Which of the following has/have value equal to zero?
(Multiple Choice Correct Answer)
(a) $\left( \begin{matrix}
8 & 2 & 7 \\
12 & 3 & 5 \\
16 & 4 & 3 \\
\end{matrix} \right)$
(b) $\left( \begin{matrix}
\dfrac{1}{a} & {{a}^{2}} & bc \\
\dfrac{1}{b} & {{b}^{2}} & ac \\
\dfrac{1}{c} & {{c}^{2}} & ab \\
\end{matrix} \right)$
(c) $\left( \begin{matrix}
a+b & 2a+b & 3a+b \\
2a+b & 3a+b & 4a+b \\
4a+b & 5a+b & 6a+b \\
\end{matrix} \right)$
(d) $\left( \begin{matrix}
2 & 43 & 6 \\
7 & 35 & 4 \\
3 & 17 & 2 \\
\end{matrix} \right)$
Answer
Verified591.9k+ views
Hint: A square matrix is called a zero matrix if all the elements in it have value equal to zero. Now, to find the value of a matrix, we always take its determinant. Determinant of a given square matrix A is calculated as
$\begin{align}
& \left| A \right|=\left| \begin{matrix}
{{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
{{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
{{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right| \\
& \left| A \right|={{a}_{11}}\left( {{a}_{22}}{{a}_{33}}-{{a}_{23}}{{a}_{32}} \right)-{{a}_{12}}\left( {{a}_{21}}{{a}_{33}}-{{a}_{23}}{{a}_{31}} \right)+{{a}_{13}}\left( {{a}_{21}}{{a}_{32}}-{{a}_{22}}{{a}_{31}} \right) \\
\end{align}$
Here, we will calculate the determinant of all the matrices given in the option, either by direct application of formula or through row and column transformations and that matrix will have a zero value whose determinant is equal to zero.
Complete step by step answer:
Considering the matrix
$A=\left( \begin{matrix}
8 & 2 & 7 \\
12 & 3 & 5 \\
16 & 4 & 3 \\
\end{matrix} \right)$
Here, we will directly apply the formula to calculate its determinant
$\begin{align}
& A=8\left( 9-20 \right)-2\left( 36-80 \right)+7\left( 48-48 \right) \\
& A=\left\{ 8\times \left( -11 \right) \right\}+\left\{ -2\times \left( -44 \right) \right\}+\left\{ 7\times 0 \right\} \\
& A=-88+88+0 \\
& A=0 \\
\end{align}$
Here, since the determinant value of the matrix A is equal to zero, hence the given matrix has zero value.
Considering the matrix
$B=\left( \begin{matrix}
\dfrac{1}{a} & {{a}^{2}} & bc \\
\dfrac{1}{b} & {{b}^{2}} & ac \\
\dfrac{1}{c} & {{c}^{2}} & ab \\
\end{matrix} \right)$
Here, it would be convenient and easy if we use row and column transformations to find the value of the matrix,
We can see that if we subtract row 1 $\left( {{R}_{1}} \right)$ and row 2 $\left( {{R}_{2}} \right)$,we will get $\left( b-a \right)$ as a common factor which we can take out as common from the matrix. Similarly, we can subtract row 2 $\left( {{R}_{2}} \right)$ and row 3 $\left( {{R}_{3}} \right)$ to get a common factor of $\left( c-b \right)$ .
