\begin{vmatrix}sin^2 a & sina \,cosa & COS^2 a\\sin^2 b & SINB\, cosb & cos^2 b\\sin^2 c & sinc\, cosc & cos^2 c \end {vmatrix}\)

Now, multiplying by 2 in both numerator and denominator of column 2 and C1 = C1 + C3 we get,

1/2 \(\begin{vmatrix}sin^2 a + cos^2 a & 2sina\, cosa & cos^2 a\\sin^2 b + cos^2 b & 2sinb\, cosb & cos^2 b\\sin^2 c + cos^2 c & 2sinc\, cosc & cos^2 c \end {vmatrix}\)

= 1/2 \(\begin{vmatrix}sin^2 a + cos^2 a & sin2a & cos^2 a\\sin^2 b + cos^2 b & SIN2B & cos^2 b\\sin^2 c + cos^2 c & sin2c & cos^2 c \end {vmatrix}\)

= 1/2 \(\begin{vmatrix}1 & sin2a & cos^2 a\\1 & sin2b & cos^2 b\\1 & sin2c & cos^2 c \end {vmatrix}\)

Solving further,

= 1/2 \(\begin{vmatrix}1 & sin2a & cos^2⁡a \\0 & sin2b-sin2a & cos^2⁡ b-cos^2 ⁡a \\0 & sin2c-sin2a & cos^2 ⁡c-cos^2 ⁡a \end {vmatrix}\)

= 1/2 [(sin2b – sin2a)(cos^2⁡c – cos^2⁡a) – (cos^2 b – cos^2a)(sin2c – sin2a)]

Now, since, [cos^2 ⁡A + cos^2 B = sin(A + B) * sin(B – A)]

So, 1/2 [2 cos(a + b) sin(b – a) * sin(c + a)sin(a – c) – sin(a + b)sin(a – b) * 2 cos(a + c)sin(c – a)]

= sin(a – b)sin(c – a)[sin(c + a)cos(a + b) – cos(c + a) sin(a + b)]

= sin(a – b) sin(c – a) sin(c + a – a – b)

= -sin(a – b) sin(b – c) sin(c – a)

The correct option is (d) -3

To ELABORATE: Consider the element -3 in Δ=\(\begin{vmatrix}1&8&-6\\2&-3&4\\-7&9&5\end{vmatrix}\)

The cofactor of the element -3 is GIVEN by

A22=(-1)^2+2 M22

M22=\(\begin{vmatrix}1&-6\\-7&5\end{vmatrix}\)=1(5)-(-6)(-7)=5-42=-37

A22=(-1)^2+2 (-37)=-37.

The CORRECT ANSWER is (b) \(\begin{bmatrix}-3&-1\\0&2\end{bmatrix}\)

The explanation is: Consider the matrix \(\begin{bmatrix}-3&-1\\0&2\end{bmatrix}\)

adj A=\(\begin{bmatrix}2&1\\0&-3\end{bmatrix}\)

|A|=-6

∴A^-1=\(\frac{1}{|A|}\) adj A=\(\frac{1}{-6}\begin{bmatrix}2&1\\0&-3\end{bmatrix}\).

Correct choice is (d) 1

Explanation: The area of triangle FORMED by collinear points is zero.

Δ=\(\FRAC{1}{2}\) \(\begin{Vmatrix}1&2&1\\3&0&1\\2&K&1\end{Vmatrix}\)=0

Expanding along C2, we get

\(\frac{1}{2}\){-2(3-2)+0-k(1-3)}=0

\(\frac{1}{2}\) {-2+2k}=0

∴k=1

Correct option is (d) \(\frac{1}{2}\) sq.units

To elaborate: The area of the triangle with vertices (0,1), (0,2), (1,5) is GIVEN by

Δ=\(\frac{1}{2}\begin{Vmatrix}0&1&1\\0&2&1\\1&5&1\end{Vmatrix}\)

Expanding along C1, we get

Δ=\(\frac{1}{2}\){(0-0+1(1-2)}=\(\frac{1}{2}\) |-1|=\(\frac{1}{2}\) sq.units.

