JEE Mathematics Determinants MCQs Set 15

Answer: (b) 5

Explanation:
1. Take \( x,y,z \) common from \( R_1,R_2,R_3 \) respectively. Every row then has the sum \( 3+\frac1x+\frac1y+\frac1z \).
2. Apply \( C_1\to C_1+C_2+C_3 \) and take \( 3+\frac1x+\frac1y+\frac1z \) common from \( C_1 \), which becomes all 1s.
3. Apply \( R_2\to R_2-R_1,\ R_3\to R_3-R_1 \). The determinant becomes \( \begin{vmatrix}1&0&0\\ *&1&-1\\ *&0&2\end{vmatrix}=2 \).
4. So the value is \( 2xyz\left(3+\frac1x+\frac1y+\frac1z\right) \), giving \( p=2,\ q=3 \) and \( p+q=5 \).

Teacher's Note:
1. Factor out the variables that appear in the entries so the row sums become equal, then use \( C_1\to C_1+C_2+C_3 \).
2. Check with \( x=y=z=1 \): the determinant is 12, and \( 2\cdot1\cdot(3+3)=12 \).

Answer: (b) \( a,b,c \) are in G.P.

Explanation:
1. A homogeneous system has a non-trivial solution exactly when the coefficient determinant is zero: \( \begin{vmatrix}a\alpha+b&a&b\\ b\alpha+c&b&c\\0&a\alpha+b&b\alpha+c\end{vmatrix}=0 \).
2. Apply \( R_3\to R_3-\alpha R_1-R_2 \). The third row becomes \( (-(a\alpha^2+2b\alpha+c),\,0,\,0) \).
3. Expanding along \( R_3 \): \( \Delta=-(a\alpha^2+2b\alpha+c)(ac-b^2)=0 \).
4. So either \( b^2=ac \) (a, b, c in G.P.) or \( \alpha \) is a root of \( ax^2+2bx+c=0 \). Only the G.P. condition among the options guarantees this; option (d) has the wrong sign of the middle term.

Teacher's Note:
1. Reduce a row using the other rows and a multiple of the unknown parameter to expose a factor.
2. Check each option against the factored condition, watching for sign changes such as \( -2b \) versus \( +2b \).

Answer: (d) 0

Explanation:
1. The condition can be written as \( (ax-b)^2+(bx-c)^2+(cx-d)^2\le0 \). A sum of squares is non-positive only if each term is zero.
2. Hence \( \frac ba=\frac cb=\frac dc=x \), so \( a,b,c,d \) are in G.P. and \( \log a,\log b,\log c \) are in A.P.
3. Now every column of the determinant is in A.P. (for example \( p,q,r \) with \( q-p=r-q \)), so \( R_1-2R_2+R_3 \) gives a zero row.
4. Therefore the determinant equals 0.

Teacher's Note:
1. Writing a quadratic inequality as a sum of squares is a standard route to equalities among the coefficients.
2. Whenever the entries of columns (or rows) form A.P.s, a suitable combination of rows gives a zero row.

Answer: (a) 10

Explanation:
1. Only the first column depends on \( n \), so by linearity the sum is the same determinant with the first column replaced by its sums: \( \sum1=k \), \( \sum2n=k(k+1) \), \( \sum(2n-1)=k^2 \).
2. So \( \sum U_n=\begin{vmatrix}k&k&k\\k(k+1)&k^2+k+1&k^2+k\\k^2&k^2&k^2+k+1\end{vmatrix} \).
3. Apply \( C_2\to C_2-C_1 \): the second column becomes \( (0,1,0) \). Expanding along it: \( k(k^2+k+1)-k\cdot k^2=k^2+k \).
4. \( k(k+1)=110 \Rightarrow k=10 \) (k is positive).

Teacher's Note:
1. If only one column depends on the summation index, the sum of determinants is the determinant of the summed column.
2. After forming the sum, use column operations to create a column with a single non-zero entry.

Answer: (a) \( f(5)=3720 \)

Explanation:
1. Use \( {}^{n}P_n=n! \) and \( {}^{n}C_n=1 \). The determinant is \( \begin{vmatrix}n&n+1&n+2\\n!&(n+1)!&(n+2)!\\1&1&1\end{vmatrix} \).
2. Apply \( C_3\to C_3-C_2 \) and \( C_2\to C_2-C_1 \): we get \( \begin{vmatrix}n&1&1\\n!&n\cdot n!&(n+1)(n+1)!\\1&0&0\end{vmatrix} \).
3. Expanding along the last row: \( f(n)=(n+1)(n+1)!-n\cdot n!=n!\,(n^2+n+1) \).
4. \( f(5)=120\times31=3720 \), while \( f(4)=24\times21=504 \), so only option (a) is true.

