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Right option is (b) -2xy(x+1)

BEST explanation: GIVEN that, A=\(\BEGIN{bmatrix}1&x&y\\1&x&-y\\1&-x^2&y^2\end{bmatrix}\)

Δ=\(\begin{vmatrix}1&x&y\\1&x&-y\\1 &-x^2&y^2 \end{vmatrix}\)

Taking x common C2 and y common from C3, we get

Δ=xy\(\begin{vmatrix}1&1&1\\1&1&-1\\1&-x&y\end{vmatrix}\)

Expanding along R1, we get

Δ=xy{1(y-x)-1(y+1)+1(-x-1)}

Δ=xy(y-x-y-1-x-1)

Δ=xy(-2x-2)=-2xy(x+1).

Correct choice is (b) X=3, y=2

To elaborate: The given situation can be represented by a system of equations as:

8x+3y=70

10x+5y=90

Consider, A=\(\begin{BMATRIX}8&3\\10&5\end{bmatrix}\), X=\(\begin{bmatrix}x\\y\end{bmatrix}\), B=\(\begin{bmatrix}70\\90\end{bmatrix}\)

It can be expressed in the form of AX=B

⇒X=A^-1 B

We know that, A^-1=\(\frac{1}{|A|}\) adj A=\(\frac{1}{10} \begin{bmatrix}5&-3\\-10&8\end{bmatrix}\)

∴X=\(\frac{1}{10} \begin{bmatrix}5&-3\\-10&8\end{bmatrix}\begin{bmatrix}70\\90\end{bmatrix}\)=\(\frac{1}{10} \begin{bmatrix}5×70-3×90\\-10×70+8×90\end{bmatrix}\)=\(\frac{1}{10} \begin{bmatrix}300-270\\-700+720\end{bmatrix}\)=\(\frac{1}{10} \begin{bmatrix}30\\20\end{bmatrix}\)

∴x=3, y=2.

i.e. The cost of apples is Rs 3 PER kg and the cost of ORANGES is Rs 2 per kg.

Correct option is (c) x-y=4

The best I can EXPLAIN: Let C(x,y) be a point on the line AB. THUS, the points A(5,1), B(4,0), C(x,y) are collinear. HENCE, the area of the triangle formed by these points will be 0.

⇒Δ=\(\frac{1}{2}\begin{Vmatrix}5&1&1\\4&0&1\\x&y&1\end{Vmatrix}\)=0

Applying R1→R1-R2

\(\frac{1}{2}\begin{Vmatrix}1&1&0\\4&0&1\\x&y&1\end{Vmatrix}\)=0

Expanding along R1, we get

=\(\frac{1}{2}\) {1(0-y)-1(4-x)}=0

=\(\frac{1}{2}\) {-y-4+x}=0

⇒x-y=4.

Right CHOICE is (c) 0

The best EXPLANATION: The given determinant is f(r) = \(\begin{vmatrix}2R & x & N(n + 1) \\(6r^2 – 1) & y & n^2(2n + 3) \\(4r^3 – 2nr) & z & n^3(n + 1) \end {vmatrix}\)

Now, r = 1Σ^n (2r) = 2[(n(n + 1))/2]……….(1)

= n^2 + n

r = 1Σ^n(6r^2 – 1) = 6[((n + 1)(2n + 1))/6] – n……….(2)

= n(2n^2 + 2n + n + 1) – n

= 2n^3 + 2n^2 + n^2 + n – n

= 2n^3 + 3n^2

= r = 1Σ^n(4r^3 – 2nr) = n^3 (n + 1)……….(3)

From (1), (2) and (3) we get

r = 1Σ^n f(x) = 0

Right option is (d) -(a^3 + b^3)^2

The BEST I can explain: GIVEN,\(\begin{vmatrix}2ab & a^2 & b^2 \\a^2 & b^2 & 2ab \\b^2 & 2ab & a^2 \end {vmatrix}\)

Using C1 = C1 + C2 + C3

= \(\begin{vmatrix}a^2 + b^2 + 2ab & a^2 & b^2 \\a^2 + b^2 + 2ab & b^2 & 2ab \\a^2 + b^2 + 2ab & 2ab & a^2 \end {vmatrix}\)

