Q.If $\begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} = 0$ then show that $\dfrac{1}{1-a}+\dfrac{1}{1-b}+\dfrac{1}{1-c}=1$

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: …

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 $n=3$:

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

…

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. Doi …

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:

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

…

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 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:

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

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-in …

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 $k$. …

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 $\det(kA)=k^3\det(A)$. …