JEE Mathematics Matrices MCQs Set 10

Answer: (c) Both (a) and (b) are correct

Explanation:
1. \( \operatorname{adj}(AB)=\operatorname{adj}B\ \operatorname{adj}A \), so \( |\operatorname{adj}(AB)|=|\operatorname{adj}B||\operatorname{adj}A| \).
2. Real numbers commute, so this equals \( |\operatorname{adj}A||\operatorname{adj}B| \) too; both (a) and (b) hold.

Teacher's Note:
1. The adjoint of a product reverses the order of the factors, but determinants are numbers and multiply commutatively.
2. Also |adj(AB)| = |AB| to the (n - 1) for an n x n matrix.

Answer: (a) \( |A|A \)

Explanation:
1. \( A^2\operatorname{adj}A=A\,(A\operatorname{adj}A)=A\,|A|I=|A|A \).

Teacher's Note:
1. Group the factors so that A times adj A appears and replace it by |A| I.
2. The result is a scalar multiple of A, not of I.

Answer: (b) \( d^{n-1} \)

Explanation:
1. \( A\operatorname{adj}A=dI\Rightarrow|A||\operatorname{adj}A|=d^n \).
2. \( d\cdot|\operatorname{adj}A|=d^n\Rightarrow|\operatorname{adj}A|=d^{n-1} \) (for \( d\ne0 \), and by continuity of the identity also when \( d=0 \)).

Teacher's Note:
1. Take determinants on both sides of A adj A = dI and remember that |dI| = d to the n.
2. For n = 3 the result is the square of the determinant.

Answer: (d) \( \operatorname{adj}A=I \)

Explanation:
1. From \( A\operatorname{adj}A=|A|I \) we get \( \operatorname{adj}A=|A|A^{-1} \) (a is true).
2. Inverting gives \( (\operatorname{adj}A)^{-1}=\frac1{|A|}A \) (b is true), and (c) is the standard determinant property.
3. The adjoint equals the identity only for special matrices, so (d) is not true in general.

Teacher's Note:
1. Check each option against the basic relation A adj A = |A| I.
2. An option of the form 'always equals the identity' is almost never true.

Answer: (c) \( \Delta^2 \)

Explanation:
1. The cofactor matrix is the transpose of the adjoint, so its determinant equals \( |\operatorname{adj}A|=\Delta^{3-1}=\Delta^2 \).

Teacher's Note:
1. Transposing does not change a determinant.
2. For a 3 x 3 matrix the determinant of the adjoint is the square of the determinant.

Answer: (c) \( \pm1 \)

Explanation:
1. The conditions say that the columns are orthonormal, so \( A^TA=I \).
2. \( |A|^2=1\Rightarrow|A|=\pm1 \).

Teacher's Note:
1. Orthonormal rows or columns mean the matrix is orthogonal.
2. The determinant of an orthogonal matrix is +1 or -1 (rotation or reflection).

Answer: (b) \( AB=O\Rightarrow|A|=0\ \text{or}\ |B|=0 \)

Explanation:
1. \( AB=O\Rightarrow|A||B|=0 \), so at least one of \( |A| \), \( |B| \) is zero; (b) is true and (a) is too strong.
2. (c) is false in general, and \( \operatorname{adj}(AB)=\operatorname{adj}B\operatorname{adj}A \) has the opposite order, so (d) is false.

Teacher's Note:
1. The determinant of a product is the product of the determinants, so the product is zero if either factor is zero.
2. Only one of the two matrices needs to be singular, so the statement with and is wrong.

Answer: (a) both are singular

Explanation:
1. If A were non-singular, \( A^{-1}AB=A^{-1}O\Rightarrow B=O \), contradicting \( B\ne O \).
2. Similarly B cannot be non-singular because \( ABB^{-1}=O\Rightarrow A=O \).
3. Hence both are singular.

Teacher's Note:
1. A non-singular factor can be cancelled, which would force the other matrix to be zero.
2. The given non-zero conditions therefore force both matrices to be singular.

Answer: (b) \( A^T \)

Explanation:
1. Orthogonal means \( AA^T=A^TA=I \).
2. Comparing with \( AA^{-1}=I \) gives \( A^{-1}=A^T \).

Teacher's Note:
1. The inverse of an orthogonal matrix is just its transpose, which is why such matrices are easy to invert.
2. A rotation matrix is the standard example.

