JEE Mathematics Determinants MCQs Set 13

Answer: (a) independent of \( \theta \)

Explanation:
1. Expanding along the first row: \( x(-x^2-1)-\sin\theta(-x\sin\theta-\cos\theta)+\cos\theta(-\sin\theta+x\cos\theta) \).
2. The \( \theta \) terms: \( x\sin^2\theta+\sin\theta\cos\theta-\sin\theta\cos\theta+x\cos^2\theta=x \), so the value is \( -x^3-x+x=-x^3 \), free of \( \theta \).

Teacher's Note:
1. Expand and use \( \sin^2\theta+\cos^2\theta=1 \).

Answer: (d) 1

Explanation:
1. Expanding along the second row: \( \sin\beta\,(\cos^2\alpha\sin\beta+\sin^2\alpha\sin\beta)+\cos\beta\,(\cos^2\alpha\cos\beta+\sin^2\alpha\cos\beta) \).
2. \( =\sin^2\beta+\cos^2\beta=1 \).

Teacher's Note:
1. This is a rotation matrix, so its determinant is 1.

Answer: (c) \( \pm6 \)

Explanation:
1. Left side \( =2x^2-40 \) and right side \( =18+14=32 \).
2. So \( 2x^2=72 \), \( x^2=36 \) and \( x=\pm6 \).

Teacher's Note:
1. Expand both determinants and equate.

Answer: (d) \( \frac ab \) is one of the cube roots of \( -1 \)

Explanation:
1. Expanding: \( a(a^2)-b(0-b^2)=a^3+b^3 \).
2. So \( a^3+b^3=0 \), that is \( \left(\frac ab\right)^3=-1 \), and \( \frac ab \) is a cube root of \( -1 \).

Teacher's Note:
1. The determinant equals \( a^3+b^3 \).

Answer: (c) \( |3A|=27|A| \)

Explanation:
1. For a \( 3\times3 \) matrix, \( |kA|=k^3|A| \) because each of the three rows is multiplied by k.
2. So \( |3A|=27|A| \); indeed \( |A|=4 \) and \( |3A|=108 \).

Teacher's Note:
1. Each row contributes a factor k.

Answer: (c) 0

Explanation:
1. Add column 2 to column 3: the third column becomes \( ab+bc+ca \) in every row.
2. Columns 1 and 3 are then proportional, so the determinant is 0.

Teacher's Note:
1. Use \( C_3\to C_3+C_2 \).

Answer: (c) \( 4a^2b^2c^2 \)

Explanation:
1. Take out a, b, c from the rows: \( abc\begin{vmatrix}-a&b&c\\a&-b&c\\a&b&-c\end{vmatrix} \); then take a, b, c from the columns to get \( a^2b^2c^2\begin{vmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{vmatrix} \).
2. The last determinant equals 4, so the value is \( 4a^2b^2c^2 \).

Teacher's Note:
1. Factor a, b, c out of the rows and then the columns.

Answer: (b) \( (x-y)(y-z)(z-x)(xy+yz+zx) \)

Explanation:
1. The determinant vanishes when any two of x, y, z are equal, so \( (x-y)(y-z)(z-x) \) is a factor. The determinant has degree 5, so the other factor is a symmetric expression of degree 2.
2. Put \( x=0,y=1,z=2 \): the determinant is \( \begin{vmatrix}0&0&2\\1&1&0\\2&4&0\end{vmatrix}=4 \) and \( (x-y)(y-z)(z-x)=2 \), so the other factor is 2 here, which equals \( xy+yz+zx \) (and not \( x+y+z=3 \)).
3. Hence the value is \( (x-y)(y-z)(z-x)(xy+yz+zx) \).

Teacher's Note:
1. An alternating determinant contains the product of differences.

Answer: (c) \( k^2(3y+k) \)

Explanation:
1. Add all rows to the first row: the first row becomes \( 3y+k \) in every entry.
2. Taking out \( (3y+k) \) and subtracting the first row from the others leaves \( (3y+k)k^2 \).

Teacher's Note:
1. Add rows, then factor out \( 3y+k \).

Answer: (d) \( (a+b+c)^3 \)

Explanation:
1. Add all rows to the first row: the first row becomes \( a+b+c \) in every entry. Take out \( (a+b+c) \).
2. Subtract column 1 from columns 2 and 3 and expand: the remaining determinant is \( (a+b+c)^2 \), so the value is \( (a+b+c)^3 \).

