Let A be the set of all 3 \times 3 symmetric matrices all of whose entries are either 0 or 1. Five of these entries are 1 and four of them are 0. The number of matrices A in A for which the system of linear equations A[ \beginmatrix x y z \endmatrix ] = [ \beginmatrix 1 0 0 \endmatrix ] has a unique solution, is

Matrices and Determinants PYQ — JEE Advanced 2009 · Shift 1 · Q39 with Solution

View written solutionFree

Correct answer: B

We need the number of $3\times 3$ symmetric matrices with entries in $\{0,1\}$ , having exactly five 1's and four 0's, such that $A\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\0\\0\end{bmatrix}$ has a unique solution.

A linear system $A\mathbf{x}=\mathbf{b}$ has a unique solution iff $A$ is invertible, i.e. iff $\det A\ne 0$ .

So we must count such symmetric matrices with exactly five 1's and nonzero determinant.

1. General form of a symmetric $3\times 3$ binary matrix

A symmetric $3\times 3$ matrix has the form

a&d&e\\ d&b&f\\ e&f&c \end{bmatrix},$$ where each of $a,b,c,d,e,f\in\{0,1\}$. Because of symmetry, the total number of 1's in the full matrix is $$a+b+c+2(d+e+f).$$ We are told this total is 5, so $$a+b+c+2(d+e+f)=5.$$ Let - $k=$ number of 1's among diagonal entries $a,b,c$, - $m=$ number of 1's among off-diagonal variables $d,e,f$. Then $$k+2m=5,$$ with $0\le k\le 3$, $0\le m\le 3$. The only possibility is $$k=3,\quad m=1.$$ Indeed: - if $m=0$, then $k=5$ impossible; - if $m=1$, then $k=3$ possible; - if $m=2$, then $k=1$ possible? $1+4=5$ yes, but then diagonal has exactly one 1; - if $m=3$, then $k=-1$ impossible. Wait carefully: $k=1,m=2$ is also possible. So there are **two cases**. --- ## 2. Case I: all three diagonal entries are 1 and exactly one off-diagonal pair is 1 Then the matrices are: $$\begin{bmatrix}1&1&0\\1&1&0\\0&0&1\end{bmatrix},\quad \begin{bmatrix}1&0&1\\0&1&0\\1&0&1\end{bmatrix},\quad \begin{bmatrix}1&0&0\\0&1&1\\0&1&1\end{bmatrix}.$$ Each of these has two identical rows/columns in the $2\times 2$ block, so determinant is 0. For example, $$\det\begin{bmatrix}1&1&0\\1&1&0\\0&0&1\end{bmatrix}=0.$$ Thus **none** of these are invertible. So Case I contributes $0$ matrices. --- ## 3. Case II: exactly one diagonal entry is 1 and exactly two off-diagonal pairs are 1 We choose: - which one diagonal entry is 1: $\binom31=3$ ways, - which two of the three off-diagonal variables $d,e,f$ are 1: $\binom32=3$ ways. Hence there are $$3\times 3=9$$ matrices in this case. We now determine which of these are invertible. --- ## 4. Write all 9 matrices and test determinant Let the diagonal 1 be at position $(1,1)$ first, so $(a,b,c)=(1,0,0)$. ### (i) $d=e=1, f=0$ $$A_1=\begin{bmatrix}1&1&1\\1&0&0\\1&0&0\end{bmatrix}.$$ Rows 2 and 3 are equal, so $\det A_1=0$. ### (ii) $d=f=1, e=0$ $$A_2=\begin{bmatrix}1&1&0\\1&0&1\\0&1&0\end{bmatrix}.$$ Compute determinant: $$\det A_2= 1\begin{vmatrix}0&1\\1&0\end{vmatrix} -1\begin{vmatrix}1&1\\0&0\end{vmatrix} +0\begin{vmatrix}1&0\\0&1\end{vmatrix} =1(0-1)-1(0-0)=-1\ne 0.$$ So $A_2$ is invertible. ### (iii) $e=f=1, d=0$ $$A_3=\begin{bmatrix}1&0&1\\0&0&1\\1&1&0\end{bmatrix}.$$ Determinant: $$\det A_3= 1\begin{vmatrix}0&1\\1&0\end{vmatrix} -0\begin{vmatrix}0&1\\1&0\end{vmatrix} +1\begin{vmatrix}0&0\\1&1\end{vmatrix} =-1+0+0=-1\ne 0.$$ So $A_3$ is invertible. Thus for diagonal pattern $(1,0,0)$, exactly **2** matrices are invertible. --- By symmetry, when the single diagonal 1 is at $(2,2)$ or $(3,3)$, the count will be the same. Hence total invertible matrices in Case II: $$3\times 2=6.$$ --- ## 5. Total count - Case I contributed $0$. - Case II contributed $6$. Therefore, the required number of matrices is $$6.$$ --- ## 6. Match with options The number is $6$, which lies in the range: - at least 4 but less than 7. So the correct option is $$\boxed{\text{B}}.$$ --- ## 7. Comparison with stored answer Stored correct answer: $\text{B}$. Our derived answer is also $\text{B}$, so they agree.

