# Find the value of matrix X, if $X + \left| {\begin{array}{*{20}{c}}

4&6 \\

{ - 3}&7

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

3&{ - 6} \\

5&{ - 8}

\end{array}} \right|$

Last updated date: 24th Mar 2023

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Answer

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Hint: In this question first take the R.H.S matrix into L.H.S then apply the property of subtraction of matrix so, use these concepts to reach the solution.

Given equation is

$X + \left| {\begin{array}{*{20}{c}}

4&6 \\

{ - 3}&7

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

3&{ - 6} \\

5&{ - 8}

\end{array}} \right|$

Now take R.H.S matrix into L.H.S we have

$X = \left| {\begin{array}{*{20}{c}}

3&{ - 6} \\

5&{ - 8}

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

4&6 \\

{ - 3}&7

\end{array}} \right|$

Now apply subtraction of matrix we have

As we know if matrix \[\left| {\begin{array}{*{20}{c}}

a&c \\

b&d

\end{array}} \right|\] is subtract from matrix \[\left| {\begin{array}{*{20}{c}}

e&h \\

f&g

\end{array}} \right|\] so the matrix is

\[\left| {\begin{array}{*{20}{c}}

e&h \\

f&g

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

a&c \\

b&d

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

{e - a}&{h - c} \\

{f - b}&{g - d}

\end{array}} \right|\], so using this property above matrix X becomes,

\[X = \left| {\begin{array}{*{20}{c}}

{3 - 4}&{ - 6 - 6} \\

{5 - \left( { - 3} \right)}&{ - 8 - 7}

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

{ - 1}&{ - 12} \\

8&{ - 15}

\end{array}} \right|\]

Therefore the matrix X becomes \[\left| {\begin{array}{*{20}{c}}

{ - 1}&{ - 12} \\

8&{ - 15}

\end{array}} \right|\].

So, this is the required answer.

Note: In such types of questions we can directly apply the method of subtraction of matrix which is stated above, so first take L.H.S matrix into R.H.S, they apply the property as above, we will get the required matrix X as above, which is the required answer.

Given equation is

$X + \left| {\begin{array}{*{20}{c}}

4&6 \\

{ - 3}&7

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

3&{ - 6} \\

5&{ - 8}

\end{array}} \right|$

Now take R.H.S matrix into L.H.S we have

$X = \left| {\begin{array}{*{20}{c}}

3&{ - 6} \\

5&{ - 8}

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

4&6 \\

{ - 3}&7

\end{array}} \right|$

Now apply subtraction of matrix we have

As we know if matrix \[\left| {\begin{array}{*{20}{c}}

a&c \\

b&d

\end{array}} \right|\] is subtract from matrix \[\left| {\begin{array}{*{20}{c}}

e&h \\

f&g

\end{array}} \right|\] so the matrix is

\[\left| {\begin{array}{*{20}{c}}

e&h \\

f&g

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

a&c \\

b&d

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

{e - a}&{h - c} \\

{f - b}&{g - d}

\end{array}} \right|\], so using this property above matrix X becomes,

\[X = \left| {\begin{array}{*{20}{c}}

{3 - 4}&{ - 6 - 6} \\

{5 - \left( { - 3} \right)}&{ - 8 - 7}

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

{ - 1}&{ - 12} \\

8&{ - 15}

\end{array}} \right|\]

Therefore the matrix X becomes \[\left| {\begin{array}{*{20}{c}}

{ - 1}&{ - 12} \\

8&{ - 15}

\end{array}} \right|\].

So, this is the required answer.

Note: In such types of questions we can directly apply the method of subtraction of matrix which is stated above, so first take L.H.S matrix into R.H.S, they apply the property as above, we will get the required matrix X as above, which is the required answer.

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