JEE Mathematics Matrices MCQs Set 11

Answer: (b) \( \begin{bmatrix} 1/2 & -1/2 \\ 2 & 1 \end{bmatrix} \)

Explanation:
1. \( a_{11}=\frac{3-2}{2}=\frac12 \), \( a_{12}=\frac{3-4}{2}=-\frac12 \)
2. \( a_{21}=\frac{6-2}{2}=2 \), \( a_{22}=\frac{6-4}{2}=1 \)
3. So \( A=\begin{bmatrix} \frac{1}{2} & -\frac{1}{2} \\ 2 & 1 \end{bmatrix} \).

Teacher's Note:
1. Fill the matrix entry by entry using the rule, taking i as the row and j as the column.
2. Keep an eye on the sign of 3i-2j.

Answer: (b) 1

Explanation:
1. Equal matrices have equal corresponding entries: \( x+3=0 \) gives \( x=-3 \).
2. \( 2y+x=-7 \) gives \( y=-2 \); \( z-1=3 \) gives \( z=4 \); \( 4a-z=2a \) gives \( a=2 \).
3. \( x+y+z+a=-3-2+4+2=1 \).

Teacher's Note:
1. Equate entries one by one and solve in the order x, y, z, a.
2. Check by substituting back into the last entry: \( 4(2)-4=4=2a \).

Answer: (a) \( \begin{bmatrix} 1 & 3/2 \\ 1 & 5/2 \end{bmatrix} \)

Explanation:
1. \( 2X=\begin{bmatrix} 3-1 & 5-2 \\ 5-3 & 9-4 \end{bmatrix}=\begin{bmatrix} 2 & 3 \\ 2 & 5 \end{bmatrix} \).
2. X is half of this: \( \begin{bmatrix} 1 & \frac32 \\ 1 & \frac52 \end{bmatrix} \).

Teacher's Note:
1. Isolate 2X first, then divide every entry by 2.
2. Add 2X back to the first matrix to verify.

Answer: (a) \( \begin{bmatrix} -1 & -4 & 7 \\ 3 & -2 & -5 \\ -2 & -3 & 1 \end{bmatrix} \)

Explanation:
1. The additive inverse of A is -A, obtained by changing the sign of every entry.
2. Changing every sign gives \( \begin{bmatrix} -1 & -4 & 7 \\ 3 & -2 & -5 \\ -2 & -3 & 1 \end{bmatrix} \).

Teacher's Note:
1. Additive inverse always exists for any matrix; do not confuse it with the multiplicative inverse.
2. Every single entry changes sign, including entries that look like they are already negative.

Answer: (a) \( \begin{bmatrix} -16 & -14 \\ -21 & -32 \end{bmatrix} \)

Explanation:
1. \( 3A=\begin{bmatrix} 27 & 3 \\ 12 & 9 \end{bmatrix} \) and \( 5B=\begin{bmatrix} 5 & 25 \\ 30 & 55 \end{bmatrix} \), so \( 3A+5B=\begin{bmatrix} 32 & 28 \\ 42 & 64 \end{bmatrix} \).
2. \( 2X=-(3A+5B) \), so \( X=-\begin{bmatrix} 16 & 14 \\ 21 & 32 \end{bmatrix}=\begin{bmatrix} -16 & -14 \\ -21 & -32 \end{bmatrix} \).

Teacher's Note:
1. Move 3A+5B to the other side with a negative sign before halving.
2. The zero matrix on the right side means 2X is the additive inverse of 3A+5B.

Answer: (d) \( 1\times1 \)

Explanation:
1. (1x3)(3x3) gives a 1x3 matrix.
2. (1x3)(3x1) gives a 1x1 matrix.

Teacher's Note:
1. Track the orders: the inner numbers cancel and the outer numbers give the order.
2. A 1x1 matrix is just a single expression, here the quadratic form.

Answer: (a) \( \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \)

Explanation:
1. A is diagonal, so \( A^2=\operatorname{diag}(i^2,(-i)^2) \).
2. \( i^2=-1 \) and \( (-i)^2=-1 \), so \( A^2=-I \).

Teacher's Note:
1. Powers of a diagonal matrix are obtained by powering each diagonal entry.
2. Remember that \( (-i)^2 = i^2 = \) -1.

Answer: (d) 81A

Explanation:
1. \( A=3I \), so \( A^5=3^5 I=243I \).
2. Since \( A=3I \), we have \( 243I=81(3I)=81A \).

