Q.Prove by Mathematical Induction that $(A')^n = (A^n)'$, where $n \in \mathbb{N}$ for any square matrix $A$.

We are given

and told that $A A^T = I_3$. This is a classic orthogonality condition: the rows of $A$ are orthonormal (each row has length 1, and distinct rows are orthogonal). The matrix $A^T$ is the transpose, so $AA^T$ gives the matrix of dot products of rows.

Why this approach works

Instead of blindly multiplying matrices, we think geometrically: $AA^T = I$ means the rows of $A$ form an orthonormal basis. That gives us three equations (one for each row’s norm) and three equations for pairwise dot products. Since the matrix has symbolic entries, these equations become algebraic relations among $p, q, r$. Solving them will reveal which of the given options holds.

Step-by-step reasoning

Let the rows of $A$ be:

• Row 1: $R_1 = (o,\; 2q,\; r)$
• Row 2: $R_2 = (p,\; q,\; -r)$
• Row 3: $R_3 = (p,\; -q,\; r)$

The condition $AA^T = I_3$ means:

1. Each row has norm 1

Notice (2) and (3) are identical, so they give the same equation.

2. Distinct rows are orthogonal

Now we have a system of equations (1)–(6). Let’s solve them.

From (6):

From (2) and (3):

Substitute (7) into this:

Then from (7), $p^2 = \frac12$.

Now use (4) and (5):

Add (4) and (5):

Concept and Intuition

For any matrix $A$, the entry $(AA^T)_{ij}$ equals (row $i$ of $A$) $\cdot$ (row $j$ of $A$), because transposing turns the columns of $A^T$ back into the rows of $A$. This is much faster than multiplying full matrices out mechanically, and it also guarantees $AA^T$ is symmetric — a useful self-check.

Step-by-Step Solution

1. Rows of $A$: $R_1=(1,-1,2)$, $R_2=(-2,3,-3)$, $R_3=(4,-4,5)$.
2. Compute all pairwise dot products: $R_1\cdot R_1 = 1+1+4=6$ $R_1\cdot R_2 = -2-3-6=-11$ $R_1\cdot R_3 = 4+4+10=18$ $R_2\cdot R_2 = 4+9+9=22$ $R_2\cdot R_3 = -8-12-15=-35$ $R_3\cdot R_3 = 16+16+25=57$ So $AA^T=\begin{pmatrix}6&-11&18\\-11&22&-35\\18&-35&57\end{pmatrix}$ (symmetric, as expected).
3. Compute $A^T=\begin{pmatrix}1&-2&4\\-1&3&-4\\2&-3&5\end{pmatrix}$, then $A+A^T$ entrywise: $(1,1):1+1=2$, $(1,2):-1-2=-3$, $(1,3):2+4=6$ $(2,2):3+3=6$, $(2,3):-3-4=-7$, $(3,3):5+5=10$ So $A+A^T=\begin{pmatrix}2&-3&6\\-3&6&-7\\6&-7&10\end{pmatrix}$.
4. Subtract entrywise: $AA^T-(A+A^T)$: $(1,1):6-2=4$, $(1,2):-11-(-3)=-8$, $(1,3):18-6=12$ $(2,2):22-6=16$, $(2,3):-35-(-7)=-28$ $(3,3):57-10=47$ …

Concept and Intuition

If $AA^T = I$, then $A$ is an orthogonal matrix: its rows form an orthonormal set (each row has unit length, and any two distinct rows have zero dot product). Scaling every entry by 3 (as in $3A$) scales row lengths by 3 (so squared length becomes 9) but preserves orthogonality (dot products of different rows of $3A$ are still zero, since they're $9\times$ the original zero dot products).

Step-by-Step Solution

1. Rows of $3A$: $R_1=(1,2,2)$, $R_2=(2,1,-2)$, $R_3=(a,2,b)$.
2. Check $R_1\cdot R_1 = 1+4+4=9$ ✓ and $R_2\cdot R_2=4+1+4=9$ ✓ (consistent with each row having squared length 9).
3. $R_3\cdot R_3 = a^2+4+b^2 = 9 \Rightarrow a^2+b^2=5$.
4. Orthogonality $R_1\cdot R_3=0$: $1\cdot a + 2\cdot 2 + 2\cdot b = a+4+2b=0 \Rightarrow a+2b=-4$.
5. Orthogonality $R_2\cdot R_3=0$: $2\cdot a+1\cdot 2+(-2)\cdot b = 2a+2-2b=0 \Rightarrow a-b=-1 \Rightarrow a=b-1$.
6. Substitute into $a+2b=-4$: $(b-1)+2b=-4 \Rightarrow 3b=-3 \Rightarrow b=-1$, then $a=b-1=-2$.
7. Verify: $a^2+b^2 = 4+1=5$ ✓ (matches step 3). …

Concept and Intuition

$(kA)^T = kA^T$ for a scalar $k$, and matrix squaring/transposing here is just direct $2\times2$ arithmetic — no shortcut is needed beyond careful multiplication.