Right answer is (a) -1(1+m+N+q)

For explanation: GIVEN that, Δ=\(\begin{vmatrix}1+m&n&q\\m&1+n&q\\n&m&1+q\end{vmatrix}\)

Applying C1→C1+C2+C3

Δ=\(\begin{vmatrix}1+m+n+q&n&q\\1+m+n+q&1+n&q\\1+m+n+q&m&1+q\end{vmatrix}\)=(1+m+n+q)\(\begin{vmatrix}1&n&q\\1&1+n&q\\1&m&1+q\end{vmatrix}\)

Applying R1→R2-R1

Δ=(1+m+n+q)\(\begin{vmatrix}0&1&0\\1&1+n&q\\1&m&1+q\end{vmatrix}\)

Expanding along the first row, we get

Δ=(1+m+n+q)(0-1(1+q-q)+0)

Δ=-1(1+m+n+q).

Correct answer is (a) 0

For explanation I WOULD say: Δ=\(\BEGIN{vmatrix}-a&b&c\\-2a+4x&2b-4y&2c+4z\\x&-y&z\end{vmatrix}\)

Using the properties of determinants, the given determinant can be expressed as a SUM of two determinants.

Δ=\(\begin{vmatrix}-a&b&c\\-2a&2b&2c\\x&-y&z\end{vmatrix}\)+\(\begin{vmatrix}-a&b&c\\4x&-4y&4z\\x&-y&z\end{vmatrix}\)

Δ=2\(\begin{vmatrix}-a&b&c\\-a&b&c\\x&-y&z\end{vmatrix}\)+4\(\begin{vmatrix}-a&b&c\\x&-y&z\\x&-y&z\end{vmatrix}\)

Since two ROWS are similar in each of the determinants, the determinant is 0.

The correct answer is (a) 0

The best I can explain: Δ=\(\begin{vmatrix}x^2&x^3&x^4\\x&y&z\\x^2&x^3&x^4 \end{vmatrix}\)

If the elements of any two rows or columns are identical, then the VALUE of determinant is ZERO. Here, the elements of ROW 1 and row 3 are identical. Hence, its determinant is 0.

Right answer is (d) 151

The best I can explain: GIVEN that, A=\(\begin{BMATRIX}7&4\\5&5\end{bmatrix}\)

|A|=(7(5)-5(4))=35-20=15

|A|^2-5|A|+1=(15)^2-5(15)+1=225-75+1=151.

The correct choice is (a) X=1, –\(\frac{1}{3}\)

For EXPLANATION I would say: GIVEN that \(\begin{vmatrix}3&x\\2&x^2 \end{vmatrix}\)=\(\begin{vmatrix}5&3\\3&2\end{vmatrix}\)

⇒3x^2-2x=5(2)-3(3)

⇒3x^2-2x=1

Solving for x, we get

x=1, –\(\frac{1}{3}\).

The CORRECT choice is (a) -34

To elaborate: The minor of element 5 in the DETERMINANT Δ=\(\BEGIN{vmatrix}1&5&4\\2&3&6\\7&9&4\end{vmatrix}\) is the determinant obtained by deleting the row and COLUMN containing element 5.

∴M12=\(\begin{vmatrix}2&6\\7&4\end{vmatrix}\)=2(4)-7(6)=-34.

The correct ANSWER is (d) 3 sq.units

The EXPLANATION is: The area of the triangle (0,2), (0, 0), (3, 0) with vertices is given by

Δ=\(\frac{1}{2}\begin{Vmatrix}0&2&1\\0&0&1\\3&0&1\end{Vmatrix}\)

Expanding ALONG R3, we get

Δ=\(\frac{1}{2}\) {0-0+3(2-0)}

Δ=3 sq.units.

