Determinants Test

Result

Determinants Test - 68

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Weekly Quiz Competition

Question 1
1 / -0

If $$A=\begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix}$$, then $$adj(3A^{2}+12A)$$ is equal to:

A
$$\begin{bmatrix} 72 & -63 \\ -84 & 51 \end{bmatrix}$$

B
$$\begin{bmatrix} 72 & -84 \\ -63 & 51 \end{bmatrix}$$

C
$$\begin{bmatrix} 51 & 63 \\ 84 & 72 \end{bmatrix}$$

D
$$\begin{bmatrix} 51 & 84 \\ 63 & 72 \end{bmatrix}$$

Solution
Question 2
1 / -0

$$P = \left[ {\begin{array}{*{20}{c}}1&\alpha &3\\1&3&3\\2&4&4\end{array}} \right]$$ is the adjoint of a $$3 \times 3$$ matrix A and $$\left| A \right| = 4,$$ then $$\alpha $$ is equal to

A
$$4$$

B
$$11$$

C
$$5$$

D
$$0$$

Solution
Question 3
1 / -0

If $$A = \left( \begin{array} { l l } { 1 } & { 2 } \\ { 3 } & { 5 } \end{array} \right), $$ then the value of the determinant $$\left| A ^ { 2009 } - 5 A ^ { 2008 } \right|$$ is

A
$$- 6$$

B
$$- 5$$

C
$$- 4$$

D
$$4$$

E
$$6$$

Solution
Question 4
1 / -0

The adjoining of the matrix $$\begin{bmatrix} 1 & -2 & 3 \\ 0 & 2 & -1 \\ -4 & 5 & 2 \end{bmatrix}$$, is

A
$$\begin{bmatrix} 9 & 19 & -4 \\ 4 & 14 & 1 \\ 8 & 3 & 2 \end{bmatrix}$$

B
$$ \begin{bmatrix} 9 & 4 & 8 \\ 19 & 14 & 3 \\ -4 & 1 & 2 \end{bmatrix}$$

C
$$\begin{bmatrix} 9 & -19 & -4 \\ -4 & 14 & -1 \\ 8 & -3 & 2 \end{bmatrix}$$

D
None of these

Solution
Question 5
1 / -0

There are 12 points in a plane of which 5 are collinear. The maximum number of distinct quadrilaterals which can be formed with vertices at these points is

A
$$2.^{ 7 }{ { { p } } }_{ 3 }$$

B
$$^{ 7 }{ { { p } } }_{ 3 }$$

C
$$6.^{ 7 }{ { { C } } }_{ 3 }$$

D
420

Solution
Question 6
1 / -0

If $$A=A=\left[ \begin{matrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{matrix} \right]$$,then $$\left| A \right| \left| AdjA \right|$$ is equal to

A
$${a}^{9}$$

B
$${a}^{27}$$

C
$${a}^{61}$$

D
$$none \ of \ these$$

Solution
Question 7
1 / -0

If $$A=\begin{bmatrix} { a }_{ 1 } & { a }_{ 2 } & { a }_{ 3 } \\ { b }_{ 1 } & { b }_{ 2 } & { b }_{ 3 } \\ { c }_{ 1 } & { c }_{ 2 } & { c }_{ 3 } \end{bmatrix}$$ and $$A_i,B_i,C_i$$ are cofactors of $$a_i,b_i,c_i$$ then $$a_1B_1+a_2B_2+a_3B_3=$$

A
0

B
|A|

C
$$|A|^2$$

D
2|A|

Solution
Question 8
1 / -0

If $$A$$ is a square matrix $$(adj \,A)' - (adj \,A')$$

A
$$2A$$

B
$$2 \,adj \,A$$

C
Unit matrix

D
Null matrix
Question 9
1 / -0

If $$A$$ is singular matrix, then $$A.(adj\,A)$$ is

A
$$singular$$

B
$$non-singular$$

C
$$symmetric$$

D
$$not \,defined$$

Solution
Question 10
1 / -0

Let $${\Delta _{\text{o}}} = $$ $$\left[ \begin{matrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \\ { a }_{ 31 } & { a }_{ 32 } & { a }_{ 33 } \end{matrix} \right] $$ and let $${\Delta _1}$$ denote the determinant formed by the cofactors of elements of $${\Delta _0}$$ and $${\Delta _2}$$ denote the determinant formed by the cofactor of $${\Delta _1},$$ similarly $${\Delta _n}$$ denotes the determinant formed by the cofactors of $${\Delta _{n - 1}}$$ then the determinant value of $${\Delta _n}$$ is

A
$${\Delta _0}^{2n}\;$$

B
$${\Delta _0}^{{2^n}}\;$$

C
$${\Delta _0}^{{n^2}}\;$$

D
$${\Delta ^2}_0\;$$

Solution

Let $$A$$ be a matrix of order $$m\times m$$ and $$C$$ be its co-factor matrix

We know that $$A.adj{\left(A\right)}=\left|A\right|.I$$

$$\Rightarrow\,\left|A\right|\left|adj{\left(A\right)}\right|={\left|A\right|}^{m}.\left|I\right|$$

$$\Rightarrow\,\left|A\right|\left|adj{\left(A\right)}\right|={\left|A\right|}^{m}$$ ......$$(1)$$ since $$\left|I\right|=1$$

But $$adj{\left(A\right)}={C}^{T}$$

$$\therefore\,\left|adj{\left(A\right)}\right|=\left|{C}^{T}\right|=\left|C\right|$$ .......$$(2)$$ since $$\left|C\right|=\left|{C}^{T}\right|$$

From $$(1)$$ and $$(2)$$ we have

$$\left|A\right|\left|C\right|={\left|A\right|}^{m}$$

$$\Rightarrow\,\left|C\right|={\left|A\right|}^{m-1}$$

Let $${\Delta}_{i}$$ be the cofactor matrix then

$${\Delta}_{i}={\left({\Delta}_{i-1}\right)}^{m-1}$$ where $$m=3$$

$$\Rightarrow\,{\Delta}_{i}={\left({\Delta}_{i-1}\right)}^{3-1}$$

$$\Rightarrow\,{\Delta}_{i}={\left({\Delta}_{i-1}\right)}^{2}$$

$$\Rightarrow\,{\Delta}_{n}={\left({\Delta}_{n-1}\right)}^{2}={\left({\left({\Delta}_{n-2}\right)}^{2}\right)}^{2}={\left({\left({\left({\Delta}_{n-3}\right)}^{2}\right)}^{2}\right)}^{2}$$ and so on.

$$\Rightarrow\,{\Delta}_{n-1}={\left({\Delta}_{n-2}\right)}^{2}$$

$$\Rightarrow\,{\Delta}_{n-2}={\left({\Delta}_{n-3}\right)}^{2}$$
and so on.

Hence $${\Delta}_{0}={\left({\Delta}_{0}\right)}^{{2}^{n}}$$