Teacher's Note:
1. A scalar matrix kI behaves like the number k under multiplication.
2. Both 243I and 81A name the same matrix; only 81A appears in the options.

Answer: (a) \( A^2=A \) and \( B^2=B \)

Explanation:
1. \( A^2=A\cdot A=(AB)A=A(BA)=AB=A \).
2. \( B^2=B\cdot B=(BA)B=B(AB)=BA=B \).
3. So both A and B are idempotent.

Teacher's Note:
1. Replace one factor using the given relation and regroup using associativity.
2. Such A and B are idempotent matrices.

Answer: (d) \( \begin{bmatrix} 12 & 4 \\ 12 & 4 \end{bmatrix} \)

Explanation:
1. \( A^2=\begin{bmatrix} 4+3 & 2 \\ 6 & 3 \end{bmatrix}=\begin{bmatrix} 7 & 2 \\ 6 & 3 \end{bmatrix} \).
2. \( 2A=\begin{bmatrix} 4 & 2 \\ 6 & 0 \end{bmatrix} \).
3. Sum with I: \( \begin{bmatrix} 7+4+1 & 2+2 \\ 6+6 & 3+0+1 \end{bmatrix}=\begin{bmatrix} 12 & 4 \\ 12 & 4 \end{bmatrix} \).

Teacher's Note:
1. Compute A squared carefully using row-by-column products.
2. Do not forget that the constant 1 becomes the identity matrix I.

Answer: (a) a^3 I+3a^2 b E

Explanation:
1. \( E^2=O \), so all powers of E beyond the first vanish.
2. Since I commutes with E, \( (aI+bE)^3=a^3I+3a^2bE+3ab^2E^2+b^3E^3 \).
3. The terms with \( E^2 \) and \( E^3 \) are zero, leaving \( a^3I+3a^2bE \).

Teacher's Note:
1. Use the binomial theorem; it is valid because I commutes with every matrix.
2. Nilpotent matrices such as E make the expansion terminate quickly.

Answer: (d) \( \begin{bmatrix} 2 & -1 & 5 \\ 6 & -3 & 15 \\ 8 & -4 & 20 \end{bmatrix} \)

Explanation:
1. P is 3x1 and Q is 1x3, so PQ is 3x3.
2. Row i of PQ is p_i times (2, -1, 5): \( [2,-1,5] \), \( [6,-3,15] \), \( [8,-4,20] \).

Teacher's Note:
1. A column times a row gives a full square matrix (outer product).
2. The product QP would instead be the 1x1 matrix \( [2-3+20]=[19] \).

Answer: (a) O

Explanation:
1. Row 1 of A times column 1 of B: \( 0\cdot a^2+c\cdot ab-b\cdot ac=0 \), and similarly for every entry.
2. Equivalently \( B=vv^T \) with \( v=(a,b,c)^T \) and \( Av=(cb-bc \), -ca+ac, \( ba-ab)^T=0 \).
3. So \( AB=O \).

Teacher's Note:
1. Writing B as a column times a row shows that AB \( = \) (Av)(v transpose).
2. A is the cross-product matrix of v, and v cross v is the zero vector.

Answer: (b) 5

Explanation:
1. Trace is linear, so \( Tr[(2-i)A]=(2-i)Tr(A) \).
2. \( (2-i)(2+i)=4+1=5 \).

Teacher's Note:
1. A scalar factor comes straight out of the trace.
2. (a-ib)(a+ib) is always \( a^2+b^2 \).

Answer: (d) -4

Explanation:
1. \( Tr(A-2B)=Tr(A)-2Tr(B)=8-12=-4 \).

Teacher's Note:
1. Trace is additive and homogeneous (linear).
2. It does not need A and B to commute.

Answer: (c) even

Explanation:
1. \( A^2=\begin{bmatrix} 4-3 & -2+2 \\ 6-6 & -3+4 \end{bmatrix}=I \).
2. So \( A^n=I \) when n is even; for odd n, \( A^n=A \), which is not I.

Teacher's Note:
1. Look for the smallest power that gives I.
2. Here A is an involutory matrix.

