第12类NCERT解决方案–数学第I部分–第4章行列式–练习4.1

评估以下问题的决定因素。

问题1 ${\begin{vmatrix}2&4\\-5&-1\end{vmatrix}}$

解决方案：

The determinant of a 2 x 2 matrix ${\begin{vmatrix}a&b\\c&d\end{vmatrix}}$
$= ad - bc\\$

Hence, ${\begin{vmatrix}2&4\\-5&-1\end{vmatrix}}$
$\\ = (2)(-1) - (4)(-5)\\ = -2 + 20\\ = 18\\$

问题2.(i) ${\begin{vmatrix}cosθ&-sinθ\\sinθ&cosθ\end{vmatrix}}$

解决方案：

${\begin{vmatrix}cosθ&-sinθ\\sinθ&cosθ\end{vmatrix}}$

$= (cosθ)(cosθ) - (-sinθ)(sinθ)\\ = cos^2θ + sin^2θ\\ = 1$ from trigonometric identities

(ii) ${\begin{vmatrix}x^2-x+1&x-1\\x+1&x+1\end{vmatrix}}$

解决方案：

${\begin{vmatrix}x^2-x+1&x-1\\x+1&x+1\end{vmatrix}}$

$= (x^2-x+1)(x+1) - (x-1)(x+1)\\ = (x^3-x^2+x+x^2-x+1) - (x^2-1)\\ = (x^3+1) - (x^2-1)\\ = x^3 - x^2 + 2\\$

问题3。 $A = {\begin{vmatrix}1&2\\4&2\end{vmatrix}}$显示$|2A| = 4|A|$

解决方案：

LHS=>

Matrix, $2A\\ = 2{\begin{pmatrix}1&2\\4&2\end{pmatrix}}\\ = {\begin{pmatrix}2*1&2*2\\2*4&2*2\end{pmatrix}}\\ = {\begin{pmatrix}2&4\\8&4\end{pmatrix}}\\$

Hence, determinant, $|2A|\\ = {\begin{vmatrix}2&4\\8&4\end{vmatrix}}\\ = (2)(4) - (4)(8)\\ = -24$

RHS=>

Determinant, $|A|\\ = {\begin{vmatrix}1&2\\4&2\end{vmatrix}}\\ = (1)(2) - (2)(4)\\ = -6$

Now, $4|A|\\ = 4(-6)\\ = -24$

Hence, proved, LHS = RHS

问题4。 $A = {\begin{vmatrix}1&0&1\\0&1&2\\0&0&4\end{vmatrix}}$然后显示| $|3A| = 27|A|$

解决方案：

LHS=>

Matrix, $3A\\ = 3{\begin{pmatrix}1&0&1\\0&1&2\\0&0&4\end{pmatrix}}\\ = {\begin{pmatrix}3*1&3*0&3*1\\3*0&3*1&3*2\\3*0&3*0&3*4\end{pmatrix}}\\ = {\begin{pmatrix}3&0&3\\0&3&6\\0&0&12\end{pmatrix}}\\$

Hence, determinant, $|3A|\\ = {\begin{vmatrix}3&0&3\\0&3&6\\0&0&12\end{vmatrix}}\\ = 3{\begin{vmatrix}3&6\\0&12\end{vmatrix}} - 0{\begin{vmatrix}0&6\\0&12\end{vmatrix}} + 3{\begin{vmatrix}0&3\\0&0\end{vmatrix}} \\ = 3(36) - 0(0) + 3(0)\\ = 108$

RHS =>

Determinant, $|A|\\ = {\begin{vmatrix}1&0&1\\0&1&2\\0&0&4\end{vmatrix}}\\ = 1{\begin{vmatrix}1&2\\0&4\end{vmatrix}} - 0{\begin{vmatrix}0&2\\0&4\end{vmatrix}} + 1{\begin{vmatrix}0&1\\0&0\end{vmatrix}} \\ = 1(4) - 0(0) + 1(0)$

Now, $27|A|\\ = 27(4)\\ = 108$

Hence, proved, LHS = RHS

问题5.评估决定因素

(一世) ${\begin{vmatrix}3&-1&-2\\0&0&-1\\3&-5&0\end{vmatrix}}\\$

解决方案：

Since the maximum number of zeroes are in the second row, we will expand the determinant along row 2.