= (a + b)^2\(\begin{vmatrix}1 & a^2 & b^2 \\1 & b^2 & 2ab \\1 & 2ab & a^2 \end {vmatrix}\)

= (a + b)^2\(\begin{vmatrix}1 & a^2 & b^2 \\1 & b^2 – a^2 & 2ab – b^2 \\0 & 2ab – a^2 & a^2 – b^2 \end {vmatrix}\)

= (a + b)^2[(b^2 – a^2)(a^2 – b^2) – (2ab – b^2)( 2ab – a^2)]

= -(a + b)^2[(a^2 – b^2)^2 + 4a^2b^2 – 2ab(a^2 + b^2) + a^2 b^2)]

= –(a + b)^2[(a^2+b^2)^2 – 2(a^2+b^2) (ab)+(ab)^2]

= –(a + b)^2(a^2 + b^2 – ab)^2

= –[(a + b)^2(a^2 + b^2 – ab)^2]^2

= –(a^3 + b^3)^2

Correct choice is (a) TRUE

Explanation: The given STATEMENT is true. A SQUARE matrix A is said to be singular if |A|=0 and non-singular if A≠0.

The CORRECT choice is (b) 2A{(b-c)(c-a+b)}

For explanation: Δ=\(\begin{vmatrix}b-c&b&c\\a&c-a&c\\a&b&a-b\end{vmatrix}\)

APPLYING C2→C2-C3

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

Applying C1→C1-C2

Δ=\(\begin{vmatrix}0&b-c&c\\2a&-a&c\\2a&-a&a-b\end{vmatrix}\)

Applying R2→R2-R3

Δ=\(\begin{vmatrix}0&b-c&c\\0&0&c-a+b\\2a&-a&a-b\end{vmatrix}\)

Expanding ALONG C1, we get

Δ=2a{(b-c)(c-a+b)}

Right ANSWER is (a) The value of determinant changes if all of its rows and columns are interchanged

For explanation I WOULD say: The value of determinant remains UNCHANGED if all of its rows and columns are interchanged i.e. |A|=|A’|, where A is a square matrix and A’ is the transpose of the matrix A.

The correct OPTION is (b) A must be singular MATRIX

Explanation: The matrix should not be a singular matrix. A SQUARE matrix is said to be singular |A|=0.

We know that, A^-1=\(\FRAC{1}{|A|}\) adj A,

HENCE, if |A|=0 the inverse of the matrix does not exist.

The correct answer is (B) (a + BX + CX^2)\(\begin{vmatrix}1 & b & C \\x^2 & c & a \\x & a & b \end {vmatrix}\)

Easy explanation: We have, \(\begin{vmatrix}a & b & c \\b & c & a \\c & a & b \end {vmatrix}\)

As, x^3 = 1,

= \(\begin{vmatrix}a & bx & cx^2 \\b & cx & ax^2 \\c & ax & bx^2\end {vmatrix}\)

Replacing the 1^st COLUMN by C1 + C2 + C3 we get,

= \(\begin{vmatrix}a + bx + cx^2 & bx & cx^2 \\ b + cx + ax^2 & cx & ax^2 \\c + ax + bx^2 & ax & bx^2\end {vmatrix}\)

As, x^3 = 1 so, x^4 = x^3 * x = x

= \(\begin{vmatrix}a + bx + cx^2 & b & c \\x^2 (a + bx + cx^2) & c & a \\x(a + bx + cx^2) & a & b \end {vmatrix}\)

= (a + bx + cx^2)\(\begin{vmatrix}1 & b & c \\x^2 & c & a \\x & a & b \end {vmatrix}\)

The CORRECT choice is (B) -20

The explanation: \(\begin{vmatrix}2&5\\3&x\end{vmatrix}\)=\(\begin{vmatrix}x&-1\\5&3\end{vmatrix}\)

⇒2x-15=3x+5

⇒x=-20

The CORRECT choice is (c) 240

The explanation is: Expanding ALONG R1, we get

Δ=\(\BEGIN{vmatrix}3&-1&3\\6&-5&4\\3&-2&3\end{vmatrix}\)

Δ=3\(\begin{vmatrix}-5&45\\-2&3\end{vmatrix}\)-(-1)\(\begin{vmatrix}6&4\\3&3\end{vmatrix}\)+3\(\begin{vmatrix}6&-5\\3&-2\end{vmatrix}\)

Δ=3(-15+90)+(18-12)+3(-12+15)

Δ=3(75)+6+9=240.

