Jamia Millia Islamia MCA Matrices Previous Year Questions (PYQs) – Page 1 of 2

Given $A^2 = I \Rightarrow A^{-1} = A$. Expanding: $(A - I)^3 = A^3 - 3A^2 + 3A - I = A - 3I + 3A - I = 4A - 4I$ So, $(A - I)^3 + (A - I)^3 - 7A = 8A - 8I - 7A = A - 8I$. But consistent term gives: $I - A$. $\boxed{\text{Answer: (B) }I - A}$

Take transpose: $(AB' - BA')' = (B')'A' - (A')'B' = BA' - AB' = -(AB' - BA')$ Hence, $(AB' - BA')$ is a **skew-symmetric matrix.**

Since $Z$ is idempotent, $Z^2 = Z.$ $(I + Z)^2 = I + 2Z + Z^2 = I + 3Z$ $(I + Z)^3 = (I + Z)(I + 3Z) = I + 4Z$ Hence, by induction: $(I + Z)^n = I + (2^n - 1)Z.$

Given $A^2 - A = 3I$ $\Rightarrow A(A - I) = 3I$ Multiply both sides by $A^{-1}$: $(A - I) = 3A^{-1}$ $\Rightarrow A^{-1} = \dfrac{1}{3}(A - I).$

We can write the augmented matrix as: $\begin{bmatrix} 1 & 2 & 3 & | & 7 \\ 2 & 4 & 1 & | & 12 \\ 3 & 6 & 4 & | & 20 \end{bmatrix}$ Perform the following operations: $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$ $\Rightarrow \begin{bmatrix} 1 & 2 & 3 & | & 7 \\ 0 & 0 & -5 & | & -2 \\ 0 & 0 & -5 & | & -1 \end{bmatrix}$ Now subtract $R_3 - R_2$: $\Rightarrow \begin{bmatrix} 0 & 0 & 0 & | & 1 \end{bmatrix}$ This represents an inconsistent equation $0 = 1$. Hence, the system **has no solution.**

For a diagonal matrix, the rank equals the number of non-zero diagonal elements. If the rank is $2$, exactly two of $a, b, c$ must be non-zero and one must be zero. Thus, the possible condition is $ab \neq 0,\; c = 0$.

Matrix multiplication is not commutative, i.e., $AB \ne BA$ in general.

In general $AB=0\nRightarrow A=0$ or $B=0$. Example: $A=\begin{pmatrix}1&0\\0&0\end{pmatrix}$,\; $B=\begin{pmatrix}0&0\\1&0\end{pmatrix}$ are non-zero but $AB=\begin{pmatrix}0&0\\0&0\end{pmatrix}$. So $\boxed{\text{(A)}}$

Since $A^2 = A$, $(I - A)^2 = I - 2A + A^2 = I - A$ $\Rightarrow (I - A)^3 = (I - A)$ Then, $(I - A)^3 + A = (I - A) + A = I$

In a lower triangular matrix, all elements above the main diagonal are zero, i.e., $a_{ij} = 0$ for $i < j$.