Question

The matrix $\left[ \begin{matrix} \lambda & 7 & -2 \\\ 4 & 1 & 3 \\\ 2 & -1 & 2 \\\ \end{matrix} \right]$ is a singular matrix if $\lambda$ is

& \left( \text{A} \right)\dfrac{2}{5} \\\ & \left( \text{B} \right)\dfrac{5}{2} \\\ & \left( \text{C} \right)\text{-5} \\\ & \left( \text{D} \right)\text{none of these} \\\ \end{aligned}$$

Answer 1

These types of problems are pretty straight forward and are very easy to solve. For the given type of problems we first need to understand what a singular matrix means. We also need to remember how to find the determinant of a given matrix. A singular matrix is a matrix whose value of the determinant is $0$ . We already know that the determinant of a matrix is defined if and only if the given matrix is a square matrix.

Complete step-by-step answer:
Now we start off with our solution as,
From the given matrix in our problem we can very easily find the order of the matrix as $3\times 3$ , as the matrix contains $3$ rows and $3$ columns. Now we need to find out the determinant of the given square matrix and then equate it to zero. The determinant is calculated as,

\lambda & 7 & -2 \\\ 4 & 1 & 3 \\\ 2 & -1 & 2 \\\ \end{matrix} \right|$$ The first term is positive, the second term is negative and the third term is positive. Finding by using the first row, we write, $$\lambda \left( 1\cdot 2-\left( -1\cdot 3 \right) \right)-7\left( 4\cdot 2-3\cdot 2 \right)+\left( -2 \right)\left( 4\cdot \left( -1 \right)-1\cdot 2 \right)$$ Now, taking out the negative signs we get, $$\Rightarrow \lambda \left( \left( 1\cdot 2 \right)+\left( 1\cdot 3 \right) \right)-7\left( \left( 4\cdot 2 \right)-\left( 3\cdot 2 \right) \right)-2\left( \left( 4\cdot -1 \right)-\left( 1\cdot 2 \right) \right)$$ Now doing the necessary multiplications we get, $$\Rightarrow \lambda \left( 2+3 \right)-7\left( 8-6 \right)-2\left( -4-2 \right)$$ Adding and Subtracting, we get, $$\Rightarrow \lambda \left( 5 \right)-7\left( 2 \right)-2\left( -6 \right)$$ Now, evaluating it, $$\begin{aligned} & \Rightarrow 5\lambda -14+12 \\\ & \Rightarrow 5\lambda -2 \\\ \end{aligned}$$ Now, equating the above intermediate equation to zero (as it is given to be a singular matrix) we get, $$5\lambda -2=0$$ Evaluating for the value of $$\lambda $$ we get, $$\begin{aligned} & 5\lambda =2 \\\ & \Rightarrow \lambda =\dfrac{2}{5} \\\ \end{aligned}$$ Now, from the options given to our problem, we see that our found out answer matches to the first option i.e. option (A). **So, the correct answer is “Option A”.** **Note:** We must always remember that a matrix should always have the same number of rows and columns, or in other words the matrix should be a square matrix, to be able to find the determinant of the matrix. We should also never forget the steps to calculate the determinant of a square matrix. For singular matrices, we should keep in mind that the determinant always evaluates to zero.