Teacher's Note:
1. Recognise that \( {}^{m}C_m=1 \) and \( {}^{m}P_m=m! \) before expanding.
2. Evaluate each option separately; one wrong numerical value (here \( f(4) \)) can eliminate an option.

Answer: (a) \( (x-1)^{n-1}(x+n-1) \)

Explanation:
1. Every row has the same sum \( x+n-1 \). Apply \( C_1\to C_1+C_2+\dots+C_n \) and take \( (x+n-1) \) out of \( C_1 \).
2. Now \( C_1 \) is all 1s. Apply \( R_i\to R_i-R_1 \) for \( i\ge2 \); the matrix becomes upper triangular with diagonal \( 1,(x-1),\dots,(x-1) \).
3. The determinant is \( (x+n-1)(x-1)^{n-1} \).

Teacher's Note:
1. Equal row sums are the signal to add all columns into one.
2. Test with \( n=2 \): \( x^2-1=(x-1)(x+1) \), which agrees with the formula.

Answer: (d) All of (i), (ii), (iii) and (iv)

Explanation:
1. Simplify the argument: \( (x+1)(x-2)-(x-1)^2=x^2-x-2-x^2+2x-1=x-3 \), so the first row is \( f(x-3),\,f(x+5),\,f(x-3) \).
2. Expanding along the first row: \( 90f(x-3)-100f(x+5)+10f(x-3)=0 \), that is \( f(x+5)=f(x-3) \).
3. Replacing \( x \) by \( x+3 \): \( f(x+8)=f(x) \), so \( f \) is periodic with period 8.
4. The differences 40, 120, 16 and 8 are all multiples of 8, so all four statements are true.

Teacher's Note:
1. Simplify the awkward argument inside the function first; it often collapses to something already present.
2. A relation \( f(x+k)=f(x) \) means every pair of inputs differing by a multiple of \( k \) gives equal values.

Answer: (b) (i), (ii) and (iv) only

Explanation:
1. Add all columns to \( C_1 \) and take \( (x+3a) \) common; then subtract \( R_1 \) from the other rows to get a triangular form.
2. \( f(x)=(x+3a)(x-a)^3 \).
3. \( f(2a)=5a\cdot a^3=5a^4 \) (true); \( f(3a)=6a\cdot8a^3=48a^4 \) (true); \( f(4a)=7a\cdot27a^3=189a^4\ne64a^4 \) (false); \( f(-a)=2a\cdot(-2a)^3=-16a^4 \) (true).

Teacher's Note:
1. For a matrix with equal off-diagonal entries, the determinant factorises as \( (x+(n-1)a)(x-a)^{n-1} \).
2. Substitute each given value in the factorised form instead of expanding.

Answer: (a) 7

Explanation:
1. Multiply \( R_1,R_2,R_3 \) by \( c,a,b \) respectively and divide by \( abc \): the determinant becomes \( \frac1{abc}\begin{vmatrix}a^2+b^2&c^2&c^2\\a^2&b^2+c^2&a^2\\b^2&b^2&c^2+a^2\end{vmatrix} \).
2. Apply \( R_1\to R_1-(R_2+R_3) \): the first row becomes \( (0,-2b^2,-2a^2) \).
3. Apply \( R_2\to R_2+\frac12R_1,\ R_3\to R_3+\frac12R_1 \) to get \( \frac1{abc}\begin{vmatrix}0&-2b^2&-2a^2\\a^2&c^2&0\\b^2&0&c^2\end{vmatrix} \).
4. Expanding: \( \frac1{abc}\left[2b^2(a^2c^2)-2a^2(-b^2c^2)\right]=\frac{4a^2b^2c^2}{abc}=4abc \), so \( \alpha=4 \) and \( [\alpha]+3=7 \).

Teacher's Note:
1. Clear denominators by multiplying each row by a suitable variable, and compensate outside the determinant.
2. Use a combination such as \( R_1-(R_2+R_3) \) to make zeros before expanding.

Answer: (a) 2

Explanation:
1. \( |A^{-1}|=\frac1{|A|} \) and \( |\operatorname{adj}M|=|M|^{n-1}=|M|^2 \) for a \( 3\times3 \) matrix.
2. \( |\operatorname{adj}(B^{-1})|=\frac1{|B|^2} \) and \( |\operatorname{adj}(2A^{-1})|=|2A^{-1}|^2=\left(\frac8{|A|}\right)^2=\frac{64}{|A|^2} \).
3. The product is \( \frac{64}{|A|^3|B|^2}=\frac{64}{(-8)(4)}=-2 \), so its absolute value is 2.

Teacher's Note:
1. Use \( |kM|=k^n|M| \), \( |M^{-1}|=1/|M| \) and \( |\operatorname{adj}M|=|M|^{n-1} \) for an \( n\times n \) matrix.
2. Take determinants of each factor separately and multiply.