The correct option is (b) abc(a^3+b^3+c^3-abc)

For explanation I WOULD say: GIVEN that, A=\(\begin{bmatrix}c^2&cb&ca\\ab&a^2&-ac\\ab&bc&-b^2\end{bmatrix}\)

Taking c a, b common from R1, R2, R3 respectively, we get

Δ=abc\(\begin{bmatrix}c&b&a\\b&a&-c\\a&c&-b\end{bmatrix}\)

Δ=abc{(c(-ab+c^2)-b(-b^2+ac)+a(bc-a^2)

Δ=abc(-abc+c^3+b^3-abc+abc-a^3)

Δ=abc(a^3+b^3+c^3-abc).

Right answer is (a) 352

To explain: Given that, A=\(\begin{bmatrix}2&5&9\\6&1&3\\4&8&2\end{bmatrix}\)

⇒|A|=\(\begin{vmatrix}2&5&9\\6&1&3\\4&8&2\end{vmatrix}\)

Evaluating along the first row, we get

∆=2\(\begin{vmatrix}1&3\\8&2\end{vmatrix}\)-5\(\begin{vmatrix}6&3\\4&2\end{vmatrix}\)+9\(\begin{vmatrix}6&1\\4&8\end{vmatrix}\)

∆=2(2-24)-5(12-12)+9(48-4)

∆=2(-22)-0+9(44)

∆=-44+9(44)=44(-1+9)=352

The CORRECT answer is (B) 7

Explanation: Consider the element 7 in the determinant Δ=\(\begin{vmatrix}2&8\\4&7\end{vmatrix}\)

The minor of the element 7 can be obtained by DELETING R2 and C2

∴M22=2

Hence, the minor of the element 7 is 2.

Correct answer is (b) Infinitely many solutions

The EXPLANATION: SOLVING the given system of equation by Cramer’s rule, we get,

x = D1/D, y = D2/D, z = D3/D where,

D = \(\begin{vmatrix}2 & -3 & 4 \\3 & -4 & 5 \\4 & -5 & 6 \end {vmatrix}\)

D = –\(\begin{vmatrix}2 & 3 & 4 \\3 & 4 & 5 \\4 & 5 & 6 \end {vmatrix}\)

Now, performing, C3 = C3 – C2 and C2 = C2 – C1 we get,

D = –\(\begin{vmatrix}2 & 1 & 1 \\3 & 1 & 1 \\4 & 1 & 1 \end {vmatrix}\)

As two columns have identical values, so,

D = 0

Similarly,

D1 = \(\begin{vmatrix}7 & -3 & 4 \\8 & -4 & 5 \\9 & -5 & 6 \end {vmatrix}\)

Now, performing, C1 = C1 – C3

D1 = –\(\begin{vmatrix}3 & -3 & 4 \\3 & -4 & 5 \\3 & -5 & 6 \end {vmatrix}\)

Now, performing, C3 = C3 – C2

D1 = -3\(\begin{vmatrix}1 & -3 & 1 \\1 & -4 & 1 \\1 & -5 & 1 \end {vmatrix}\)

As two columns have identical values, so,

D1 = 0

D2 = \(\begin{vmatrix}2 & 7 & 4 \\3 & 9 & 5 \\4 & 8 & 6 \end {vmatrix}\)

Now, performing,

D2 = –\(\begin{vmatrix}2 & 3 & 2 \\3 & 3 & 2 \\4 & 3 & 2 \end {vmatrix}\)

Now, performing, C2 = C2 – C3 and C3 = C3 – C1

D2 = 6\(\begin{vmatrix}2 & 1 & 1 \\3 & 1 & 1 \\4 & 1 & 1 \end {vmatrix}\)

As two columns have identical values, so,

D2 = 0

D3 = \(\begin{vmatrix}2 & -3 & 7 \\3 & -4 & 9 \\4 & -5 & 6 \end {vmatrix}\)

D3 = –\(\begin{vmatrix}2 & 3 & 4 \\3 & 4 & 4 \\4 & 5 & 4 \end {vmatrix}\)

Now, performing, C2 = C2 – C2 and C3 = C3 – C2

D3 = -4\(\begin{vmatrix}2 & 1 & 1 \\3 & 1 & 1 \\4 & 1 & 1 \end {vmatrix}\)

As two columns have identical values, so,

D3 = 0

SINCE, D = D1 = D2 = D3 = 0, thus, it has infinitely many solutions.

