Question

If D = \left| {\begin{array}{*{20}{c}} \alpha &\beta \\\ \gamma &\delta \end{array}} \right| then what is the value of \left| {\begin{array}{*{20}{c}} {2\alpha }&{2\beta } \\\ {2\gamma }&{2\delta } \end{array}} \right|?

Answer 1

In this problem, the determinant is given as D = \left| {\begin{array}{*{20}{c}} \alpha &\beta \\\ \gamma &\delta \end{array}} \right| and we need to find the value of another determinant in which $2$ is multiplied with every element of $D$. For this, we will use the scalar multiple property of determinant. Scalar multiple property says that if $A$ is any $n \times n$ matrix and $k$ is non-zero constant then the determinant of matrix $kA$ is ${k^n}$ times the determinant of $A$. That is, $\left| {kA} \right| = {k^n}\left| A \right|$. Note that in the given problem $n = 2$ is order of determinant.

Complete step-by-step solution:
Let A = \left[ {\begin{array}{*{20}{c}} \alpha &\beta \\\ \gamma &\delta \end{array}} \right] be the matrix. The determinant of matrix $A$ is denoted by $\det \left( A \right)$ or $\left| A \right|$. In this problem, it is given that D = \left| {\begin{array}{*{20}{c}} \alpha &\beta \\\ \gamma &\delta \end{array}} \right|.
Let us find $2A$. Then, 2A = \left[ {\begin{array}{*{20}{c}} {2\alpha }&{2\beta } \\\ {2\gamma }&{2\delta } \end{array}} \right]. The determinant of matrix $2A$ is given by \left| {2A} \right| = \left| {\begin{array}{*{20}{c}} {2\alpha }&{2\beta } \\\ {2\gamma }&{2\delta } \end{array}} \right|.
Note that there are two rows and two columns in \left| {\begin{array}{*{20}{c}} {2\alpha }&{2\beta } \\\ {2\gamma }&{2\delta } \end{array}} \right|. Therefore, we can say that the order is $n = 2$. We need to find the value of \left| {\begin{array}{*{20}{c}} {2\alpha }&{2\beta } \\\ {2\gamma }&{2\delta } \end{array}} \right|. For this, we will use the scalar multiple property of determinant. Scalar multiple property says that if $A$ is any $n \times n$ matrix and $k$ is non-zero constant then determinant of matrix $kA$ is ${k^n}$ times the determinant of $A$. That is, $\left| {kA} \right| = {k^n}\left| A \right|$.
Let us compare $\left| {2A} \right|$ with $\left| {kA} \right|$ then we can say that $k = 2$.
Now we have scalar multiple is $k = 2$ and the order of determinant is $n = 2$. Now we are going to use the property $\left| {kA} \right| = {k^n}\left| A \right|$. Therefore, $\left| {2A} \right| = {2^2}\left| A \right|$
\Rightarrow \left| {\begin{array}{*{20}{c}} {2\alpha }&{2\beta } \\\ {2\gamma }&{2\delta } \end{array}} \right| = 4\left| {\begin{array}{*{20}{c}} \alpha &\beta \\\ \gamma &\delta \end{array}} \right|
\Rightarrow \left| {\begin{array}{*{20}{c}} {2\alpha }&{2\beta } \\\ {2\gamma }&{2\delta } \end{array}} \right| = 4D\quad \quad \left[ {\because D = \left| {\begin{array}{*{20}{c}} \alpha &\beta \\\ \gamma &\delta \end{array}} \right|} \right]
Therefore, if D = \left| {\begin{array}{*{20}{c}} \alpha &\beta \\\ \gamma &\delta \end{array}} \right| then the value of \left| {\begin{array}{*{20}{c}} {2\alpha }&{2\beta } \\\ {2\gamma }&{2\delta } \end{array}} \right| is $4D$.

Note: We can find the value of determinant if the matrix is square. The square matrix is the matrix in which the number of rows and columns are equal. If we have the matrix A = \left[ {\begin{array}{*{20}{c}} a&b; \\\ c&d; \end{array}} \right] then the determinant of matrix $A$ is \left| {\begin{array}{*{20}{c}} a&b; \\\ c&d; \end{array}} \right| = ad - bc. For example, if we have the matrix A = \left[ {\begin{array}{*{20}{c}} 1&2 \\\ 3&4 \end{array}} \right] then the value of determinant is \left| {\begin{array}{*{20}{c}} 1&2 \\\ 3&4 \end{array}} \right| = \left( {1 \times 4} \right) - \left( {2 \times 3} \right) = 4 - 6 = - 2. If all the elements of at least one row (or at least one column) are zero then the determinant will be zero. Determinant of the matrix $A$ and ${A^T}$ are the same where ${A^T}$ is the matrix obtained by interchanging the rows and columns of the matrix $A$.