Answer: (a) 7

Explanation:
1. In an A.P. with common difference \( d \), the differences \( a_1-a_3,\ a_2-a_4 \) equal \( -2d \), so the row operations \( R_1\to R_1-R_3,\ R_2\to R_2-R_3 \) turn the entries into multiples of \( d \).
2. Writing \( a_n=a_1+(n-1)d \) and expanding, every term containing \( a_1 \) cancels and \( |A|=-2d^4 \); in the same way \( |B|=-2d^4 \). For \( d=2 \) each equals \( -32 \).
3. \( N=(-32)(-32)=1024 \) and the digit sum is \( 1+0+2+4=7 \).

Teacher's Note:
1. For arithmetic progressions, row operations turn the entries into multiples of the common difference.
2. Direct numerical evaluation (for example with \( a_1=1 \)) is a quick check on the symbolic result.

Answer: (a) 1

Explanation:
1. Apply \( C_1\to C_1+C_2+C_3 \): every entry of \( C_1 \) becomes \( 1+x(a^2+b^2+c^2+2)=1 \), because \( a^2+b^2+c^2=-2 \).
2. Apply \( R_2\to R_2-R_1,\ R_3\to R_3-R_1 \): the determinant becomes \( \begin{vmatrix}1&(1+b^2)x&(1+c^2)x\\0&1-x&0\\0&0&1-x\end{vmatrix}=(1-x)^2 \).
3. \( (1-x)^2=0 \) has the single distinct root \( x=1 \).

Teacher's Note:
1. Adding all columns often produces a common factor or a constant column.
2. Count distinct roots, not multiplicity: a repeated root counts once.

Answer: (a) 3

Explanation:
1. For a non-trivial solution, \( \begin{vmatrix}a&\sin b&\cos b\\1&\cos b&\sin b\\-1&\sin b&-\cos b\end{vmatrix}=0 \).
2. Expanding: \( -a+\sin2b+\cos2b=0 \), so \( a=\sin2b+\cos2b=\sqrt2\sin\left(2b+\frac\pi4\right) \).
3. Hence \( -\sqrt2\le a\le\sqrt2 \), and the integers in this interval are \( -1,0,1 \): three values.

Teacher's Note:
1. Combine \( p\sin\theta+q\cos\theta \) into one sine term to find its range.
2. Count only integers in the closed interval \( [-\sqrt2,\sqrt2] \).

Answer: (b) \( (a-b)(b-c)(c-a)(a+b+c) \)

Explanation:
1. The basic determinant \( \begin{vmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{vmatrix}=(a-b)(b-c)(c-a) \).
2. Multiply \( C_1,C_2,C_3 \) by \( a,b,c \) and divide by \( abc \): the third row becomes \( abc,abc,abc \), so the determinant equals \( \begin{vmatrix}a&b&c\\a^3&b^3&c^3\\1&1&1\end{vmatrix} \), which is the basic determinant with the degree of one row raised by 1.
3. The result is the basic determinant times a cyclic linear expression: \( (a-b)(b-c)(c-a)(a+b+c) \).
4. Check with \( (a,b,c)=(0,1,2) \): the determinant is \( \begin{vmatrix}1&1&1\\0&1&4\\2&0&0\end{vmatrix}=6 \) and \( (-1)(-1)(2)(3)=6 \).

Teacher's Note:
1. Raising the degree of one row symmetrically multiplies the basic determinant by a cyclic expression of that degree.
2. Quick numeric substitution confirms the factor.

Answer: (c) \( (a-b)(b-c)(c-a)(ab+bc+ca) \)

Explanation:
1. Multiply the columns by \( a,b,c \) and divide by \( abc \) as before: the determinant equals \( \begin{vmatrix}a^2&b^2&c^2\\a^3&b^3&c^3\\1&1&1\end{vmatrix} \) (rows reordered).
2. That raises the degree of two rows of the basic determinant by one each, so the factor is a cyclic expression of degree 2: \( k_1(a^2+b^2+c^2)+k_2(ab+bc+ca) \).
3. Substituting \( (a,b,c)=(0,1,-1) \) and \( (0,1,2) \) gives \( k_1=0,\ k_2=1 \).
4. Hence the value is \( (a-b)(b-c)(c-a)(ab+bc+ca) \).

Teacher's Note:
1. When the degree rises by 2, the extra factor is a combination of \( \sum a^2 \) and \( \sum ab \); find the constants by substituting simple values.
2. Direct expansion of such determinants is long, so the structure argument saves time.

