Q40. - Algebra 2

Question

Solve $\det \left[\begin{array}{ccc}4& x& -2\\ -x& -3& 1\\ -6& 2& 3\end{array}\right]=-3$ for x.

Step-by-Step Solution

Answer

The value of $x=\frac{5}{3}, x=-1$.

Step 1 - Calculate the determinant

Find the determinant in this case by using expansion by minors.

 

$\begin{array}{c}\det \left[\begin{array}{ccc}4& x& -2\\ -x& -3& 1\\ -6& 2& 3\end{array}\right]=-3\\ 4\left|\begin{array}{cc}-3& 1\\ 2& 3\end{array}\right|-x\left|\begin{array}{cc}-x& 1\\ -6& 3\end{array}\right|+\left(-2\right)\left|\begin{array}{cc}-x& -3\\ -6& 2\end{array}\right|=-3\end{array}$

Step 2 - Evaluate 2 × 2 determinants

The determinant of second order matrix is found by calculating the difference of the product of the two diagonals, that is.,  $\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=ad-bc$. Apply this definition to find the $2\times 2$ determinant.

 

$\begin{array}{c}4\left|\begin{array}{cc}-3& 1\\ 2& 3\end{array}\right|-x\left|\begin{array}{cc}-x& 1\\ -6& 3\end{array}\right|+\left(-2\right)\left|\begin{array}{cc}-x& -3\\ -6& 2\end{array}\right|=-3\\ 4\left(-3\left(3\right)-2\left(1\right)\right)-x\left(-x\left(3\right)-\left(-6\right)\right)-2\left(-x\left(2\right)-\left(-3\right)\left(-6\right)\right)=-3\\ 4\left(-9-2\right)-x\left(-3x+6\right)-2\left(-2x-18\right)=-3\end{array}$

Step 3 - Solve for x

Further simplify the values.

 

 $\begin{array}{c}4\left(-9-2\right)-x\left(-3x+6\right)-2\left(-2x-18\right)=-3\\ 4\left(-11\right)-x\left(-3x\right)-x\left(6\right)-2\left(-2x\right)-2\left(-18\right)=-3\\ -44+3x^2-6x+4x+36=-3\\ 3x^2-2x-5=0\\ 3x^2+3x-5x-5=0\\ 3x\left(x+1\right)-5\left(x+1\right)=0\\ \left(3x-5\right)\left(x+1\right)=0\\ x=\frac{5}{3},x=-1\end{array}$

 

Therefore, possible values for x are $x=\frac{5}{3}, x=-1$.