$\begin{align}
& {{R}_{1}}\to {{R}_{1}}-{{R}_{2}}\text{ and }{{R}_{2}}\to {{R}_{2}}-{{R}_{3}} \\
& B=\left( \begin{matrix}
\dfrac{1}{a}-\dfrac{1}{b} & {{a}^{2}}-{{b}^{2}} & bc-ac \\
\dfrac{1}{b}-\dfrac{1}{c} & {{b}^{2}}-{{c}^{2}} & ac-ab \\
\dfrac{1}{c} & {{c}^{2}} & ab \\
\end{matrix} \right) \\
\end{align}$
$B=\left( \begin{matrix}
\dfrac{b-a}{ab} & \left( a-b \right)\left( a+b \right) & c\left( b-a \right) \\
\dfrac{c-b}{bc} & \left( b-c \right)\left( b+c \right) & a\left( c-b \right) \\
\dfrac{1}{c} & {{c}^{2}} & ab \\
\end{matrix} \right)$
Taking out $\left( b-a \right)$ from ${{R}_{1}}$ and $\left( c-b \right)$ from ${{R}_{2}}$ common, we get
$B=\left( b-a \right)\left( c-b \right)\left( \begin{matrix}
\dfrac{1}{ab} & -\left( a+b \right) & c \\
\dfrac{1}{bc} & -\left( b+c \right) & a \\
\dfrac{1}{c} & {{c}^{2}} & ab \\
\end{matrix} \right)$
If we observe further, we can still take out the factor $\left( c-a \right)$ so as to simplify it further. So, subtracting row 2 $\left( {{R}_{2}} \right)$ from row 1 $\left( {{R}_{1}} \right)$, we get
$\begin{align}
& {{R}_{1}}\to {{R}_{1}}-{{R}_{2}} \\
& B=\left( b-a \right)\left( c-b \right)\left( \begin{matrix}
\dfrac{1}{ab}-\dfrac{1}{bc} & -a-b+b+c & c-a \\
\dfrac{1}{bc} & -\left( b+c \right) & a \\
\dfrac{1}{c} & {{c}^{2}} & ab \\
\end{matrix} \right) \\
\end{align}$
Taking out $\left( c-a \right)$ factor common from row 1 $\left( {{R}_{1}} \right)$,$B=\left( b-a \right)\left( c-b \right)\left( c-a \right)\left[ {} \right]$
Now taking the determinant of the above matrix,
$\left| B \right|=\left( b-a \right)\left( c-b \right)\left( c-a \right)\left[ \left\{ \dfrac{-a{{b}^{2}}-abc-a{{c}^{2}}}{abc} \right\}-\left\{ \dfrac{ab}{bc}-\dfrac{a}{c} \right\}+\left\{ \dfrac{{{c}^{2}}}{bc}+\dfrac{b+c}{c} \right\} \right]$
$\begin{align}
& \left| B \right|=\left( b-a \right)\left( c-b \right)\left( c-a \right)\left[ \left\{ \dfrac{-a{{b}^{2}}-abc-a{{c}^{2}}}{abc} \right\}-\left\{ \dfrac{a}{c}-\dfrac{a}{c} \right\}+\left\{ \dfrac{a{{c}^{2}}+a{{b}^{2}}+abc}{abc} \right\} \right] \\
& \therefore \left| B \right|=0 \\
\end{align}$
Here, since the determinant value of the matrix B is equal to zero, hence the given matrix has zero value.
Now, let us consider the third option
$C=\left( \begin{matrix}
a+b & 2a+b & 3a+b \\
2a+b & 3a+b & 4a+b \\
4a+b & 5a+b & 6a+b \\
\end{matrix} \right)$
Here, it would be easier to use column transformation in the above matrix to get a simplified value,
$\begin{align}
& {{C}_{2}}\to {{C}_{2}}-{{C}_{1}} \\
& and\text{ }{{\text{C}}_{3}}\to {{C}_{3}}-{{C}_{1}} \\
& C=\left( \begin{matrix}
a+b & 2a+b & 3a+b \\
2a+b & 3a+b & 4a+b \\
4a+b & 5a+b & 6a+b \\
\end{matrix} \right) \\
& C=\left( \begin{matrix}
a+b & a & 2a \\
2a+b & a & 2a \\
4a+b & a & 2a \\
\end{matrix} \right) \\
\end{align}$
Now, we will apply a row transformation of subtracting row 2 from row 1, this will make the two elements of row 1 zero which will make it easier to calculate its determinant.
$\begin{align}
& C=\left( \begin{matrix}
a+b & a & 2a \\
2a+b & a & 2a \\
4a+b & a & 2a \\
\end{matrix} \right) \\
& {{R}_{1}}\to {{R}_{1}}-{{R}_{2}} \\
& C=\left( \begin{matrix}
-a & 0 & 0 \\
2a+b & a & 2a \\
4a+b & a & 2a \\
\end{matrix} \right) \\
\end{align}$
Now, calculating the determinant value of matrix C, we get
$\begin{align}
& \left| C \right|=-a\left( 2{{a}^{2}}-2{{a}^{2}} \right) \\
& \therefore \left| C \right|=0 \\
\end{align}$
Here, since the determinant value of the matrix C is equal to zero, hence the given matrix has zero value.
Considering the last option,
$D=\left( \begin{matrix}
2 & 43 & 6 \\
7 & 35 & 4 \\
3 & 17 & 2 \\
\end{matrix} \right)$
Taking its determinant value by applying direct formula,
$\left| D \right|=\left[ 2\left\{ \left( 35\times 2 \right)-\left( 4\times 17 \right) \right\}-43\left\{ \left( 7\times 2 \right)-\left( 4\times 3 \right) \right\}+6\left\{ \left( 7\times 17 \right)-\left( 35\times 3 \right) \right\} \right]$
$\begin{align}
& \left| D \right|=\left[ 2\left\{ 70-68 \right\}-43\left\{ 14-12 \right\}+6\left\{ 119-105 \right\} \right] \\
& \left| D \right|=\left[ 4-86+84 \right] \\
& \left| D \right|=2 \\
& \therefore \left| D \right|\ne 0 \\
\end{align}$
Since, the determinant of matrix D has a non-zero value, hence the given matrix has a non-zero value.