Step-by-Step Solution

1. $A=\begin{bmatrix}2&-3\\-4&1\end{bmatrix} \Rightarrow A^T=\begin{bmatrix}2&-4\\-3&1\end{bmatrix}$.
2. Compute $(A^T)^2 = A^T\cdot A^T$:
   • Row1: $[2(2)+(-4)(-3),\ 2(-4)+(-4)(1)] = [4+12,\ -8-4] = [16,-12]$
   • Row2: $[-3(2)+1(-3),\ -3(-4)+1(1)] = [-6-3,\ 12+1] = [-9,13]$ So $(A^T)^2=\begin{bmatrix}16&-12\\-9&13\end{bmatrix}$.
3. Compute $(12A)^T = 12A^T = \begin{bmatrix}24&-48\\-36&12\end{bmatrix}$. …

Concept and Intuition

When a matrix-relation problem gives several plausible-looking "$S^2$ equals ___" options, it's worth checking whether one of them is actually a general algebraic identity true for any matrix, independent of the specific conditions given — sometimes exam options include exactly this kind of "always true" statement as the intended answer, especially when the other options would require additional unstated assumptions (like orthogonality) to hold. Here, $(S^T)^2 = (S^2)^T$ follows purely from the transpose-reversal rule $(XY)^T = Y^TX^T$, with no special conditions needed at all.

Step-by-Step Solution

1. Treat $A,B,C,D$ as $1\times1$ matrices (scalars) to build a concrete, valid test case (transpose of a scalar is itself).
2. The conditions become $C = DAB$ and $D = ABC$. Substituting the second into the first: $C = (ABC)(AB) = A^2B^2C$, so (for $C\ne0$) $A^2B^2=1$, i.e. $AB=\pm1$.
3. Choose $A=1, B=1$ (so $AB=1$). Then $D = ABC = C$, so set $C=D=5$ (any nonzero value works, consistent with both original relations: $C=DAB=5\cdot1\cdot1=5$ ✓, $D=ABC=1\cdot1\cdot5=5$ ✓).
4. Compute $S = ABCD = 1\cdot1\cdot5\cdot5 = 25$, so $S^2 = 625$.
5. Test each option against $S^2=625$: (A) $S=25 \ne 625$. (B) $BCD = 1\cdot5\cdot5=25 \ne 625$. (C) $S^T = S = 25$ (scalars are self-transpose) $\ne 625$. …

Concept and Intuition

A matrix satisfying $AA^T = A^TA$ is called normal; this symmetry is exactly what's needed

to let $A^TA$ and $AA^T$ be swapped freely inside an algebraic manipulation. Since $B$ is built

from $A^{-1}$ and $A^T$, testing $BB^T$ naturally produces a chain that can be simplified using

this swap, collapsing to the identity.

Step-by-Step Solution

1. $B = A^{-1}A^T$.
2. $B^T = (A^{-1}A^T)^T = (A^T)^T (A^{-1})^T = A\,(A^{-1})^T$.
3. Since $(A^{-1})^T = (A^T)^{-1}$ (transpose and inverse commute), $B^T = A(A^T)^{-1}$.
4. $BB^T = \left(A^{-1}A^T\right)\left(A(A^T)^{-1}\right) = A^{-1}\big(A^TA\big)(A^T)^{-1}$.
5. Substitute the given condition $A^TA = AA^T$: …

Concept and Intuition

Matrix multiplication $PQ$ is defined only when the number of columns of $P$ equals the number of rows of $Q$. Here $A$ is $2\times3$, so $A^T$ is $3\times2$; multiplying $A^T$ ($3\times2$) by $B$ ($2\times3$) gives a valid $3\times3$ product (inner dimensions 2 match), computed entry-by-entry as row·column dot products.

Step-by-Step Solution

1. $A=\begin{bmatrix}1&-1&2\\0&3&4\end{bmatrix}$, so transposing rows/columns: $A^T=\begin{bmatrix}1&0\\-1&3\\2&4\end{bmatrix}$ (3 rows, 2 columns).
2. $B=\begin{bmatrix}4&0&-3\\-1&-2&-3\end{bmatrix}$ (2 rows, 3 columns). Since $A^T$ has 2 columns and $B$ has 2 rows, $A^TB$ is defined and will be $3\times3$.
3. Row 1 of $A^T$ is $(1,0)$. Dot with each column of $B$: col1 $(4,-1)$: $1(4)+0(-1)=4$; col2 $(0,-2)$: $1(0)+0(-2)=0$; col3 $(-3,-3)$: $1(-3)+0(-3)=-3$. So row 1 of the product is $(4,0,-3)$.
4. Row 2 of $A^T$ is $(-1,3)$. Dot: col1: $-1(4)+3(-1)=-4-3=-7$; col2: $-1(0)+3(-2)=-6$; col3: $-1(-3)+3(-3)=3-9=-6$. Row 2: $(-7,-6,-6)$. …

Concept and Intuition

Two determinant facts make this immediate: (1) transposing a matrix never changes its determinant, $\det(M^T)=\det(M)$; (2) the determinant of a product of square matrices equals the product of their determinants, $\det(PQ)=\det(P)\det(Q)$. Combining these collapses the given expression to a simple product of the original (non-transposed) determinants.

Step-by-Step Solution

1. $\det(B^T A^T) = \det(B^T)\cdot\det(A^T)$ (product rule).
2. $\det(B^T)=\det(B)$ and $\det(A^T)=\det(A)$ (transpose rule).
3. So $\det(B^TA^T) = \det(B)\cdot\det(A)$. …