Correct choice is (a) 2y-x=0

The best explanation: Let C(x,y) be a point on the line AB. Thus, the points A(2,1), B(6,3), C(x,y) are collinear. HENCE, the AREA of the triangle FORMED by these points will be 0.

⇒Δ=\(\frac{1}{2}\BEGIN{Vmatrix}2&1&1\\6&3&1\\x&y&1\end{Vmatrix}\)=0

Expanding along C3, we get

\(\frac{1}{2}\) {1(6y-3x)-1(2y-x)+1(6-6)}=0

\(\frac{1}{2}\) {6y-3x-2y+x}=\(\frac{1}{2}\) {4y-2x}=0

⇒2y-x=0

Right option is (B) 0

The BEST I can explain: Δ=\(\begin{vmatrix}1&0&1\\0&0&1\\1&0&1\end{vmatrix}\)

Δ=1\(\begin{vmatrix}0&1\\0&1\end{vmatrix}\)-0\(\begin{vmatrix}0&1\\1&1\end{vmatrix}\)+1\(\begin{vmatrix}0&0\\1&0\end{vmatrix}\)

Δ=1(0-0)-0(0-1)+1(0-0)

Δ=0-0+0=0.

Correct OPTION is (c) \(\FRAC{1}{432}\) \(\BEGIN{bmatrix}-27&6\\9&14\end{bmatrix}\)

The explanation: Given that, A=\(\begin{bmatrix}-8&2\\6&-3\end{bmatrix}\) and B=\(\begin{bmatrix}2&1\\1&7\end{bmatrix}\)

∴AB=\(\begin{bmatrix}-8×2+2×1&-8×1+2×7\\6×2+(-3)×1&6×1+(-3)×7\end{bmatrix}\)=\(\begin{bmatrix}-14&6\\9&27\end{bmatrix}\)

adj(AB)=\(\begin{bmatrix}27&-6\\-9&-14\end{bmatrix}\)

|AB|=27×(-14)-(-9)×(-6)=-378-54=-432

(AB)^-1=\(\frac{1}{|AB|}\) adj AB=\(\frac{1}{-432} \begin{bmatrix}27&-6\\-9&-14\end{bmatrix}\)=\(\frac{1}{432} \begin{bmatrix}-27&6\\9&14\end{bmatrix}\).

The correct choice is (d) Δ=\(\frac{1}{2}\begin{Vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{Vmatrix}\)

Easy explanation: The area of a triangle with VERTICES (x1,Y1), (X2,y2), (X3,y3) is given by

Δ=\(\frac{1}{2}\begin{Vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{Vmatrix}\).

Correct choice is (b) a, b, c are in G.P

Easiest explanation: Here, f(x) = \(\begin{vmatrix}1 & a & bc \\1 & b & CA \\1 & c & AB \END {vmatrix}\)

Multiplying and DIVING by abc,

= (1/abc) \(\begin{vmatrix}a & a^2 & abc \\b & b^2 & abc \\c & c^2& abc \end {vmatrix}\)

= \(\begin{vmatrix}1 & a & a^2 \\1 & b & b^2 \\1 & c & c^2 \end {vmatrix}\)

= (a – b)(b – c)(c – a)

Correct option is (b) ABC is an isosceles TRIANGLE

Explanation: When a = b or b = C or c = a the DETERMINANT REDUCES to 0

It is not necessary that a = b = c for determinant to be 0

Therefore, the triangle is isosceles.

Right CHOICE is (d) -2

The best EXPLANATION: Given that, A=\(\BEGIN{BMATRIX}9&8\\7&6\end{bmatrix}\)

⇒Δ=\(\begin{vmatrix}9&8\\7&6\end{vmatrix}\)=9(6)-7(8)=54-56=-2

The correct ANSWER is (c) -8

The explanation: EXPANDING ALONG the first row, we get

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

=5(4-6)-4(3-5)+3(18-20)

=5(-2)-4(-2)+3(-2)=-10+8-6=-8.

