**\[ M=\left[\ \begin{matrix}7&x\\-8&-7\\\end{matrix}\ \right]\]**

The aim of the article is to find the **value of the variable** $x$ within the given **matrix** for which it will be equal to its inverse **matrix**.

The basic concept behind this question is the understanding of the** Matrix**, how to find the **determinant** of a **matrix**, and the **inverse **of a **matrix**.

For a **matrix** $A$, the **inverse **of its **matrix** is represented by the following formula:

\[A^{ -1} = \dfrac{1}{det\space A} Adj\ A\]

Where:

$A^{ -1} = inverse \space of \space matrix$

$det\space A = Determinant \space of \space matrix$

$Adj\ A= Adjoint \space of \space matrix$

**Expert Answer**

Let us suppose the given **matrix** is $M$:

\[ M=\left[\ \begin{matrix}7&x\\-8&-7\\\end{matrix}\ \right]\]

For the **given condition** in the question, we know that the **matrix** should be equal to its **inverse **so we can write it as follows:

\[M = M^{-1 }\]

We know that the **inverse **of a **matrix** is determined by the following formula:

\[M^{ -1} = \dfrac{1}{det\space M} Adj\ M\]

Now first to find out the **determinant** of **matrix** $M$:

\[ det\ M = 7(-7) -x (-8)\]

\[ det\ M = -49 +8x \]

\[ det\ M = 8x -49 \]

Now we will find the **Adjoint** of the **matrix** $M$ as follows:

\[ M=\left[\ \begin{matrix}7&x\\-8&-7\\\end{matrix}\ \right] \]

\[ Adj\ M\ = \left[\ \begin{matrix} -7&-x\\8&7\\\end{matrix}\ \right] \]

To find the **inverse **of the **matrix,** we will put the values of its **determinant** and **adjoint** in the following formula:

\[M^{ -1} = \dfrac{1}{det\space M} Adj\ M\]

\[M^{ -1} = \dfrac{1}{8x -49} \times \left[\ \begin{matrix} -7&-x\\8&7\\\end{matrix}\ \right] \]

\[M^{ -1} = \left[\ \begin{matrix}\dfrac{-7}{8x -49} &\dfrac{-x}{8x -49}\\\dfrac{8}{8x -49}&\dfrac{7}{8x -49}\\\end{matrix}\ \right] \]

According to the condition given in the question, we have:

\[M = M^{-1 }\]

Putting the **matrix** $M$ and its **inverse **here, we have:

\[ \left[\ \begin{matrix}7&x\\-8&-7\\\end{matrix}\ \right] = \left[\ \begin{matrix}\dfrac{-7}{8x -49} &\dfrac{-x}{8x -49}\\\dfrac{8}{8x -49}&\dfrac{7}{8x -49}\\\end{matrix}\ \right] \]

Now **compare the matrices** on both sides so that we can find out the value of $x$. For this put any of the four equations equal to the equation in the other **matrix** in the same position. We have chosen the **first equation**, so we get:

\[ 7 = \dfrac{-7}{8x-49} \]

\[ 7 (8x-49) = -7 \]

\[ 56x-343 = -7 \]

\[ 56x = 343 -7 \]

\[ 56x = 336 \]

\[ x = \dfrac {336}{56} \]

\[ x = 6 \]

So the value of $x$ for which the **matrix** will be equal to its **inverse **is $x=6$.

**Numerical Results**

For the given **matrix** $\left[\ \begin{matrix}7&x\\-8&-7\\\end{matrix}\ \right]$ it will be equal to its **inverse **when the value of $x$ will be:

\[ x = 6 \]

**Example**

For the given **matrix** $\left[\ \begin{matrix}2&x\\-8&-2\\\end{matrix}\ \right]$ find the **determinant** and **adjoint**.

**Solution**

Let us suppose the given **matrix** is $Y$:

\[Y=\left[\ \begin{matrix}2&x\\-8&-2\\\end{matrix}\ \right]\]

Now first to find out the **determinant** of **matrix** $Y$:

\[det\ Y=2(-2) -x (-8)\]

\[det\ Y=-4 +8x\]

\[det\ Y=8x -4\]

**Adjoint** of the **matrix** $Y$:

\[Y=\left[ \begin{matrix}2&x\\-8&-2\\\end{matrix}\ \right]\]

\[Adj\ Y=\left[ \begin{matrix} -2&-x\\8&2\\\end{matrix}\ \right]\]