Right answer is (c) the solution does not exist

The explanation is: If A is a SINGULAR MATRIX, then |A|=0

In this case, if (ADJ A)B≠O, then solution does not exist and the system of EQUATIONS is called inconsistent.

Right OPTION is (b) –\(\frac{8}{3}\)

Explanation: Given that the vertices are (1,2), (3,5), (k,0)

Therefore, the area of the TRIANGLE with vertices (1,2), (3,5), (k,0) is given by

Δ=\(\frac{1}{2}\begin{Vmatrix}1&2&1\\3&5&1\\k&0&1\end{Vmatrix}\)=\(\frac{7}{2}\)

Expanding ALONG R3, we get

\(\frac{1}{2}\) {k(2-5)-0+1(5-6)}=\(\frac{1}{2}\) {-3k-1}=\(\frac{7}{2}\)

⇒-\(\frac{1}{2}\) (3k+1)=\(\frac{7}{2}\)

3k=-8

k=-\(\frac{8}{3}\).

The correct option is (a) 8 ±√37

For explanation: Here, we have, \(\begin{vmatrix}2-x & 2 & 3 \\2 & 5-x & 6 \\3 & 4 & 10-x \end {vmatrix}\) = 0

Now, replacing C3 = C3 – 3C1, we get,

\(\begin{vmatrix}2-x & 2 & 3-6 + 3x \\2 & 5-x & 6-6 \\3 & 4 & 10-x -9 \end {vmatrix}\) = 0

\(\begin{vmatrix}2-x & 2 & 3(x-1) \\2 & 5-x & 0 \\3 & 4 & -(x-1) \end {vmatrix}\) = 0

Or, (x – 1)\(\begin{vmatrix} 2-x & 2 & 3 \\2 & 5-x & 0 \\3 & 4 & -1 \end {vmatrix}\) = 0

Now, replacing R1 = R1 + 3R3, we get,

Or, (x – 1)\(\begin{vmatrix}11-x & 14 & 0 \\2 & 5-x & 0 \\3 & 4 & -1 \end {vmatrix}\) = 0

Or, (x – 1)[-1 {(11 – x)(5 – x) – 28}] = 0

Or, -(x – 1)(55 – 11x – 5x + x^2 – 28)

Or, (x – 1)(x^2 – 16x + 27) = 0

Thus, either x – 1 = 0 i.e. x = 1 or x^2 – 16x + 27 = 0

THEREFORE, solving x^2 – 16x + 27 = 0 further, we get,

x = 8 ± √37

Right answer is (C) 2α = 2nπ ± π/4 + π/4

The explanation is: The given system have non-trivial SOLUTION if \(\begin{vmatrix}\beta & sin \ALPHA & cos \alpha \\1 & cos \alpha & sin \alpha \\ -1 & sin \alpha & -cos \alpha \end {vmatrix}\) = 0

On opening the DETERMINANT we get β = sin 2α + cos 2 α

Therefore, -√2 ≤ β ≤ √2

Now, for β = 1,

sin 2α + cos 2 α = 1

=> (1/√2)sin 2α + (1/√2) cos 2α = (1/√2)

Or, cos(2α – π/4) = 1/√2 = cos(2nπ ± π/4)