Answer: (a) only the solution \( x=1,y=1,z=1 \)

Explanation:
1. The coefficient determinant is \( \begin{vmatrix}1&1&1\\2&1&2\\1&-1&3\end{vmatrix}=1(3+2)-1(6-2)+1(-2-1)=5-4-3=-2\ne0 \).
2. A non-zero determinant means a unique solution.
3. Substituting \( x=y=z=1 \): \( 3=3,\ 5=5,\ 3=3 \). So the only solution is \( x=y=z=1 \).

Teacher's Note:
1. If \( |A|\ne0 \), the system \( AX=B \) has exactly one solution.
2. Verify by substitution when the solution looks simple.

Answer: (c) no solution

Explanation:
1. The coefficient determinant is \( \begin{vmatrix}1&1&1\\2&2&2\\1&-1&3\end{vmatrix}=0 \) because \( R_2=2R_1 \).
2. The first two equations imply \( 2(x+y+z)=6 \) but the second states \( 2(x+y+z)=7 \), a contradiction.
3. (Equivalently, \( (\operatorname{adj}A)B\ne0 \).) So the system is inconsistent and has no solution.

Teacher's Note:
1. If \( |A|=0 \) and \( (\operatorname{adj}A)B\ne0 \), the system is inconsistent.
2. Look for proportional left-hand sides with non-proportional right-hand sides.

Answer: (b) \( 2+\sin2\theta \)

Explanation:
1. Apply \( C_1\to C_1+C_2 \): the first column becomes \( (1+1,\,1+1,\,1) \) since \( \sin^2\theta+\cos^2\theta=1 \), i.e. \( (2,2,1) \).
2. Apply \( R_2\to R_2-R_1,\ R_3\to R_3-R_1 \): \( \begin{vmatrix}2&\cos^2\theta&\sin2\theta\\0&1&0\\-1&0&1\end{vmatrix} \).
3. Expanding along \( R_2 \): \( 1\cdot(2\cdot1-(\sin2\theta)(-1))=2+\sin2\theta \).

Teacher's Note:
1. Use \( \sin^2\theta+\cos^2\theta=1 \) to simplify a column sum.
2. Zeros created by row operations make the expansion a one-line calculation.

Answer: (a) \( \cos2\theta \)

Explanation:
1. \( \begin{bmatrix}1&\tan\theta\\-\tan\theta&1\end{bmatrix}^{-1}=\frac1{1+\tan^2\theta}\begin{bmatrix}1&-\tan\theta\\ \tan\theta&1\end{bmatrix} \).
2. The product is \( \frac1{1+\tan^2\theta}\begin{bmatrix}1-\tan^2\theta&-2\tan\theta\\2\tan\theta&1-\tan^2\theta\end{bmatrix}=\begin{bmatrix}\cos2\theta&-\sin2\theta\\ \sin2\theta&\cos2\theta\end{bmatrix} \).
3. Comparing entries: \( a=\cos2\theta \) and \( b=\sin2\theta \).

Teacher's Note:
1. Use \( \frac{1-\tan^2\theta}{1+\tan^2\theta}=\cos2\theta \) and \( \frac{2\tan\theta}{1+\tan^2\theta}=\sin2\theta \).
2. Rotation matrices multiply by adding angles; here the result is a rotation by \( 2\theta \).

Answer: (a) \( \sin2\theta \)

Explanation:
1. The inverse is \( \frac1{1+\tan^2\theta}\begin{bmatrix}1&-\tan\theta\\ \tan\theta&1\end{bmatrix} \).
2. The product equals \( \begin{bmatrix}\cos2\theta&-\sin2\theta\\ \sin2\theta&\cos2\theta\end{bmatrix} \) using the double-angle formulas in terms of \( \tan\theta \).
3. Comparing with \( \begin{bmatrix}a&-b\\b&a\end{bmatrix} \) gives \( b=\sin2\theta \).

Teacher's Note:
1. Write the double-angle formulas in terms of \( \tan\theta \) to match the entries.
2. Compare the matching positions, paying attention to signs.

Answer: (a) \( (3\cos\theta-\sin\theta)^2 \)

Explanation:
1. Apply \( R_3\to R_3-R_1 \): the third row becomes \( (0,\ \sin\theta-3\cos\theta,\ 0) \).
2. Expanding along \( R_3 \) the only term is \( (\sin\theta-3\cos\theta)\times(-1)^{3+2}\begin{vmatrix}1&1\\ \sin\theta&3\cos\theta\end{vmatrix} \).
3. The minor equals \( 3\cos\theta-\sin\theta \), so the value is \( -(\sin\theta-3\cos\theta)(3\cos\theta-\sin\theta)=(3\cos\theta-\sin\theta)^2 \).

Teacher's Note:
1. A row difference that leaves a single non-zero entry reduces the determinant to a \( 2\times2 \) minor.
2. Check the sign factor \( (-1)^{i+j} \) carefully.