Hence, out of all the options, matrices A, B and C have value equal to zero. So, options (a), (b) and (c) are the correct options for this question.
Note: While calculating the determinant of a matrix, just remember the correct formula and apply it carefully. Any mistake in calculation will lead to the wrong answer.
It is easier to first simplify the given matrix using consecutive row and column transformations wherever possible to turn certain elements zero. This helps to calculate the determinant of that matrix in a much convenient way.
$\begin{align}
& \left| A \right|=\left| \begin{matrix}
{{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
{{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
{{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right| \\
& \left| A \right|={{a}_{11}}\left( {{a}_{22}}{{a}_{33}}-{{a}_{23}}{{a}_{32}} \right)-{{a}_{12}}\left( {{a}_{21}}{{a}_{33}}-{{a}_{23}}{{a}_{31}} \right)+{{a}_{13}}\left( {{a}_{21}}{{a}_{32}}-{{a}_{22}}{{a}_{31}} \right) \\
\end{align}$
Here, we will calculate the determinant of all the matrices given in the option, either by direct application of formula or through row and column transformations and that matrix will have a zero value whose determinant is equal to zero.
Complete step by step answer:
Considering the matrix
$A=\left( \begin{matrix}
8 & 2 & 7 \\
12 & 3 & 5 \\
16 & 4 & 3 \\
\end{matrix} \right)$
Here, we will directly apply the formula to calculate its determinant
$\begin{align}
& A=8\left( 9-20 \right)-2\left( 36-80 \right)+7\left( 48-48 \right) \\
& A=\left\{ 8\times \left( -11 \right) \right\}+\left\{ -2\times \left( -44 \right) \right\}+\left\{ 7\times 0 \right\} \\
& A=-88+88+0 \\
& A=0 \\
\end{align}$
Here, since the determinant value of the matrix A is equal to zero, hence the given matrix has zero value.
Considering the matrix
$B=\left( \begin{matrix}
\dfrac{1}{a} & {{a}^{2}} & bc \\
\dfrac{1}{b} & {{b}^{2}} & ac \\
\dfrac{1}{c} & {{c}^{2}} & ab \\
\end{matrix} \right)$
Here, it would be convenient and easy if we use row and column transformations to find the value of the matrix,
We can see that if we subtract row 1 $\left( {{R}_{1}} \right)$ and row 2 $\left( {{R}_{2}} \right)$,we will get $\left( b-a \right)$ as a common factor which we can take out as common from the matrix. Similarly, we can subtract row 2 $\left( {{R}_{2}} \right)$ and row 3 $\left( {{R}_{3}} \right)$ to get a common factor of $\left( c-b \right)$ .
$\begin{align}
& {{R}_{1}}\to {{R}_{1}}-{{R}_{2}}\text{ and }{{R}_{2}}\to {{R}_{2}}-{{R}_{3}} \\
& B=\left( \begin{matrix}
\dfrac{1}{a}-\dfrac{1}{b} & {{a}^{2}}-{{b}^{2}} & bc-ac \\
\dfrac{1}{b}-\dfrac{1}{c} & {{b}^{2}}-{{c}^{2}} & ac-ab \\
\dfrac{1}{c} & {{c}^{2}} & ab \\
\end{matrix} \right) \\
\end{align}$
$B=\left( \begin{matrix}
\dfrac{b-a}{ab} & \left( a-b \right)\left( a+b \right) & c\left( b-a \right) \\
\dfrac{c-b}{bc} & \left( b-c \right)\left( b+c \right) & a\left( c-b \right) \\
\dfrac{1}{c} & {{c}^{2}} & ab \\
\end{matrix} \right)$
Taking out $\left( b-a \right)$ from ${{R}_{1}}$ and $\left( c-b \right)$ from ${{R}_{2}}$ common, we get
$B=\left( b-a \right)\left( c-b \right)\left( \begin{matrix}
\dfrac{1}{ab} & -\left( a+b \right) & c \\
\dfrac{1}{bc} & -\left( b+c \right) & a \\
\dfrac{1}{c} & {{c}^{2}} & ab \\
\end{matrix} \right)$
If we observe further, we can still take out the factor $\left( c-a \right)$ so as to simplify it further. So, subtracting row 2 $\left( {{R}_{2}} \right)$ from row 1 $\left( {{R}_{1}} \right)$, we get
$\begin{align}
& {{R}_{1}}\to {{R}_{1}}-{{R}_{2}} \\
& B=\left( b-a \right)\left( c-b \right)\left( \begin{matrix}
\dfrac{1}{ab}-\dfrac{1}{bc} & -a-b+b+c & c-a \\
