Find values of $x$ for which $\left|\begin{array}{ll}3 & x \\ x & 1\end{array}\right|=\left|\begin{array}{ll}3 & 2 \\ 4 & 1\end{array}\right|$.

Step 1: Calculate the determinant of the left-hand side matrix

To find the determinant of a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, we use the formula $ad - bc$. Applying this to the left-hand side matrix, we multiply the elements on the main diagonal ($3 \times 1$) and subtract the product of the elements on the anti-diagonal ($x \times x$).

Step 2: Calculate the determinant of the right-hand side matrix

Similarly, for the right-hand side matrix, we apply the same determinant formula. We multiply $3 \times 1$ and subtract $2 \times 4$, which simplifies to $3 - 8$, resulting in $-5$.

Step 3: Equate the determinants and solve for x

Now we set the two calculated determinants equal to each other, as given in the problem statement. This forms an algebraic equation that we can solve for $x$.

Step 4: Isolate x squared

To solve for $x$, we first isolate the $x^2$ term. We subtract 3 from both sides of the equation, which gives us $-x^2 = -8$. Then, we multiply both sides by $-1$ to get $x^2 = 8$.

Step 5: Find the values of x

Finally, to find $x$, we take the square root of both sides of the equation $x^2 = 8$. Remember that taking the square root yields both positive and negative solutions. We can simplify $\sqrt{8}$ as $\sqrt{4 \times 2}$, which is $2\sqrt{2}$.

More from this chapter

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