Q.Find values of $x$ for which $\begin{vmatrix} 3 & x \\ x & 1 \end{vmatrix} = \begin{vmatrix} 3 & 2 \\ 4 & 1 \end{vmatrix}$.

We are given that $x = 4$ satisfies

The determinant of a $2\times2$ matrix $\begin{pmatrix}a & b \\ c & d\end{pmatrix}$ is $ad - bc$. So the equation becomes

1. Simplify the determinant equation

Bring all terms to one side:

2. Recognize the quadratic

   The equation $x^2 - 2x - 8 = 0$ is a quadratic in $x$. Its two roots are the values of $x$ that make the original determinant equal to $5$. We are told one root is $4$.

3. Use Vieta’s formulas

   For a quadratic $x^2 + bx + c = 0$, the sum of the roots is $-b$ and the product is $c$. Here $b = -2$ and $c = -8$.

   Let the roots be $r_1 = 4$ and $r_2$. Then

   $r_1 + r_2 = 2 \quad\Longrightarrow\quad 4 + r_2 = 2 \quad\Longrightarrow\quad r_2 = -2.$ …

Step 1 — Exploit the structure of the determinant.

Every entry of the second column is $x$. A determinant is linear in each column, so $x$ can be taken out of that column:

Step 2 — Expand the reduced determinant along the first column.

Hence

Step 3 — Divide by $x^{2}$ and take the limit.

$\frac{D}{x^{2}}=\frac{x\sin x-x^{2}\cos x}{x^{2}}=\frac{\sin x}{x}-\cos x$ …

Concept & Intuition

A singular matrix is one that does not have an inverse — its determinant is exactly zero. For a 3×3 matrix, the determinant is a polynomial in its entries. Here, the matrix contains the variable $x$ in three places, so setting $\det(A) = 0$ gives an equation we can solve for $x$. The trick is to compute the determinant carefully, especially noticing the zeros in the first column — they will simplify the expansion.

Step-by-step solution

1. Write down the matrix

2. Expand the determinant along the first column (because it has two zeros, making the calculation quick). The first column entries are: $a_{11}=0$, $a_{21}=x$, $a_{31}=0$. Only the term from $a_{21}$ survives. The sign factor for row 2, column 1 is $(-1)^{2+1} = -1$. So:

where $M_{21}$ is the minor (determinant of the submatrix after removing row 2 and column 1).

3. Find the minor $M_{21}$ Remove row 2 and column 1:

The determinant of this 2×2 matrix is:

4. Assemble the full determinant

5. Set the determinant to zero (singular condition) $$$$

Concept: a matrix is singular iff its determinant is zero.

det = | 2+x 3 4 ; 1 -1 2 ; x 1 -5 |

Expand along the first row:

(2+x) * [(-1)(-5) - (2)(1)] - 3 * [(1)(-5) - (2)(x)] + 4 * [(1)(1) - (-1)(x)]

= (2+x)(5 - 2) - 3(-5 - 2x) + 4(1 + x)

= 3(2 + x) + 15 + 6x + 4 + 4x …

$A=\begin{bmatrix}k+1&2\\4&k-1\end{bmatrix}$ singular means $\det A=0$:

$(k+1)(k-1)-(2)(4)=0$ …

Concept: A matrix is singular iff its determinant is zero.

A = [[2 - k, 2], [1, 3 - k]]

|A| = (2 - k)(3 - k) - (2)(1) = 6 - 2k - 3k + k^2 - 2 = k^2 - 5k + 4 …

Step 1 — What scalar multiplication does.

$5A$ means every one of the nine entries is multiplied by $5$. In particular, each of the 3 rows is scaled by $5$.

Step 2 — The determinant property.

A determinant is a multilinear function of its rows: if one row is multiplied by $k$, the determinant is multiplied by $k$ (this is the standard property $|R_i \to kR_i| = k|A|$). Applying it once per row for a $3\times3$ matrix:

Step 3 — General rule and evaluation.

$|5A| = 5^{3}|A| = 125\,|A|$ …