In this matrix problem, a square matrix of order $3$ is given with $9$ entries.

${\begin{vmatrix} a^2+2a & 2a+1 & 1 \\ 2a+1 & a+2 & 1 \\ 3 & 3 & 1 \\ \end{vmatrix}}$

In this matrix, four entries are algebraic expressions and the remaining elements are numbers.

- In the first row, $a^2+2a$, $2a+1$ and $1$ are the entries in the first, second and third columns.
- In the second row, $2a+1$, $a+2$ and $1$ are the elements in the first, second and third columns.
- In the third row, $3$, $3$ and $1$ are the entries in the first, second and third columns.

The determinant of this matrix should be evaluated in this problem. So, let’s learn how to find the determinant of the given $3 \times 3$ matrix.

In this matrix, the elements in the third column are $1$. Subtracting the entries of the second row from the entries of the first row makes an element to become $0$ in the third column. So, perform this operation and substitute their differences in the first row.

$R_1-R_2 \,\to\, R_1$

The subtraction of the entries can be written in matrix as follows.

$=\,\,\,$ ${\begin{vmatrix} a^2+2a-(2a+1) & 2a+1-(a+2) & 1-1 \\ 2a+1 & a+2 & 1 \\ 3 & 3 & 1 \\ \end{vmatrix}}$

Now, simplify the expressions in the first row of each column.

$=\,\,\,$ ${\begin{vmatrix} a^2+2a-2a-1 & 2a+1-a-2 & 0 \\ 2a+1 & a+2 & 1 \\ 3 & 3 & 1 \\ \end{vmatrix}}$

$=\,\,\,$ ${\begin{vmatrix} a^2+\cancel{2a}-\cancel{2a}-1 & 2a-a+1-2 & 0 \\ 2a+1 & a+2 & 1 \\ 3 & 3 & 1 \\ \end{vmatrix}}$

$=\,\,\,$ ${\begin{vmatrix} a^2-1 & a-1 & 0 \\ 2a+1 & a+2 & 1 \\ 3 & 3 & 1 \\ \end{vmatrix}}$

In the first row, $a-1$ is an entry in the second column and it is also a factor in the first column. Hence, the entry $a^2-1$ can be written as the difference of squares of the terms.

$=\,\,\,$ ${\begin{vmatrix} a^2-1^2 & a-1 & 0 \\ 2a+1 & a+2 & 1 \\ 3 & 3 & 1 \\ \end{vmatrix}}$

Now, write the difference of squares of the terms in factor form in the first row and the first column.

$=\,\,\,$ ${\begin{vmatrix} (a+1)(a-1) & a-1 & 0 \\ 2a+1 & a+2 & 1 \\ 3 & 3 & 1 \\ \end{vmatrix}}$

In the first row, $a-1$ is a factor in both first and second columns but there is no factor in the third column. However, the entry $0$ can be written as a product of $0$ and $a-1$.

$=\,\,\,$ ${\begin{vmatrix} (a+1)(a-1) & 1 \times (a-1) & 0 \times (a-1) \\ 2a+1 & a+2 & 1 \\ 3 & 3 & 1 \\ \end{vmatrix}}$

In the first row, each entry has $a-1$ as a factor. So, the common factor $a-1$ can be taken out from the expressions.

$=\,\,\,$ $(a-1){\begin{vmatrix} a+1 & 1 & 0 \\ 2a+1 & a+2 & 1 \\ 3 & 3 & 1 \\ \end{vmatrix}}$

The matrix contains $1$ as an entry in third column of both second and third rows. Just like in first step, an entry in the third column can make to become zero by the subtraction.

$R_2-R_3 \,\to\, R_2$

Now, subtract the entries of third row from the entries of second row. Then, substitute the difference of the elements in the respective positions of the second row.

$=\,\,\,$ $(a-1){\begin{vmatrix} a+1 & 1 & 0 \\ 2a+1-3 & a+2-3 & 1-1 \\ 3 & 3 & 1 \\ \end{vmatrix}}$

It is time to simplify the expressions in each column of second row.

$=\,\,\,$ $(a-1){\begin{vmatrix} a+1 & 1 & 0 \\ 2a-2 & a-1 & 0 \\ 3 & 3 & 1 \\ \end{vmatrix}}$

$=\,\,\,$ $(a-1){\begin{vmatrix} a+1 & 1 & 0 \\ 2a-2\times 1 & a-1 & 0 \\ 3 & 3 & 1 \\ \end{vmatrix}}$

$=\,\,\,$ $(a-1){\begin{vmatrix} a+1 & 1 & 0 \\ 2(a-1) & a-1 & 0 \\ 3 & 3 & 1 \\ \end{vmatrix}}$

In second row, $a-1$ is an entry in the second column and also a factor in the first column but there is no factor or term like this in the third column of second row. However, the entry $0$ can be written as a product of $0$ and $a-1$ for our convenience.

$=\,\,\,$ $(a-1){\begin{vmatrix} a+1 & 1 & 0 \\ 2 \times (a-1) & 1 \times (a-1) & 0 \times (a-1) \\ 3 & 3 & 1 \\ \end{vmatrix}}$

Each entry in the second row contains $a-1$ as a factor commonly. So, take each factor common from all entries of this row.

$=\,\,\,$ $(a-1) \times (a-1){\begin{vmatrix} a+1 & 1 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \\ \end{vmatrix}}$

$=\,\,\,$ $(a-1)^2{\begin{vmatrix} a+1 & 1 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \\ \end{vmatrix}}$

The entries are $1$ in second column of both first and second rows. The difference of the two rows can make the entry in the second column to become zero.

$R_1-R_2 \,\to\, R_1$

Now, subtract the entries of second row from the elements of first row, and then substitute their differences in the respective positions of the first row.

$=\,\,\,$ $(a-1)^2{\begin{vmatrix} a+1-2 & 1-1 & 0-0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \\ \end{vmatrix}}$

$=\,\,\,$ $(a-1)^2{\begin{vmatrix} a-1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \\ \end{vmatrix}}$

There is nothing to simplify in the matrix. So, the determinant of the $3 \times 3$ square matrix can be calculated by the determinant formula of the square matrix of order $3$.

$=\,\,\,$ $(a-1)^2$ $\times$ $\Big((a-1) \times (1 \times 1-0 \times 3)\Big)$

$=\,\,\,$ $(a-1)^2$ $\times$ $\Big((a-1) \times (1-0)\Big)$

$=\,\,\,$ $(a-1)^2$ $\times$ $\Big((a-1) \times 1\Big)$

$=\,\,\,$ $(a-1)^2 \times (a-1)$

$=\,\,\,$ $(a-1)^3$

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