Answer

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Hint: In this question we will use row and column transformation arithmetic operations to simplify the matrix and after simplification we will calculate the determinant and after further simplification we will get our solution.

Given that:

\[ \Rightarrow \left( {\begin{array}{*{20}{c}}

{x + y}&x&x \\

x&{x + y}&x \\

x&x&{x + y}

\end{array}} \right) = 16(3x + 4)\]

Taking L.H.S, we will proceed further

\[ = \left( {\begin{array}{*{20}{c}}

{x + y}&x&x \\

x&{x + y}&x \\

x&x&{x + y}

\end{array}} \right)\]

We will apply arithmetic operation on row second

$ro{w_2} \to ro{w_2} - ro{w_3}$

\[

= \left( {\begin{array}{*{20}{c}}

{x + y}&x&x \\

{x - x}&{x + y - x}&{x - (x + y)} \\

x&x&{x + y}

\end{array}} \right) \\

= \left( {\begin{array}{*{20}{c}}

{x + y}&x&x \\

0&y&{ - y)} \\

x&x&{x + y}

\end{array}} \right) \\

\]

Now, we will do arithmetic operation third row

$ro{w_3} \to ro{w_3} - ro{w_1}$

\[

= \left( {\begin{array}{*{20}{c}}

{x + y}&x&x \\

0&y&{ - y)} \\

{x - (x + y)}&{x - x}&{x + y - x}

\end{array}} \right) \\

= \left( {\begin{array}{*{20}{c}}

{x + y}&x&x \\

0&y&{ - y)} \\

{ - y}&0&y

\end{array}} \right) \\

\]

After simplification we will try to find out its determinant using first column

$

= (x + y)\left| {\begin{array}{*{20}{c}}

y&{ - y} \\

0&y

\end{array}} \right| - 0\left| {\begin{array}{*{20}{c}}

x&x \\

0&y

\end{array}} \right| - y\left| {\begin{array}{*{20}{c}}

x&x \\

y&{ - y}

\end{array}} \right| \\

= (x + y)({y^2}) - y( - xy - yx) \\

= x{y^2} + {y^3} + 2x{y^2} \\

= {y^2}(y + 3x) \\

$

As we know

L.H.S=R.H.S

$

\Rightarrow {y^2}(y + 3x) = 16(3x + y) \\

\\

$

Solving for the value of y, we will get

$

\Rightarrow {y^2} = 16 \\

\Rightarrow y = \pm 4 \\

$

Hence, the value of $y = \pm 4$

Note: This problem is a combination of matrix and determinant. This problem can be directly solved by calculating the determinant of the given matrix and then comparing and simplifying but it takes a lot of time. It becomes simple to calculate the determinant when the matrix is reduced using arithmetic operations and one or more columns or rows are zero.

Given that:

\[ \Rightarrow \left( {\begin{array}{*{20}{c}}

{x + y}&x&x \\

x&{x + y}&x \\

x&x&{x + y}

\end{array}} \right) = 16(3x + 4)\]

Taking L.H.S, we will proceed further

\[ = \left( {\begin{array}{*{20}{c}}

{x + y}&x&x \\

x&{x + y}&x \\

x&x&{x + y}

\end{array}} \right)\]

We will apply arithmetic operation on row second

$ro{w_2} \to ro{w_2} - ro{w_3}$

\[

= \left( {\begin{array}{*{20}{c}}

{x + y}&x&x \\

{x - x}&{x + y - x}&{x - (x + y)} \\

x&x&{x + y}

\end{array}} \right) \\

= \left( {\begin{array}{*{20}{c}}

{x + y}&x&x \\

0&y&{ - y)} \\

x&x&{x + y}

\end{array}} \right) \\

\]

Now, we will do arithmetic operation third row

$ro{w_3} \to ro{w_3} - ro{w_1}$

\[

= \left( {\begin{array}{*{20}{c}}

{x + y}&x&x \\

0&y&{ - y)} \\

{x - (x + y)}&{x - x}&{x + y - x}

\end{array}} \right) \\

= \left( {\begin{array}{*{20}{c}}

{x + y}&x&x \\

0&y&{ - y)} \\

{ - y}&0&y

\end{array}} \right) \\

\]

After simplification we will try to find out its determinant using first column

$

= (x + y)\left| {\begin{array}{*{20}{c}}

y&{ - y} \\

0&y

\end{array}} \right| - 0\left| {\begin{array}{*{20}{c}}

x&x \\

0&y

\end{array}} \right| - y\left| {\begin{array}{*{20}{c}}

x&x \\

y&{ - y}

\end{array}} \right| \\

= (x + y)({y^2}) - y( - xy - yx) \\

= x{y^2} + {y^3} + 2x{y^2} \\

= {y^2}(y + 3x) \\

$

As we know

L.H.S=R.H.S

$

\Rightarrow {y^2}(y + 3x) = 16(3x + y) \\

\\

$

Solving for the value of y, we will get

$

\Rightarrow {y^2} = 16 \\

\Rightarrow y = \pm 4 \\

$

Hence, the value of $y = \pm 4$

Note: This problem is a combination of matrix and determinant. This problem can be directly solved by calculating the determinant of the given matrix and then comparing and simplifying but it takes a lot of time. It becomes simple to calculate the determinant when the matrix is reduced using arithmetic operations and one or more columns or rows are zero.

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