Answer: (c) \( \begin{bmatrix} 1+2p & -4p \\ p & 1-2p \end{bmatrix} \)

Explanation:
1. \( A=I+N \) with \( N=\begin{bmatrix} 2 & -4 \\ 1 & -2 \end{bmatrix} \) and \( N^2=\begin{bmatrix} 4-4 & -8+8 \\ 2-2 & -4+4 \end{bmatrix}=O \).
2. \( A^p=I+pN=\begin{bmatrix} 1+2p & -4p \\ p & 1-2p \end{bmatrix} \).

Teacher's Note:
1. Split A as I plus a nilpotent matrix, then use the binomial theorem.
2. Test with \( p=1 \) to confirm the formula returns A.

Answer: (a) \( C(A+B^T) \)

Explanation:
1. \( A+B^T \) is 2x3, so \( C(A+B^T) \) needs (3x3)(2x3), which is not defined.
2. The other three are defined: CB is 3x2 and \( A^T \) is 3x2; BAC is (3x2)(2x3)(3x3); \( C(A+B^T)^T \) is (3x3)(3x2).

Teacher's Note:
1. Write each order and check that the inner numbers match at every product.
2. Sums need identical orders.

Answer: (a) \( 5\times7 \)

Explanation:
1. \( A^T \) is 3x4, so \( A^T \) B is 3x5 and its transpose is 5x3.
2. \( C^T \) is 3x7, so the product is \( (5x3)(3x7)=5x7 \).

Teacher's Note:
1. Transpose swaps the order; track each step.
2. The inner dimensions (3 and 3) match, so the product exists.

Answer: (c) \( (AB)^T=A^TB^T \)

Explanation:
1. The transpose of a product reverses the order: \( (AB)^T=B^T A^T \).
2. So \( (AB)^T=A^T B^T \) is false in general; the other three are true.

Teacher's Note:
1. The reversal law is the one most often tested.
2. It extends to any number of factors.

Answer: (b) (0,0)

Explanation:
1. For a skew-symmetric matrix, the diagonal entries are zero \( (a_ii=-a_ii) \).
2. So \( x=0 \) and \( y=0 \).

Teacher's Note:
1. Skew-symmetric means \( A^T=-A \).
2. The off-diagonal entries -7 and 7 are already negatives of each other.

Answer: (a) a nilpotent matrix

Explanation:
1. \( A^2 \) has (1,1) entry \( \lambda ²\mu ²-\lambda ²\mu ²=0 \) and (1,2) entry \( \lambda \mu ³-\lambda \mu ³=0 \).
2. Similarly the second row gives zeros, so \( A^2=O \) and A is nilpotent.

Teacher's Note:
1. Compute A squared; zero entries appear because the matrix has zero trace and zero determinant.
2. Check: trace 0 and determinant \( -\lambda ²\mu ²+\lambda ²\mu ²=0 \).

Answer: (b) \( \begin{bmatrix} 6 & -4 \\ -2 & 1 \end{bmatrix} \)

Explanation:
1. For a 2x2 matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), adj \( = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \).
2. So adj P \( = \begin{bmatrix} 6 & -4 \\ -2 & 1 \end{bmatrix} \).

Teacher's Note:
1. Swap the diagonal entries and change the sign of the off-diagonal entries.
2. Check: \( P\cdot adj \) P \( = \) (6-8)I \( = \) -2I.

Answer: (d) 25

Explanation:
1. For order n, \( |\operatorname{adj}A|=|A|^{n-1} \).
2. With n=3, \( |\operatorname{adj}A|=5^2=25 \).

Teacher's Note:
1. Remember the exponent is n-1, not n.
2. The adjoint exists for every square matrix; only the inverse needs a non-zero determinant.

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

Explanation:
1. \( A(\operatorname{adj}A)=|A|I \), so \( |A|=k \).
2. For a 3x3 matrix \( |\operatorname{adj}A|=|A|^2=k^2 \).

Teacher's Note:
1. Compare the identity for A times its adjoint with the given diagonal matrix.
2. Then apply the adjoint determinant formula.

Answer: (a) \( \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix} \)

Explanation:
1. The determinant is \( 7(1)+3(-1)-3(0+1)=7-3-3=1 \).
2. The adjoint (transpose of the cofactor matrix) is \( \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix} \).
3. So the inverse is \( \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix} \); multiplying it by the matrix gives I.

Teacher's Note:
1. When the determinant is 1 the inverse equals the adjoint.
2. Verify by multiplying the first row of the matrix with the first column of the answer: \( 7-3-3=1 \).