Answer
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Hint: Every square matrix A is associated with a number called a determinant and it is denoted by $\det \left( A \right)$ or $\left| A \right|$ . Only square matrices have determinants. Matrices which are not square will not have any determinants. Square matrices are those which have an equal number of rows and columns. We usually see first order determinant, second order determinant and third order determinant. But there are many more.
Complete step-by-step answer:
Let’s first how to find the determinant of a square matrix of order 1 :
$\Rightarrow $ If A = $\left[ a \right]$ then $\det \left( A \right)=\left| A \right|=a$ . It has only one element in it which is $a$ .
Elements are the numbers written inside the matrix.
Now let’s see how to find the determinant of a square matrix of order 2 :
$\Rightarrow $ If A = $\left[ \begin{matrix}
{{a}_{11}} & {{a}_{12}} \\
{{a}_{21}} & {{a}_{22}} \\
\end{matrix} \right]$ then $\det \left( A \right)=\left| A \right|={{a}_{11}}{{a}_{22}}-{{a}_{21}}{{a}_{12}}$
In the question , we can clearly see that we got a square matrix of order 2 .
Let’s apply the formula which we have just seen.
Let’s consider the given matrix to be B.
B = $\left| \begin{matrix}
4 & -1 \\
-2 & 3 \\
\end{matrix} \right|$ . Let’s compare this with the standard matrix A.
Upon comparing, we conclude the following :
$\begin{align}
& \Rightarrow {{a}_{11}}=4 \\
& \Rightarrow {{a}_{12}}=-1 \\
& \Rightarrow {{a}_{21}}=-2 \\
& \Rightarrow {{a}_{22}}=3 \\
\end{align}$ $$ $$
Now let’s apply the formula which we have written in the above statements.
\[\Rightarrow \det \left( B \right)=\left| B \right|={{a}_{11}}{{a}_{22}}-{{a}_{21}}{{a}_{12}}\]
Upon applying the formula ,we get the following :
\[\Rightarrow \det \left( B \right)=\left| B \right|=4\times 3-\left( -1 \right)\times \left( -2 \right)=12-2=10\]
Det value cannot be negative.
$\therefore $ Hence $\det \left( B \right)=\left| B \right|=10$
Note: Please be careful while comparing the elements with the standard form . And we should also be able to identify the order of the given square matrix as the det formula for each square matrix of different order varies.
.
Complete step-by-step answer:
Let’s first how to find the determinant of a square matrix of order 1 :
$\Rightarrow $ If A = $\left[ a \right]$ then $\det \left( A \right)=\left| A \right|=a$ . It has only one element in it which is $a$ .
Elements are the numbers written inside the matrix.
Now let’s see how to find the determinant of a square matrix of order 2 :
$\Rightarrow $ If A = $\left[ \begin{matrix}
{{a}_{11}} & {{a}_{12}} \\
{{a}_{21}} & {{a}_{22}} \\
\end{matrix} \right]$ then $\det \left( A \right)=\left| A \right|={{a}_{11}}{{a}_{22}}-{{a}_{21}}{{a}_{12}}$
In the question , we can clearly see that we got a square matrix of order 2 .
Let’s apply the formula which we have just seen.
Let’s consider the given matrix to be B.
B = $\left| \begin{matrix}
4 & -1 \\
-2 & 3 \\
\end{matrix} \right|$ . Let’s compare this with the standard matrix A.
Upon comparing, we conclude the following :
$\begin{align}
& \Rightarrow {{a}_{11}}=4 \\
& \Rightarrow {{a}_{12}}=-1 \\
& \Rightarrow {{a}_{21}}=-2 \\
& \Rightarrow {{a}_{22}}=3 \\
\end{align}$ $$ $$
Now let’s apply the formula which we have written in the above statements.
\[\Rightarrow \det \left( B \right)=\left| B \right|={{a}_{11}}{{a}_{22}}-{{a}_{21}}{{a}_{12}}\]
Upon applying the formula ,we get the following :
\[\Rightarrow \det \left( B \right)=\left| B \right|=4\times 3-\left( -1 \right)\times \left( -2 \right)=12-2=10\]
Det value cannot be negative.
$\therefore $ Hence $\det \left( B \right)=\left| B \right|=10$
Note: Please be careful while comparing the elements with the standard form . And we should also be able to identify the order of the given square matrix as the det formula for each square matrix of different order varies.
.
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