=> 2α = 2nπ ± π/4 + π/4

Right answer is (b) -9

Easy EXPLANATION: GIVEN that,

\(\begin{vmatrix}x & 3 & 6 \\3 & 6 & x \\6 & x & 3 \end {vmatrix}\) = \(\begin{vmatrix}2 & x & 7 \\x & 7 & 2\\7 & 2 & x \end {vmatrix}\) = \(\begin{vmatrix}4 & 5 & x \\ 5 & x & 4 \\x & 4 & 6 \end {vmatrix}\)

So, by circular DETERMINANT property,

Sum of the elements of a row = 0

So, x + 3 + 6 = 2 + x + 7 = 4 + 5 + x = 0

=> x = -9

The CORRECT answer is (b) The area of the triangle is independent on m

Explanation: The area o the triangle with the given POINT as vertices is,

1/2 \(\begin{vmatrix}m(m+1) & (m+1) & 1 \\(m+1)(m+2) & (m+2) & 1 \\(m+2)(m+3) & (m+3) & 1 \end {vmatrix}\)

 = 1/2 \(\begin{vmatrix}m^2 + m & (m+1) & 1 \\m^2 + 3m + 2 & (m+2) & 1 \\m^2 + 5m + 6 & (m+3) & 1 \end {vmatrix}\)

Now, by performing the row operation R2 = R2 – R1 and R3 = R3 – R2

= 1/2 \(\begin{vmatrix}m^2 + m & (m+1) & 1 \\2m + 2 & 1 & 0 \\2m + 4 & 1 & 0 \end {vmatrix}\)

Now, breaking the DETERMINANT we get,

= 1/2 (2m + 2 – 2m – 4)

= -1

Thus, it is independent of m.

The correct choice is (d) sin⁡θ-cos^2⁡⁡θ

The explanation is: Δ=\(\begin{vmatrix}cos⁡θ&-cos⁡θ&1\\sin^2⁡θ&cos^2⁡θ&1\\sin⁡θ&-sin⁡θ&1\end{vmatrix}\)

APPLYING C1→C1+C2

Δ=\(\begin{vmatrix}cos⁡θ-cos⁡θ&-cos⁡θ&1\\sin^2⁡θ+cos^2⁡θ&cos^2⁡θ&1\\sinθ-sin⁡θ&-sin⁡θ&1\end{vmatrix}\)=\(\begin{vmatrix}0&-cos⁡θ&1\\1&cos^2⁡θ&1\\0&-sin⁡θ&1\end{vmatrix}\)

Expanding alongC1, we get

0-1(cos^2⁡⁡θ+sinθ)=sin⁡θ-cos^2⁡⁡θ.

Right option is (d) 0

Explanation: The above matrix is a skew symmetric matrix and its order is odd

And we KNOW that for any skew symmetric matrix with odd order has determinant = 0

Therefore, the value of the GIVEN determinant = 0

Right CHOICE is (c) 1

Best EXPLANATION: We have,\(\BEGIN{vmatrix}1 & ab & (\FRAC{1}{a} + \frac{1}{b}) \\1 & bc & (\frac{1}{b} + \frac{1}{c}) \\1 & ca & (\frac{1}{c} + \frac{1}{a})\end {vmatrix}\)

= (1/ABC)\(\begin{vmatrix}1 & ab & \frac{b + a}{ab} * abc \\1 & bc & \frac{b + c}{bc} * abc \\1 & ca & \frac{c + a}{ac} * abc \end {vmatrix}\)

= (1/abc)\(\begin{vmatrix}1 & ab & bc + ac \\1 & bc & ac + ab \\1 & ca & ab + bc \end {vmatrix}\)

Operating, C3 = C3 + C2

= (1/abc)\(\begin{vmatrix}1 & ab & ab + bc + ac \\1 & bc & ab + bc + ac \\1 & ca & ab + bc + ac \end {vmatrix}\)

= ((ab + bc + ac)/abc)\(\begin{vmatrix}1 & ab & 1 \\1 & bc & 1 \\1 & ca & 1 \end {vmatrix}\)

= 0

Right OPTION is (a) x=2, –\(\frac{1}{3}\)

The explanation: Given that, \(\BEGIN{vmatrix}1&-1\\3&-5\end{vmatrix}\)=\(\begin{vmatrix}x&x^2\\3&5\end{vmatrix}\)

-5—(-3)=5x-3x^2

-2=5x-3x^2

3x^2-5x-2=0

Solving for x, we get

x=2, –\(\frac{1}{3}\).