Answer: (b) a symmetric matrix

Explanation:
1. \( (A^{-1})^T=(A^T)^{-1}=A^{-1} \) for symmetric A.
2. Hence \( A^{-1} \) is symmetric.

Teacher's Note:
1. Transposition and inversion commute.
2. The same argument shows that the inverse of a skew-symmetric matrix is skew-symmetric.

Answer: (d) does not exist

Explanation:
1. A skew-symmetric matrix of odd order has determinant \( 0 \) (because \( |A|=|A^T|=(-1)^n|A|=-|A| \)).
2. A singular matrix has no inverse, so the inverse does not exist.

Teacher's Note:
1. The determinant argument works for any odd n.
2. For even order, an inverse may exist and it is again skew-symmetric.

Answer: (c) \( |A|^8 \)

Explanation:
1. For an order-3 matrix \( X \): \( |\operatorname{adj}X|=|X|^2 \) and so \( |\operatorname{adj}(\operatorname{adj}X)|=|X|^{4} \).
2. With \( X=A^2 \): \( |X|=|A|^2 \), hence the required value is \( (|A|^2)^4=|A|^8 \).

Teacher's Note:
1. Apply the rule |adj X| = |X| to the (n - 1) twice, and then substitute X = A squared.
2. Order of the exponents: 2 for the first adjoint, 2 for the second, so 4 in total for X.

Answer: (a) a unique solution

Explanation:
1. Cramer's rule requires the coefficient determinant to be non-zero.
2. A non-zero determinant means the coefficient matrix is invertible and the solution is unique.

Teacher's Note:
1. Cramer's rule divides by the determinant of the coefficient matrix, so it fails when that determinant is zero.
2. It is a tool for square systems with a unique solution.

Answer: (d) infinitely many solutions or no solution

Explanation:
1. With \( |A|=0 \) the matrix is not invertible, so a unique solution is impossible.
2. Depending on whether B lies in the column space of A, the system is either consistent (infinitely many solutions) or inconsistent (no solution).

Teacher's Note:
1. Singular coefficient matrix means either infinitely many solutions or none.
2. Check the rank of the augmented matrix against the rank of A to decide.

Answer: (d) infinitely many solutions

Explanation:
1. A homogeneous system always has the trivial solution \( (0,0,0) \).
2. When the coefficient determinant is zero, there are also non-trivial solutions, and then the solution set is a line or a plane through the origin: infinitely many solutions.

Teacher's Note:
1. Non-trivial solutions of a homogeneous system exist exactly when the determinant is zero.
2. Any scalar multiple of a non-trivial solution is again a solution, so there are infinitely many.

Answer: (b) \( \operatorname{rank}(AB)=\operatorname{rank}(B) \)

Explanation:
1. Multiplying by an invertible matrix does not change the rank (it is a product of elementary operations).
2. Hence \( \operatorname{rank}(AB)=\operatorname{rank}(B) \).

Teacher's Note:
1. Row operations are equivalent to left multiplication by invertible matrices, and they preserve rank.
2. In general, rank(AB) is at most the smaller of rank A and rank B.

Answer: (b) 1

Explanation:
1. \( AB \) is an \( m\times n \) matrix whose every row is a multiple of the single row B.
2. All non-zero rows are proportional, and the matrix is non-zero, so its rank is 1.

Teacher's Note:
1. The product of a column and a row is called an outer product and always has rank at most 1.
2. It is non-zero here, so the rank is exactly 1.

Answer: (a) 3

Explanation:
1. A consistent system has a unique solution exactly when the rank of A equals the number of unknowns.
2. There are 3 unknowns, so the rank of A is 3.

Teacher's Note:
1. Consistent: rank A = rank [A | D]. Unique: this common rank equals the number of unknowns.
2. Rank less than the number of unknowns gives infinitely many solutions.

Answer: (c) \( r\le\min\{m,n\} \)

Explanation:
1. The rank is the maximum number of linearly independent rows (equal to that of columns).
2. There are only m rows and n columns, so \( r\le\min\{m,n\} \).

Teacher's Note:
1. Rank can equal the smaller dimension (full rank) or be less, so the inequality is not strict.
2. Equality r = min{m, n} is called full rank.