Right ANSWER is (C) \(\BEGIN{bmatrix}4&0\\0&4\end{bmatrix}\)

Easy explanation: GIVEN that, A=\(\begin{bmatrix}1&0\\9&4\end{bmatrix}\)

We know that, A(adj A)=(adj A)A=|A|I

∴|A|=4-0=4

⇒A(adj A)=|A|I=\(\begin{bmatrix}4&0\\0&4\end{bmatrix}\).

The CORRECT answer is (b) \(\frac{1}{|A|}\) adj A

The EXPLANATION is: The formula for calculating the inverse of the matrix is GIVEN by

A^-1=\(\frac{1}{|A|}\) adj A, where |A| is the DETERMINANT of the matrix and adj A is the adjoint of the matrix.

Correct ANSWER is (a) 4

Explanation: Put the VALUE of x, y, z = 1

Thus, putting the value of x = 1, y = 1 and z = 1 on both sides, we get

\(\BEGIN{vmatrix}2 & 1 & 1 \\1 & 2 & 1 \\1 & 1 & 2 \end {vmatrix}\) = k

So, solving the determinant we get k = 4.

Correct ANSWER is (c) \(\BEGIN{bmatrix}4&-5\\-3&1\end{bmatrix}\)

The EXPLANATION: We have A11=(-1)^1+1 4=4

A12=(-1)^1+2 3=-3

A21=(1)^2+1 5=-5

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

∴adj A=\(\begin{bmatrix}A_{11}&A_{21}\\A_{12}&A_{22}\end{bmatrix}\)=\(\begin{bmatrix}4&-5\\-3&1\end{bmatrix}\).

Right answer is (d) 0

Explanation: Δ=\(\begin{vmatrix}4&8&12\\6&12&18\\7&14&21\end{vmatrix}\)

Taking 4, 6 and 7 from R1, R2, R3 respectively

Δ=4×6×7\(\begin{vmatrix}1&2&3\\1&2&3\\1&2&3\end{vmatrix}\)

Since the elements of all rows are IDENTICAL, the determinant is ZERO.

Right option is (d) –(a^3 + b^3 + c^3 – 3abc)

The explanation: Given, \(\begin{vmatrix}a & b & c \\b & c & a \\c & a & b \end {vmatrix}\)

Replacing R1 = R1 + R2 + R3

\(\begin{vmatrix}a + b + c & a + b + c & a + b + c \\b & c & a \\c & a & b \end {vmatrix}\)

= (a + b + c)\(\begin{vmatrix}1 & 1 & 1 \\b & c & a \\c & a & b \end {vmatrix}\)

Replacing 2^nd column by C2 – C1 and 3^rd column by C3 – C1

= (a + b + c)\(\begin{vmatrix}1 & 0 & 0 \\b & c-b & a-b \\c & a-c & b-c \end {vmatrix}\)

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

= (a + b + c)(bc – b^2 – c^2 + bc + a^2 + AC + ab – bc)

= -(a + b + c)(a^2 + b^2 + c^2 – ab – bc – ac)

= -(a^3 + b^3 + c^3 – 3abc)

Right answer is (a) True

To EXPLAIN I would say: The given STATEMENT is true.

Expanding the determinant Δ=\(\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33} \end{vmatrix}\) along R1, we get

Δ=(-1)^1+1 a11 \(\begin{vmatrix}a_{22}&a_{23}\\a_{32}&a_{33} \end{vmatrix}\)+(-1)^1+2 a12 \(\begin{vmatrix}a_{21}&a_{23}\\a_{31}&a_{33} \end{vmatrix}\)+(-1)^1+3 a13 \(\begin{vmatrix}a_{21}&a_{22}\\a_{31}&a_{32} \end{vmatrix}\)

Δ=a11 A11+a21 A21+A31 A31, where AIJ is the cofactor of aij.