\dfrac{1}{bc} & -\left( b+c \right) & a \\
\dfrac{1}{c} & {{c}^{2}} & ab \\
\end{matrix} \right) \\
\end{align}$
Taking out $\left( c-a \right)$ factor common from row 1 $\left( {{R}_{1}} \right)$,$B=\left( b-a \right)\left( c-b \right)\left( c-a \right)\left[ {} \right]$
Now taking the determinant of the above matrix,
$\left| B \right|=\left( b-a \right)\left( c-b \right)\left( c-a \right)\left[ \left\{ \dfrac{-a{{b}^{2}}-abc-a{{c}^{2}}}{abc} \right\}-\left\{ \dfrac{ab}{bc}-\dfrac{a}{c} \right\}+\left\{ \dfrac{{{c}^{2}}}{bc}+\dfrac{b+c}{c} \right\} \right]$
$\begin{align}
& \left| B \right|=\left( b-a \right)\left( c-b \right)\left( c-a \right)\left[ \left\{ \dfrac{-a{{b}^{2}}-abc-a{{c}^{2}}}{abc} \right\}-\left\{ \dfrac{a}{c}-\dfrac{a}{c} \right\}+\left\{ \dfrac{a{{c}^{2}}+a{{b}^{2}}+abc}{abc} \right\} \right] \\
& \therefore \left| B \right|=0 \\
\end{align}$
Here, since the determinant value of the matrix B is equal to zero, hence the given matrix has zero value.
Now, let us consider the third option
$C=\left( \begin{matrix}
a+b & 2a+b & 3a+b \\
2a+b & 3a+b & 4a+b \\
4a+b & 5a+b & 6a+b \\
\end{matrix} \right)$
Here, it would be easier to use column transformation in the above matrix to get a simplified value,
$\begin{align}
& {{C}_{2}}\to {{C}_{2}}-{{C}_{1}} \\
& and\text{ }{{\text{C}}_{3}}\to {{C}_{3}}-{{C}_{1}} \\
& C=\left( \begin{matrix}
a+b & 2a+b & 3a+b \\
2a+b & 3a+b & 4a+b \\
4a+b & 5a+b & 6a+b \\
\end{matrix} \right) \\
& C=\left( \begin{matrix}
a+b & a & 2a \\
2a+b & a & 2a \\
4a+b & a & 2a \\
\end{matrix} \right) \\
\end{align}$
Now, we will apply a row transformation of subtracting row 2 from row 1, this will make the two elements of row 1 zero which will make it easier to calculate its determinant.
$\begin{align}
& C=\left( \begin{matrix}
a+b & a & 2a \\
2a+b & a & 2a \\
4a+b & a & 2a \\
\end{matrix} \right) \\
& {{R}_{1}}\to {{R}_{1}}-{{R}_{2}} \\
& C=\left( \begin{matrix}
-a & 0 & 0 \\
2a+b & a & 2a \\
4a+b & a & 2a \\
\end{matrix} \right) \\
\end{align}$
Now, calculating the determinant value of matrix C, we get
$\begin{align}
& \left| C \right|=-a\left( 2{{a}^{2}}-2{{a}^{2}} \right) \\
& \therefore \left| C \right|=0 \\
\end{align}$
Here, since the determinant value of the matrix C is equal to zero, hence the given matrix has zero value.
Considering the last option,
$D=\left( \begin{matrix}
2 & 43 & 6 \\
7 & 35 & 4 \\
3 & 17 & 2 \\
\end{matrix} \right)$
Taking its determinant value by applying direct formula,
$\left| D \right|=\left[ 2\left\{ \left( 35\times 2 \right)-\left( 4\times 17 \right) \right\}-43\left\{ \left( 7\times 2 \right)-\left( 4\times 3 \right) \right\}+6\left\{ \left( 7\times 17 \right)-\left( 35\times 3 \right) \right\} \right]$
$\begin{align}
& \left| D \right|=\left[ 2\left\{ 70-68 \right\}-43\left\{ 14-12 \right\}+6\left\{ 119-105 \right\} \right] \\
& \left| D \right|=\left[ 4-86+84 \right] \\
& \left| D \right|=2 \\
& \therefore \left| D \right|\ne 0 \\
\end{align}$
Since, the determinant of matrix D has a non-zero value, hence the given matrix has a non-zero value.
Hence, out of all the options, matrices A, B and C have value equal to zero. So, options (a), (b) and (c) are the correct options for this question.
Note: While calculating the determinant of a matrix, just remember the correct formula and apply it carefully. Any mistake in calculation will lead to the wrong answer.
It is easier to first simplify the given matrix using consecutive row and column transformations wherever possible to turn certain elements zero. This helps to calculate the determinant of that matrix in a much convenient way.
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