Teacher's Note:
1. A degree-3 symmetric result.

Answer: (b) \( (5x+4)(4-x)^2 \)

Explanation:
1. Add all rows to the first row: it becomes \( 5x+4 \) in every entry.
2. Take out \( 5x+4 \); subtracting the first row from the others leaves \( (4-x)^2 \). So the value is \( (5x+4)(4-x)^2 \).

Teacher's Note:
1. Row-sum trick again.

Answer: (b) 2

Explanation:
1. Add all columns to the first column: it becomes \( 2(x+y+z) \) in each row.
2. Take out \( 2(x+y+z) \) and simplify; the remaining determinant is \( (x+y+z)^2 \), so \( k=2 \).

Teacher's Note:
1. Column-sum trick.

Answer: (d) \( -2(x^3+y^3) \)

Explanation:
1. Add all rows to the first: it becomes \( 2(x+y) \) in every entry; take out \( 2(x+y) \).
2. The remaining determinant equals \( -(x^2-xy+y^2) \), so the value is \( -2(x+y)(x^2-xy+y^2)=-2(x^3+y^3) \).

Teacher's Note:
1. Use the identity \( x^3+y^3=(x+y)(x^2-xy+y^2) \).

Answer: (d) \( xy \)

Explanation:
1. Subtract row 1 from rows 2 and 3: the rows become \( (0,y,0) \) and \( (0,0,x) \).
2. The determinant is \( 1\cdot(y\cdot x-0)=xy \).

Teacher's Note:
1. Subtract the first row from the other two.

Answer: (d) none of the above

Explanation:
1. Add column 1 to column 3: the third column becomes \( \alpha+\beta+\gamma \) in every row.
2. Taking it out leaves a Vandermonde-type determinant equal to \( \pm(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha) \), so the value contains the extra factor \( \alpha+\beta+\gamma \).
3. Option (a) lacks this factor, and options (b) and (c) are symmetric whereas the determinant changes sign when two variables are swapped. So none of (a) to (c) is correct.

Teacher's Note:
1. The extra factor \( \alpha+\beta+\gamma \) rules out options (a) to (c).

Answer: (a) \( (1+pxyz)(x-y)(y-z)(z-x) \)

Explanation:
1. Split the third column: the determinant is \( D_1+pD_2 \), where \( D_1=\begin{vmatrix}x&x^2&1\\y&y^2&1\\z&z^2&1\end{vmatrix} \) and \( D_2=\begin{vmatrix}x&x^2&x^3\\y&y^2&y^3\\z&z^2&z^3\end{vmatrix} \).
2. Moving the last column of \( D_1 \) to the front (two swaps) gives the Vandermonde determinant \( (y-x)(z-x)(z-y)=(x-y)(y-z)(z-x) \), and \( D_2=xyz \) times the same Vandermonde.
3. So the value is \( (1+pxyz)(x-y)(y-z)(z-x) \).

Teacher's Note:
1. Split the last column and use the Vandermonde pattern.

Answer: (c) 3

Explanation:
1. Every row adds up to \( a+b+c \), so add all columns to column 1 and take out \( (a+b+c) \).
2. With the new column of ones, apply \( R_2-R_1 \) and \( R_3-R_1 \): the rows become \( (0,\,a+2b,\,a-b) \) and \( (0,\,a-c,\,a+2c) \).
3. The determinant of the remaining part is \( (a+2b)(a+2c)-(a-b)(a-c)=3(ab+bc+ca) \), so \( k=3 \).

Teacher's Note:
1. Check with simple numbers to fix k.

Answer: (c) 1

Explanation:
1. Apply \( R_2\to R_2-2R_1 \) and \( R_3\to R_3-3R_1 \): the rows become \( (0,\,1,\,2+p) \) and \( (0,\,3,\,7+3p) \).
2. Expanding along the first column: \( 1\cdot[(7+3p)-3(2+p)]=1 \).

Teacher's Note:
1. Reduce with \( R_2-2R_1 \) and \( R_3-3R_1 \).