Right answer is (c) 1

The explanation: The GIVEN matrix is, \(\begin{vmatrix}cos^2 θ & cosθ\, sinθ & -sinθ \\cosθ\, sinθ & sin^2⁡ θ & cosθ \\sinθ & -cosθ & 0 \end {vmatrix}\)

Now, performing the row OPERATIONS R1 = R1 + sinθR3 and R2 = R2 –cosθR3

=\(\begin{vmatrix}cos^2⁡ θ + sin^2⁡ θ & cosθ\, sinθ – cosθ sinθ & -sinθ \\cosθ\, sinθ – cosθ sinθ & cos^2⁡ θ + sin^2 ⁡θ & cosθ \\sinθ & -cosθ & 0 \end {vmatrix}\)

Solving further,

= \(\begin{vmatrix}1 & 0 & -sinθ \\0 & 1 & cosθ \\sinθ & -cosθ & 0 \end {vmatrix}\)

Breaking the determinant, we get,

 = 1(0 + cos^2θ) – sinθ(0 – sinθ)

= 1

The correct option is (b) 0

For explanation I would say: Δ=\(\BEGIN{vmatrix}5&0&5\\1&4&3\\0&8&6\end{vmatrix}\)

Expanding ALONG R1, we get

Δ=5\(\begin{vmatrix}4&3\\8&6\end{vmatrix}\)-0\(\begin{vmatrix}1&3\\0&6\end{vmatrix}\)+5\(\begin{vmatrix}1&4\\0&8\end{vmatrix}\)

Δ=5(24-24)-0+5(8-0)

Δ=0-0+40=40.

The CORRECT ANSWER is (d) 0

Explanation: Expanding ALONG R1, we get

∆=-i(i)-(-1)(-1)=-i^2-1=-(-1)-1=0.

The CORRECT answer is (c) 5\(\SQRT{3}\)+4

Easiest explanation: Evaluating ALONG R1, we get

∆=5(\(\sqrt{3}\))-(-4)1=5\(\sqrt{3}\)+4.

The correct choice is (B) 9

To EXPLAIN I would say: The MINOR of the ELEMENT 2 can be obtained by deleting the FIRST row and the first column

∴M11=9.

Right choice is (c) 12

To explain I would say: Here, in L.H.S we have,

\(\begin{vmatrix}2 & 5 & 8 \\1 & 4 & 7 \\6 & 4 & \alpha \end {vmatrix}\)

So, for trivial ROOTS the above VALUE is = 0

=>\(\begin{vmatrix}2 & 5 & 8 \\1 & 4 & 7 \\6 & 4 & \alpha \end {vmatrix}\) = 0

SOLVING it further we get α = 12

Right OPTION is (c) (Σab)^3

For explanation: Applying R1 –> R1 + R2 + R3

\(\begin{vmatrix}\SIGMA ab & \Sigma ab & \Sigma ab \\ab + bc & -ca & ab + bc \\bc + ca & bc + ca & -ab \end {vmatrix}\)

This is equal to,

= Σab \(\begin{vmatrix}1 & 1 & 1 \\ab + bc & -ca & ab + bc \\bc + ca & bc + ca & -ab \end {vmatrix}\)

Applying C1 –> C1 – C2 and C2 –> C2 – C3

= Σab \(\begin{vmatrix}0 & 0 & 1 \\ \Sigma ab & -\Sigma ab & ab + bc \\0 & \Sigma ca & -ab \end {vmatrix}\)

= (Σab)^3