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Matrices Quiz 2

mathematics

 If $A=\left( \begin{matrix}   -5 & -8 & 0  \\   3 & 5 & 0  \\   1 & 2 & -1  \\ \end{matrix} \right)$, then ${{A}^{2}}$ is: 

 Following the rules of matrix multiplication, ${{A}^{2}}=\left( \begin{matrix}   1 & 0 & 0  \\   0 & 1 & 0  \\   0 & 0 & 1  \\ \end{matrix} \right)=I$.

  ${{A}^{2}}=\left( \begin{matrix}   -5 & -8 & 0  \\   3 & 5 & 0  \\   1 & 2 & -1  \\ \end{matrix} \right)\left( \begin{matrix}   -5 & -8 & 0  \\   3 & 5 & 0  \\   1 & 2 & -1  \\ \end{matrix} \right) \\ \Rightarrow \left( \begin{matrix}   1 & 0 & 0  \\   0 & 1 & 0  \\   0 & 0 & 1  \\ \end{matrix} \right)$ 

 If A is a non- singular matrix of order 3 and $\left| adj.\left( adj.A \right) \right|=1296$ , which among the following matrices can be equal to A?

 First use the formula $|adj.(adj.A)|=|A{{|}^{({{(n-1)}^{2}})}}$, then solve for $\left| adj.\left( adj.A \right) \right|=1296$, to get the value of $\left| A \right|$, then consider the options for the required matrix.

  $\text{Here, }|adj.(adj.A)|=|A{{|}^{({{(n-1)}^{2}})}}, \\  \text{where A is a non – singular matrix of order } n. \\ \text{It is given that }\left| adj.\left( adj.A \right) \right|=1296 \\ \text{As }n=3,|A|^{(3-1)^{2}}=|A{{|}^{4}}=1296 \\ \therefore \left| A \right|=\pm 6 \\ \text{Identify the suitable matrix from the options} \\ \text{Which has the determinant value of} \pm{6}. \\ \text{Considering the third option ;} \\ \text{i.e. }A=\left( \begin{matrix}   0 & 1 & -2  \\   3 & 4 & 1  \\   2 & 3 & 2  \\ \end{matrix} \right)\to |A|=-6$ 

 The following matrix has a unique solution, which of the following relations is true? $ \\ \left( \begin{matrix}    1 & 2 & 4  \\   2 & 1 & 2  \\   1 & 2 & x-4  \\\end{matrix} \right)\left( \begin{matrix}   a  \\   b  \\   c  \\\end{matrix} \right)=\left( \begin{matrix}   6  \\   4  \\   x  \\\end{matrix} \right)$ 

 First do a row operation on the 3rd row to get a relation of x, which in turn gives the required relation

  $\text{The system of equation }AX=B\text{ has a solution if } \\ \text{Rank of A = Rank of }\left[ \text{A:B} \right] \\ \text{The solution is unique, } \\ \text{Then }detA\ne 0 \\ \text{So rank of A should be 3 } \\ \text{Solving by row operations we get,} \\ R3\text{ implies }R3-R1 \\ \left( \begin{matrix}   1 & 2 & 4  \\   2 & 1 & 2  \\   1 & 2 & x-8  \\ \end{matrix} \right) \\ \text{So, }x\ne 8\text{ since rank should be 3}$

 If $\sqrt{3}2i$ and 5 are the two Eigenvalues of a matrix X of order 3 by 3, what is the determinant of the X.

 First find the three roots of the matrix X from the eigenvalues, then by use the formula $\text{Determinant of X = product of eigenvalues}$, for the required determinant of X.

  $\text{Eigenvalues always occur in conjugate pairs.} \\ \text{Conjugate pair of }\sqrt{3}2i=\sqrt{3}+2i\left( i\ge -i \right) \\ \text{So all the 3 roots will be }\sqrt{3}2i,\sqrt{3}+2i,5 \\ \text{Determinant of X = product of eigenvalues}  \\ \Rightarrow \left( \sqrt{3}2i \right)\times \left( \sqrt{3}+2i \right)\times 5  \\ \Rightarrow \left( 34\times {{i}^{2}} \right)\times 5  \\ \Rightarrow \left( 3+4 \right)\times 5  \\ \Rightarrow 7\times 5  \\ \Rightarrow 35  \\ \text{The determinant of }X=35$ 

 Which of the following statements is true, if  $ \\ \left( \begin{matrix}   1 & 1 & 1  \\   1+\sin A & 1+\sin B & 1+\sin C  \\   \sin A+{{\sin }^{2}}A & \sin B+{{\sin }^{2}}B & \sin C+{{\sin }^{2}}C  \\ \end{matrix} \right)=0$.

 First do elementary row operations on the 3rd row then again on the 3rd column, for a simplified matrix, then expand along the first row.

  ${{R}_{3}}\to {{R}_{3}}-{{R}_{2}}, \\ \left( \begin{matrix}   1 & 1 & 1  \\   1+\sin A & 1+\sin B & 1+\sin C  \\   -{{\cos }^{2}}A & -{{\cos }^{2}}B & -{{\cos }^{2}}C  \\ \end{matrix} \right) \\ {{C}_{3}}\to {{C}_{3}}-{{C}_{2}},{{C}_{2}}\to {{C}_{2}}-{{C}_{1}} \\ \left( \begin{matrix}   1 & 1 & 1  \\   1+\sin A & \sin B-\sin A & \sin C-\sin B  \\   -{{\cos }^{2}}A & {{\cos }^{2}}A-{{\cos }^{2}}B & {{\cos }^{2}}B-{{\cos }^{2}}C  \\ \end{matrix} \right) \\ \text{Now, expanding along the first row, we get,} \\ \begin{array}{*{35}{l}}   \Rightarrow \left( sinB-sinA \right)\left( cos2B-cos2C \right)-\left( sinC-sinB \right)\left( cos2A-cos2B \right)=0  \\ \Rightarrow \left( sinB-sinA \right)\left( 1-si{{n}^{2}}B-1+si{{n}^{2}}C \right)-\left( sinC-sinB \right)\left( 1-si{{n}^{2}}A-1+si{{n}^{2}}B \right)=0  \\ \Rightarrow \left( sinB-sinA \right)\left( si{{n}^{2}}C-si{{n}^{2}}B \right)-\left( sinC-sinB \right)\left( si{{n}^{2}}B-si{{n}^{2}}A \right)=0  \\ \Rightarrow \left( sinB-sinA \right)\left( sinC-sinB \right)\left( sinC-sinA \right)=0  \\   \text{So, either }\left( sinB-sinA \right)=0or\left( sinC-sinB \right)=0or\left( sinC-sinA \right)=0  \\   A=B\text{ or }B=C\text{ or }C=A.  \\\end{array} \\ \text{So, triangle ABC is isosceles.}$