$= -0{\begin{vmatrix}-1&-2\\-5&0\end{vmatrix}} + 0{\begin{vmatrix}3&-2\\3&0\end{vmatrix}} -(-1){\begin{vmatrix}3&-1\\3&-5\end{vmatrix}} \\ = -(-1)(-15 + 3)\\ = -12$

(ii) ${\begin{vmatrix}3&-4&5\\1&1&-2\\2&3&1\end{vmatrix}}\\$

解决方案：

$= 3{\begin{vmatrix}1&-2\\3&1\end{vmatrix}} -(-4){\begin{vmatrix}1&-2\\2&1\end{vmatrix}} +5{\begin{vmatrix}1&1\\2&3\end{vmatrix}} \\ = 3(7) + 4(5) + 5(1)\\ = 46$

(iii) ${\begin{vmatrix}0&1&2\\-1&0&-3\\-2&3&0\end{vmatrix}}\\$

解决方案：

$= 0{\begin{vmatrix}0&-3\\3&0\end{vmatrix}} -1{\begin{vmatrix}-1&-3\\-2&0\end{vmatrix}} +2{\begin{vmatrix}-1&0\\-2&3\end{vmatrix}} \\ = 0 - 1(-6) + 2(-3)\\ = 0$

Note: This matrix is skew symmetric i.e. $A^T = -A$

For every skew symmetric matrix of “odd dimension”, the determinant vanishes i.e. determinant is zero.

(iv) ${\begin{vmatrix}2&-1&-2\\0&2&-1\\3&-5&0\end{vmatrix}}\\$

解决方案：

Since the maximum number of zeroes are in the second row, we will expand the determinant along row 2.

$= -0{\begin{vmatrix}-1&-2\\-5&0\end{vmatrix}} + 2{\begin{vmatrix}2&-2\\3&0\end{vmatrix}} - (-1){\begin{vmatrix}2&-1\\3&-5\end{vmatrix}} \\ = 0 + 2(6) +1(-7)\\ = 5$

问题6。 $A = {\begin{pmatrix}1&1&-2\\2&1&-3\\5&4&-9\end{pmatrix}}$查找| A |

解决方案：

$|A| \\ = 1{\begin{vmatrix}1&-3\\4&-9\end{vmatrix}} -1{\begin{vmatrix}2&-3\\5&-9\end{vmatrix}} + (-2){\begin{vmatrix}2&1\\5&4\end{vmatrix}} \\ = 1(3) - 1(-3) - 2(3)\\ = 0$

问题7.在以下情况下找到x的值

(一世) ${\begin{vmatrix}2&4\\5&1\end{vmatrix}} = {\begin{vmatrix}2x&4\\6&x\end{vmatrix}}$

解决方案：

Solving determinants on both sides,

$(2)(1) - (4)(5) = (2x)(x) - (4)(6)\\ -18 = 2x^2 - 24\\ x^2 = 3\\ x = ±√3$

(ii) ${\begin{vmatrix}2&3\\4&5\end{vmatrix}} = {\begin{vmatrix}x&3\\2x&5\end{vmatrix}}$

解决方案：

Solving determinants on both sides

$(2)(5) - (3)(4) = (x)(5) - (3)(2x)\\ x = 2$

问题8。 ${\begin{vmatrix}x&2\\18&x\end{vmatrix}} = {\begin{vmatrix}6&2\\18&6\end{vmatrix}}$那么x等于

(A)6(B)±6(C)-6(D)0

解决方案：

Solving determinants on both sides

$(x)(x) - (2)(18) = (6)(6) - (2)(18)\\ x^2 -36 = 0\\ x = ±6\\$

Hence, Option (B)