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

Explanation:
1. Put \( p=a^2,\ q=b^2,\ r=c^2 \) and expand along the first row: \( (q+r)\left[(r+p)(p+q)-qr\right]-p\left[q(p+q)-qr\right]+p\left[qr-r(r+p)\right] \).
2. The brackets are \( p(p+q+r) \), \( q(p+q-r) \) and \( r(q-r-p) \), so the determinant is \( p\left[(q+r)(p+q+r)-q(p+q-r)+r(q-r-p)\right] \).
3. The expression in the square bracket simplifies to \( 4qr \), so the value is \( 4pqr=4a^2b^2c^2 \).

Teacher's Note:
1. A quick check: for a = b = c = 1 the matrix is [[2,1,1],[1,2,1],[1,1,2]] with determinant 4, which matches 4 a squared b squared c squared.
2. Option (b) is dimensionally wrong: the determinant has degree 6 in a, b, c.

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

Explanation:
1. Let \( s=a+b+c \). Apply \( R_1\to R_1+R_2+R_3 \): the first row becomes \( (s,\ s,\ s) \).
2. Take out \( s \) and apply \( C_2\to C_2-C_1,\ C_3\to C_3-C_1 \): the matrix becomes \( \begin{vmatrix}1&0&0\\2b&-s&0\\2c&0&-s\end{vmatrix} \).
3. This lower triangular determinant is \( s^2 \), so the value is \( s\cdot s^2=(a+b+c)^3 \).

Teacher's Note:
1. Look for a common factor after adding all rows to one row.
2. Check with a = b = c = 1: the matrix is [[-1,2,2],[2,-1,2],[2,2,-1]] with determinant 27 = 3 cubed.

Answer: (d) \( \frac15 \)

Explanation:
1. A is upper triangular, so \( |A|=5\cdot\alpha\cdot5=25\alpha \).
2. \( |A^2|=|A|^2=625\alpha^2=25\Rightarrow\alpha^2=\tfrac1{25} \).
3. So \( |\alpha|=\tfrac15 \).

Teacher's Note:
1. The determinant of a triangular matrix is the product of its diagonal entries.
2. Use |A squared| = |A| squared.

Answer: (b) 0

Explanation:
1. Apply \( R_3\to R_3-2R_2+R_1 \): the new third row is \( (0,\ 0,\ a-2b+c) \).
2. Since a, b, c are in A.P., \( a+c=2b \), so the third row is zero and the determinant is \( 0 \).

Teacher's Note:
1. The first two columns have entries in arithmetic progression, and so does the third column when a, b, c are in A.P.
2. Rows that form an A.P. combine to a zero row under R3 - 2 R2 + R1.

Answer: (b) 0

Explanation:
1. Split the third column: the determinant is \( \begin{vmatrix}x&x^2&1\\y&y^2&1\\z&z^2&1\end{vmatrix}+\begin{vmatrix}x&x^2&x^3\\y&y^2&y^3\\z&z^2&z^3\end{vmatrix} \).
2. The second determinant is \( xyz\,V \) with \( V=\begin{vmatrix}1&x&x^2\\1&y&y^2\\1&z&z^2\end{vmatrix} \); the first equals \( V \) (it needs two column swaps to reach this form).
3. So the total is \( (1+xyz)V \), with \( V=(y-x)(z-x)(z-y)\ne0 \) since x, y, z are distinct. Hence \( 1+xyz=0 \).

Teacher's Note:
1. The factor 1 + xyz appears after splitting the column and recognising two Vandermonde determinants.
2. V is non-zero for distinct x, y, z, so the other factor must vanish.

Answer: (d) cannot be said

Explanation:
1. The trace of a product is not determined by the traces of the factors.
2. Example 1: \( A=\operatorname{diag}(1,2),\ B=\operatorname{diag}(1,4) \) has \( \operatorname{tr}A=3,\ \operatorname{tr}B=5 \) and \( \operatorname{tr}(AB)=9 \).
3. Example 2: \( A=\operatorname{diag}(1,2),\ B=\operatorname{diag}(4,1) \) has the same traces but \( \operatorname{tr}(AB)=6 \). So it cannot be said.

Teacher's Note:
1. Trace is additive and tr(AB) = tr(BA), but trace is not multiplicative.
2. Two different matrix pairs with the same traces can give different values of tr(AB).