Q.Evaluate $\begin{vmatrix}6&3&9\\2&4&8\\1&5&7\end{vmatrix}$ by first taking the common factor $3$ out of row 1.

Take $x,y,z$ common from the rows, then again from the columns, leaving a constant determinant equal to $4$.

Extracting common factors from rows/columns is a standard CBSE/NCERT Class 12 determinants property.

Take $x$ common from row 1, $y$ from row 2, $z$ from row 3:

$\begin{vmatrix} -x^2 & xy & xz \\ xy & -y^2 & yz \\ xz & yz & -z^2 \end{vmatrix} = xyz\begin{vmatrix} -x & y & z \\ x & -y & z \\ x & y & -z \end{vmatrix}.$

Now take $x$ common from column 1, $y$ from column 2, $z$ from column 3:

$= xyz\cdot xyz\begin{vmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = x^2y^2z^2\,D.$

Evaluate $D$:

$D = -1(1-1) - 1(-1-1) + 1(1+1) = 0 + 2 + 2 = 4.$

So the determinant $= 4x^2y^2z^2$, giving $\lambda = 4$.

Use $|kA|=k^n|A|$ with $n=3$ and $|A^T|=|A|$: $|2A^T|=2^3\cdot7=56$.

The scaling property of determinants ($|kA|=k^n|A|$) is a CBSE/NCERT Class 12 determinants property.

For a square matrix of order $n=3$:

$|2A^T| = 2^3\,|A^T|.$

Since a matrix and its transpose have equal determinants, $|A^T| = |A| = 7$. Therefore

$|2A^T| = 8\cdot 7 = 56.$

Multiplying every entry of an $n\times n$ matrix by $k$ scales its determinant by $k^n$.

For a square matrix $A$ of order $n$, the standard determinant property states $|kA| = k^n|A|$. This is because multiplying a matrix by a scalar $k$ multiplies EVERY row by $k$, and multiplying a single row of a determinant by a scalar multiplies the whole determinant by that scalar. Doing this once for each of the $n$ rows gives a total factor of $k^n$.

Here $A$ is order $3$, so $n=3$, giving $|kA| = k^3|A|$.

If any rows (or columns) of a determinant are identical, the determinant equals $0$, so the answer is option (b).

One of the standard properties of determinants is:

$\text{If any two rows (or columns) of a determinant are identical, then the value of the determinant is } 0.$

If three rows are identical, then in particular two of them are identical, so the property still forces the value to be zero. Intuitively, identical rows make the rows linearly dependent, so the determinant (which measures the 'volume' spanned) collapses to zero.

Multiplying all elements of one row by $K$ multiplies the determinant by $K$, so the answer is (b).

Take a general $2\times 2$ determinant to see the property clearly:

$D = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc.$

Now multiply the first row by $K$:

$\begin{vmatrix} Ka & Kb \\ c & d \end{vmatrix} = (Ka)d - (Kb)c = K(ad - bc) = K\,D.$

The common factor $K$ comes straight out of that row, so the determinant is scaled by $K$ exactly once. (The option $2K$ or $3K$ would only arise if several rows were each multiplied by $K$.)

The second determinant is obtained from $\Delta$ by interchanging rows $R_1$ and $R_2$; one row swap multiplies the value by $-1$, so the answer is $-\Delta$. This is a standard property-of-determinants question in the Tamil Nadu HSC Business Maths syllabus.

Write the rows of the given $\Delta$:

$\Delta = \begin{vmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{vmatrix},\qquad R_1=(1,2,3),\; R_2=(3,1,2),\; R_3=(2,3,1).$

Now look at the second determinant:

$\begin{vmatrix} 3 & 1 & 2 \\ 1 & 2 & 3 \\ 2 & 3 & 1 \end{vmatrix}.$

Its first row is $(3,1,2)=R_2$, its second row is $(1,2,3)=R_1$, and its third row is $(2,3,1)=R_3$. So it is exactly $\Delta$ with rows $R_1$ and $R_2$ interchanged.

Property used: if any two rows (or columns) of a determinant are interchanged, the value of the determinant changes sign. One interchange therefore gives a factor of $-1$:

$\begin{vmatrix} 3 & 1 & 2 \\ 1 & 2 & 3 \\ 2 & 3 & 1 \end{vmatrix} = -\Delta.$

A determinant with two identical rows or columns is always zero — a standard property.

If two rows (or two columns) of a determinant are identical, interchanging those two rows leaves the determinant unchanged (same matrix), but the row-interchange property says the determinant's sign must flip: $D=-D$, which forces $D=0$.

Scaling every row of an $n\times n$ matrix by $k$ multiplies the determinant by $k$ once per row; for a $3\times3$ matrix that is $3$ rows, giving the factor $k^3$.

1. $A$ is a square matrix of order $3$ (i.e. $3\times3$), and $kA$ means every entry of $A$ is multiplied by the scalar $k$ — equivalently, every one of the 3 rows is scaled by $k$.
2. The determinant property for scaling a single row by $k$ is: multiplying one row by $k$ multiplies the determinant by $k$.
3. Since all 3 rows of $A$ are scaled by $k$ in forming $kA$, the determinant is multiplied by $k$ a total of 3 times: $\det(kA) = k\cdot k\cdot k\cdot\det(A) = k^3\det(A).$
4. In general, for an $n\times n$ matrix, $\det(kA)=k^n\det(A)$; here $n=3$.
5. This matches option (a).

Scaling a matrix by $k$ scales the determinant by $k^n$; for a $3\times3$ matrix this is $k^3$.

1. If $A$ is a square matrix of order $n$, then $kA$ is obtained by multiplying every entry of $A$ by $k$.
2. A determinant is multilinear in its rows (equivalently columns): multiplying a single row by $k$ multiplies the determinant by $k$.
3. Since $kA$ has all $n$ rows scaled by $k$ (not just one), the determinant is multiplied by $k$ a total of $n$ times: $\det(kA)=k^n\det(A)$.
4. Here $A$ has order $3$, so $n=3$, giving $\det(kA)=k^3\det(A)$.
5. Distractors (b) $k^2\det(A)$, (c) $k\det(A)$, (d) $\det(A)$ correspond to scaling only $2$, $1$, or $0$ of the rows respectively, not the correct all-rows scaling implied by $kA$.