Q.Find the rank of $B=\begin{pmatrix}1&2&3\\0&1&4\\5&6&0\end{pmatrix}$.

For a $3\times3$ matrix, rank $=2$ means the determinant is $0$ but at least one $2\times2$ minor is non-zero. Setting $\det=0$ gives $\lambda=1$.

In the Tamil Nadu HSC Business Maths syllabus, the rank of a square matrix is $3$ only when its determinant is non-zero; if the rank drops to $2$, the determinant must be $0$.

Expand along the first row of $A=\begin{pmatrix} \lambda & -1 & 0 \\ 0 & \lambda & -1 \\ -1 & 0 & \lambda \end{pmatrix}$:

$\det A = \lambda\begin{vmatrix}\lambda & -1\\ 0 & \lambda\end{vmatrix} -(-1)\begin{vmatrix}0 & -1\\ -1 & \lambda\end{vmatrix} + 0$

$= \lambda(\lambda^2 - 0) + 1\,(0\cdot\lambda - (-1)(-1)) = \lambda^3 + (0-1) = \lambda^3 - 1.$

Rank $2$ needs $\det A = 0$:

$\lambda^3 - 1 = 0 \;\Rightarrow\; \lambda^3 = 1 \;\Rightarrow\; \lambda = 1.$

(At $\lambda=1$ the top-left minor $\begin{vmatrix}1 & -1\\ 0 & 1\end{vmatrix}=1 eq0$, confirming the rank is exactly $2$.)

The rank of a matrix equals the number of non-zero rows in its echelon form. A null matrix has all entries zero, so its rank is $0$.

In the TN HSC Business Maths syllabus, the rank of a matrix $A$, written $\rho(A)$, is the order of its highest-order non-zero minor — equivalently the number of non-zero rows after reducing to echelon form.

A null matrix $O$ has every element equal to $0$, so every minor of every order is $0$ and there are no non-zero rows. Therefore

$\rho(O) = 0.$

The zero matrix is in fact the only matrix whose rank is $0$; any matrix with even one non-zero entry has rank at least $1$.

$AA^{T}$ is the outer product of the non-zero column $A$ with itself, so all its rows/columns are proportional: its rank is $1$, option (d).

Form $AA^{T}$. With $A=\begin{pmatrix}1\\2\\3\end{pmatrix}$,

$AA^{T}=\begin{pmatrix}1\\2\\3\end{pmatrix}\begin{pmatrix}1&2&3\end{pmatrix}=\begin{pmatrix}1&2&3\\2&4&6\\3&6&9\end{pmatrix}.$

Observe the dependence. Row 2 $=2\times$Row 1 and Row 3 $=3\times$Row 1, so only one independent row remains.

Conclusion. The rank of a non-zero outer product $u\,u^{T}$ is always $1$.

Rank $r$ $\Rightarrow$ some order-$r$ minor $\neq 0$ (and all order-$(r+1)$ minors $=0$).

The rank of a matrix is the order of its largest non-vanishing minor. So $\rho(A)=r$ requires at least one minor of order $r$ that is non-zero. Options (b), (c), (d) are false: not all order-$r$ minors need be non-zero, and every minor of order $(r+1)$ and higher must vanish.

When every entry of a matrix equals $1$, all rows (and columns) are copies of one another, leaving a single independent row, so the rank is $1$.

This is a standard rank question in the Tamil Nadu HSC Class-12 Business Mathematics syllabus. Take the $m\times n$ matrix $J$ with $J_{ij}=1$ for all $i,j$:

$J=\begin{bmatrix}1&1&\cdots&1\\1&1&\cdots&1\\\vdots&\vdots&&\vdots\\1&1&\cdots&1\end{bmatrix}.$

• The rank is the number of non-zero rows in row-echelon form. Subtracting row $1$ from every other row ($R_i\to R_i-R_1$) turns rows $2,3,\dots,m$ into all-zero rows.

• Only one non-zero row ($R_1$) survives, so the rank is $1$.

• Equivalently, any $2\times2$ minor is $\begin{vmatrix}1&1\\1&1\end{vmatrix}=0$, while a $1\times1$ minor $=1 eq0$; hence $\rho(J)=1$.

Use the property $|\operatorname{adj}A|=|A|^{\,n-1}$; substituting the given values gives $3^{\,n-1}=3^5$, so $n=6$.

For any square matrix $A$ of order $n$, the determinant of its adjoint is:

$|\operatorname{adj}A|=|A|^{\,n-1}.$

Given $|A|=3$ and $|\operatorname{adj}A|=243$:

$3^{\,n-1}=243.$

Write $243$ as a power of $3$: $243=3^5$. Therefore

$3^{\,n-1}=3^5\;\Rightarrow\;n-1=5\;\Rightarrow\;n=6.$

$A$ is a $2\times2$ matrix whose determinant is non-zero, so its rank is the full order $2$.

In this Tamil Nadu HSC Business Mathematics question, $\rho(A)$ denotes the rank of $A$ — the order of the largest square sub-matrix (minor) whose determinant is non-zero.

$A = \begin{pmatrix} 2 & 0 \\ 0 & 8 \end{pmatrix}$

Compute the determinant of the whole matrix:

$\det A = (2)(8) - (0)(0) = 16 e 0.$

Since the largest possible minor here is the $2\times2$ matrix itself and it is non-zero, the rank equals its order.

Use $|\text{Adj }A| = |A|^{n-1}$ with $n = 2$, so $|\text{Adj }A| = |A| = 13$; then equate the given determinant to $13$ and solve for $x$. This is a very common TN HSC Class-12 Business Mathematics matrices-and-determinants question.

Step 1 — Apply the adjoint property.

For a square matrix $A$ of order $n$, $\;|\text{Adj }A| = |A|^{\,n-1}$.

The adjoint here is a $2\times 2$ determinant, so $A$ is of order $n = 2$:

$|\text{Adj }A| = |A|^{\,2-1} = |A|^{1} = |A| = 13.$

Step 2 — Evaluate the given determinant.

$\begin{vmatrix} 4 & x \\ 5 & 7 \end{vmatrix} = (4)(7) - (x)(5) = 28 - 5x.$

Step 3 — Equate and solve.

$28 - 5x = 13 \;\Rightarrow\; 5x = 15 \;\Rightarrow\; x = 3.$

The valid elementary transformations are (i) interchange of two rows/columns, (ii) multiplying a row/column by a non-zero scalar, and (iii) adding a multiple of one row/column to another row/column of the SAME type. Mixing a row with a column is not allowed — that rules out option (c).

The three permitted elementary transformations (used, for instance, when reducing a matrix to echelon form to find its rank in the TN HSC syllabus):

1. Interchange of two rows or two columns: $R_i \leftrightarrow R_j$ or $C_i \leftrightarrow C_j$.
2. Multiplying a row/column by a non-zero constant $k$.
3. Adding a scalar multiple of one row (column) to another row (column): $R_i \rightarrow R_i + kR_j$ or $C_i \rightarrow C_i + kC_j$.

Checking each option:

• (a) $C_i \rightarrow C_i + 5C_j$ — column added to a column. Valid.

• (b) $R_i \leftrightarrow R_j$ — interchange of rows. Valid.

• (c) $R_i \rightarrow 2R_i + 2C_j$ — a column $C_j$ is being added into a row $R_i$. This mixes the two types and is not permitted.

• (d) $R_i \rightarrow 2R_i - 4R_j$ — row combination. Valid.