Answer: (c) \( a^3+b^3+c^3-3abc \)

Explanation:
1. The determinant is a cubic expression in a, b, c; test it at simple values.
2. At \( a=1,b=c=0 \) it equals 1; at \( a=b=1,c=0 \) it equals 2; at \( a=b=c=1 \) it equals 0.
3. These match \( a^3+b^3+c^3-3abc \) (values 1, 2, 0) and no other option.

Teacher's Note:
1. Test with a=1, b=c=0 to distinguish the options.

Answer: (b) \( 9y^2(x+y) \)

Explanation:
1. Add all rows to the first: it becomes \( 3(x+y) \) in every entry; take out \( 3(x+y) \).
2. The remaining determinant equals \( 3y^2 \), so the value is \( 9y^2(x+y) \).

Teacher's Note:
1. Use the row-sum trick.

Answer: (c) 0

Explanation:
1. Expanding the first determinant (the Vandermonde pattern): \( \Delta=(y-x)(z-x)(z-y) \).
2. In \( \Delta_1 \) multiply columns 1, 2, 3 by x, y, z and divide by xyz. Then row 2 becomes \( xyz(1,1,1) \) and the determinant reduces to \( \begin{vmatrix}x&y&z\\1&1&1\\x^2&y^2&z^2\end{vmatrix} \).
3. Swapping the first two rows gives the Vandermonde determinant with a change of sign, so \( \Delta_1=-\Delta \) and \( \Delta+\Delta_1=0 \).

Teacher's Note:
1. The two determinants differ only by sign.

Answer: (b) \( (x-p)(x^2+px-2q^2) \)

Explanation:
1. Subtract row 2 from row 1: the first row becomes \( (x-p,\ p-x,\ 0) \); take out \( (x-p) \).
2. Expand the remaining determinant: \( \begin{vmatrix}1&-1&0\\p&x&q\\q&q&x\end{vmatrix}=x^2+px-2q^2 \).
3. So the value is \( (x-p)(x^2+px-2q^2) \).

Teacher's Note:
1. Row subtraction exposes the factor \( x-p \).

Answer: (a) 1, 3

Explanation:
1. Expanding: \( x(x^2-2)-2(x-3)+3(2-3x)=x^3-13x+12 \).
2. Factor: \( x^3-13x+12=(x+4)(x-1)(x-3) \), so the other roots are 1 and 3.

Teacher's Note:
1. Expand and factor using the known root.

Answer: (a) \( 2x^3y^3z^3 \)

Explanation:
1. Take x, y, z out of rows 1, 2, 3 respectively, then \( x^2,y^2,z^2 \) out of columns 1, 2, 3: the determinant becomes \( x^3y^3z^3\begin{vmatrix}0&1&1\\1&0&1\\1&1&0\end{vmatrix} \).
2. The last determinant equals \( 0-1(0-1)+1(1-0)=2 \), so the value is \( 2x^3y^3z^3 \).

Teacher's Note:
1. Factor each row and column to reach the 0-1 determinant.

Answer: (c) \( 3(x+y+z)(xy+yz+zx) \)

Explanation:
1. Add all rows to the first row: it becomes \( x+y+z \) in every entry; take out \( x+y+z \).
2. The remaining determinant evaluates to \( 3(xy+yz+zx) \), so the value is \( 3(x+y+z)(xy+yz+zx) \).

Teacher's Note:
1. Check with \( x=y=z=1 \): the determinant is \( 3\cdot3\cdot3=27 \).

Answer: (c) \( 0,\pm\sqrt{\frac32(a^2+b^2+c^2)} \)

Explanation:
1. Each row sums to \( a+b+c-x=-x \), so add all columns to the first column and take out \( -x \).
2. The remaining \( 2\times2 \) part gives \( x^2-3(a^2+ab+b^2) \), and \( 3(a^2+ab+b^2)=\frac32(a^2+b^2+c^2) \) when \( c=-a-b \).
3. So \( x=0 \) or \( x=\pm\sqrt{\frac32(a^2+b^2+c^2)} \).

Teacher's Note:
1. Use \( a+b+c=0 \) after taking out \( -x \).

Answer: (b) \( a^3\left(1+\frac3a\right) \)

Explanation:
1. Add all rows to the first row: it becomes \( a+3 \) in every entry.
2. Take out \( a+3 \) and subtract: the value is \( a^2(a+3)=a^3\left(1+\frac3a\right) \).

Teacher's Note:
1. Row-sum trick.