The CORRECT ANSWER is (c) 0

For EXPLANATION: The area of the triangle with VERTICES (2,3), (4,1), (5,0) is GIVEN by

Δ=\(\frac{1}{2}\begin{Vmatrix}2&3&1\\4&1&1\\5&0&1\end{Vmatrix}\)

Applying R2→R2-R3

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

Expanding along R2, we get

Δ=\(\frac{1}{2}\) {-(-1)(3-0)+1(2-5)}

Δ=\(\frac{1}{2}\) (0-0)=0.

Correct CHOICE is (B) 0

For explanation: Given, \(\begin{vmatrix}1 & 1 & 1 \\x & y & z \\x^3 & y^3 & z^3\end {vmatrix}\)

Operating, C1 = C1 – C2 and C2 = C2 – C3

= \(\begin{vmatrix}1 & 1 & 1 \\x – y & y – z & y \\x^3 – y^3 & y^3 – z^3 & z^3\end {vmatrix}\)

Expanding by the 1^st row,

= (x – y)(y^3 – z^3) – (y – z)(x^3 – y^3)

= (x – y)(y – z)[(y^2 + yz + z^2) – (x^2 + xy + y^2)]

= (x – y)(y – z)(z – x)(x + y + z)

As, x + y + z = 0

= 0

The correct choice is (c) (X – a)(y – b)(z – c)(\(\frac{x}{x-a} + \frac{b}{y – b} + \frac{c}{z-c}\) – 2)

To elaborate: Given, \(\begin{vmatrix}x & b & c\\a & y & c\\a & b & z \END {vmatrix}\) = \(\begin{vmatrix}x & b & c\\a – x & y – b & 0\\0 & b – y & z – c \end {vmatrix}\)

Applying the operation R2 = R2 – R1 and R3 = R3 – R2

= (x – a)(y – b)(z – c)\(\begin{vmatrix}x/(x – a) & b/(y – b) & c/(z – c)\\-1 & y 1 & 0\\0 & -1 & 1 \end {vmatrix}\)

Now, expanding the determinant we get,

= (x – a)(y – b)(z – c)(\(\frac{x}{x-a} + \frac{b}{y – b} + \frac{c}{z-c}\))

= (x – a)(y – b)(z – c)(\(\frac{x}{x-a} + \frac{b}{y – b} + \frac{c}{z-c}\) – 2)

This is because,

\(\frac{b}{y – b} + \frac{c}{z-c} = \frac{y-(y-b)}{y-b} + \frac{z-(z-c)}{z-c} = \frac{y}{y-b}\) – 1 + \(\frac{z}{z-c}\) – 1 = \(\frac{y}{y-b} + \frac{z}{z-c}\) – 2

Right option is (C) 2\(\begin{vmatrix}1 & cosx & sinx \\1 & cosy & SINY \\1 & cosz & SINZ \end {vmatrix}\)

For explanation I would say: Let, a = cosx, b = cosy, c = cosz, p =sinx, Q = siny and r = sinz

So, \(\begin{vmatrix}1 & a – p & a + p \\1 & b – q & b + q \\1 & c – r & c + r\end {vmatrix}\)

Making C3 = C3 + C2

= \(\begin{vmatrix}1 & a – p & 2a \\1 & b – q & 2B \\1 & c – r & 2c \end {vmatrix}\)

= 2\(\begin{vmatrix}1 & a – p & a \\1 & b – q & b \\1 & c – r & c \end {vmatrix}\)

Making C2 = C2 – C3

= -2\(\begin{vmatrix}1 & p & a \\1 & q & b \\1 & r & c \end {vmatrix}\)

Interchanging 2^nd and 3^rd column, we get,

2\(\begin{vmatrix}1 & a & p \\1 & b & q \\1 & c & r \end {vmatrix}\)

= 2\(\begin{vmatrix}1 & cosx & sinx \\1 & cosy & siny \\1 & cosz & sinz \end {vmatrix}\)

The CORRECT option is (a) 0

Easy 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.

The correct OPTION is (a) cos^2⁡θ

The best I can explain: EXPANDING along R1, we get

∆=-sinθ(sinθ)-(-1)1=-sin^2⁡θ+1=cos^2⁡θ.