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Determinants

Short Answer Type

321.

Evaluate: $[\begin{matrix} a + ib & c + id \\ - c + id & a - ib \end{matrix}]$

322.

Find the co-factor of a12 in the following:

$|\begin{matrix} 2 & - 3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & - 7 \end{matrix}|$

Long Answer Type

323.

Using properties of determinants, prove the following:

$|\begin{matrix} α & β & γ \\ α^{2} & β^{2} & γ^{2} \\ β + γ & γ + α & α + β \end{matrix}| = (α - β) (β - γ) (γ - α) (α + β + γ)$

Short Answer Type

324.

Write the value of the determinant $|\begin{matrix} 2 & 3 & 4 \\ 5 & 6 & 8 \\ 6 x & 9 x & 12 x \end{matrix}|$

Long Answer Type

325.

Using properties of determinants prove the following:

$|\begin{matrix} a & b & c \\ a - b & b - c & c - a \\ b + c & c + a & a + b \end{matrix}| = a^{3} + b^{3} + c^{3} - 3 abc$

Short Answer Type

326.

Find the minor of the element of second row and third column ( a₂₃ ) in the following determinant:

$|\begin{matrix} 2 & - 3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & - 7 \end{matrix}|$

Long Answer Type

327.
Using properties of determinants show the following:
$|\begin{matrix} {(b + c)}^{2} & ab & ca \\ ab & {(a + c)}^{2} & bc \\ ac & bc & {(a + b)}^{2} \end{matrix}| = 2 abc (a + b + c)^{3}$

Consider,

$∆ = |\begin{matrix} (b + c)^{2} & ab & ac \\ ab & (a + c)^{2} & bc \\ ac & bc & (a + b)^{2} \end{matrix}| By performing R_{1} \to a R_{1}, R_{2} \to b R_{2}, R_{3} \to c R_{3} and dividing the determinant by abc, we get ∆ = \frac{1}{abc} |\begin{matrix} a (b + c)^{2} & a^{2} b & a^{2} c \\ {ab}^{2} & b (a + c)^{2} & b^{2} c \\ {ac}^{2} & {bc}^{2} & c (a + b)^{2} \end{matrix}|$

Now, taking a, b, c common from C₁, C₂, and C₃

$∆ = \frac{abc}{abc} |\begin{matrix} (b + c)^{2} & a^{2} & a^{2} \\ b^{2} & (a + c)^{2} & b^{2} \\ c^{2} & c^{2} & (a + b)^{2} \end{matrix}| \Rightarrow ∆ = |\begin{matrix} (b + c)^{2} & a^{2} & a^{2} \\ b^{2} & (c + a)^{2} & b^{2} \\ c^{2} & c^{2} & (a + b)^{2} \end{matrix}| Applying C_{1} \to C_{1} - C_{2}, C_{2} \to C_{2} - C_{3} \Rightarrow ∆ = |\begin{matrix} (b + c)^{2} - a^{2} & 0 & a^{2} \\ b^{2} - (c + a)^{2} & (c + a)^{2} - b^{2} & b^{2} \\ 0 & c^{2} - (a + b)^{2} & (a + b)^{2} \end{matrix}| ∆ = {(a + b + c)}^{2} |\begin{matrix} b + c - a & 0 & a^{2} \\ b - c - a^{} & c + a - b & b^{2} \\ 0 & c - a - b & (a + b)^{2} \end{matrix}|$

$Applying R_{3} \to R_{3} - (R_{1} + R_{2}) ∆ = {(a + b + c)}^{2} |\begin{matrix} b + c - a & 0 & a^{2} \\ b - c - a & c + a - b & b^{2} \\ 2 a - 2 b & - 2 a & 2 ab \end{matrix}|$

$Applying C_{1} \to C_{1} + C_{2} ∆ = {(a + b + c)}^{2} |\begin{matrix} b + c - a & 0 & a^{2} \\ 0 & c + a - b & b^{2} \\ - 2 b & - 2 a & 2 ab \end{matrix}| Applying C_{3} \to C_{3} + {bC}_{2} ∆ = {(a + b + c)}^{2} |\begin{matrix} b + c - a & 0 & a^{2} \\ 0 & c + a - b & bc + ab \\ - 2 b & - 2 a & 0 \end{matrix}|$

$Applying C_{1} \to {aC}_{1} and C_{2} \to {bC}_{2} ∆ = \frac{{(a + b + c)}^{2}}{ab} |\begin{matrix} ab + ac - a^{2} & 0 & a^{2} \\ 0 & bc + ab - b^{2} & bc + ab \\ - 2 ab & - 2 ab & 0 \end{matrix}| Applying C_{1} \to C_{1} - C_{2} ∆ = \frac{{(a + b + c)}^{2}}{ab} |\begin{matrix} ab + ac - a^{2} & 0 & a^{2} \\ - bc - ab + b^{2} & bc + ab - b^{2} & bc + ab \\ 0 & - 2 ab & 0 \end{matrix}|$

Expanding along R₃

$= \frac{{(a + b + c)}^{2}}{ab} (2 ab ({ab}^{2} c + a^{2} b^{2} + {abc}^{2} + a^{2} bc - a^{2} bc - a^{3} b + a^{2} bc + a^{3} b - a^{2} b^{2})) = 2 (a + b + c)^{2} ({ab}^{2} c + {abc}^{2} + a^{2} bc) = 2 (a + b + c)^{3} abc = 2 abc (a + b + c)^{3} = R . H . S .$

328.

Using properties of determinants, prove that

$|\begin{matrix} - a^{2} & ab & ac \\ ba & - b^{2} & bc \\ ca & cb & - c^{2} \end{matrix}| = 4 a^{2} b^{2} c^{2}$

329.

Using matrix method, solve the following system of equations:

$\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4, \frac{4}{x} - \frac{6}{y} + \frac{5}{z}, \frac{6}{x} + \frac{9}{y} - \frac{20}{z}; x, y, z \neq 0$

Short Answer Type

330.

If $∆ = |\begin{matrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{matrix}|$ , white the cofactor of the element a₃₂.

Previous Year Papers

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Short Answer Type

Long Answer Type

Short Answer Type

Long Answer Type

Short Answer Type

Long Answer Type

327. Using properties of determinants show the following: b + c 2 ab caab a + c 2 bcac bc a + b 2 = 2abc ( a + b + c )3

Short Answer Type

327.
Using properties of determinants show the following:
$|\begin{matrix} {(b + c)}^{2} & ab & ca \\ ab & {(a + c)}^{2} & bc \\ ac & bc & {(a + b)}^{2} \end{matrix}| = 2 abc (a + b + c)^{3}$