More from Matrices and Determinants

1. General form of a symmetric

3\times 3

binary matrix

A symmetric

3\times 3

matrix has the form

a&d&e\\ d&b&f\\ e&f&c \end{bmatrix},$$ where each of $a,b,c,d,e,f\in\{0,1\}$. Because of symmetry, the total number of 1's in the full matrix is $$a+b+c+2(d+e+f).$$ We are told this total is 5, so $$a+b+c+2(d+e+f)=5.$$ Let - $k=$ number of 1's among diagonal entries $a,b,c$, - $m=$ number of 1's among off-diagonal variables $d,e,f$. Then $$k+2m=5,$$ with $0\le k\le 3$, $0\le m\le 3$. The only possibility is $$k=3,\quad m=1.$$ Indeed: - if $m=0$, then $k=5$ impossible; - if $m=1$, then $k=3$ possible; - if $m=2$, then $k=1$ possible? $1+4=5$ yes, but then diagonal has exactly one 1; - if $m=3$, then $k=-1$ impossible. Wait carefully: $k=1,m=2$ is also possible. So there are **two cases**. --- ## 2. Case I: all three diagonal entries are 1 and exactly one off-diagonal pair is 1 Then the matrices are: $$\begin{bmatrix}1&1&0\\1&1&0\\0&0&1\end{bmatrix},\quad \begin{bmatrix}1&0&1\\0&1&0\\1&0&1\end{bmatrix},\quad \begin{bmatrix}1&0&0\\0&1&1\\0&1&1\end{bmatrix}.$$ Each of these has two identical rows/columns in the $2\times 2$ block, so determinant is 0. For example, $$\det\begin{bmatrix}1&1&0\\1&1&0\\0&0&1\end{bmatrix}=0.$$ Thus **none** of these are invertible. So Case I contributes $0$ matrices. --- ## 3. Case II: exactly one diagonal entry is 1 and exactly two off-diagonal pairs are 1 We choose: - which one diagonal entry is 1: $\binom31=3$ ways, - which two of the three off-diagonal variables $d,e,f$ are 1: $\binom32=3$ ways. Hence there are $$3\times 3=9$$ matrices in this case. We now determine which of these are invertible. --- ## 4. Write all 9 matrices and test determinant Let the diagonal 1 be at position $(1,1)$ first, so $(a,b,c)=(1,0,0)$. ### (i) $d=e=1, f=0$ $$A_1=\begin{bmatrix}1&1&1\\1&0&0\\1&0&0\end{bmatrix}.$$ Rows 2 and 3 are equal, so $\det A_1=0$. ### (ii) $d=f=1, e=0$ $$A_2=\begin{bmatrix}1&1&0\\1&0&1\\0&1&0\end{bmatrix}.$$ Compute determinant: $$\det A_2= 1\begin{vmatrix}0&1\\1&0\end{vmatrix} -1\begin{vmatrix}1&1\\0&0\end{vmatrix} +0\begin{vmatrix}1&0\\0&1\end{vmatrix} =1(0-1)-1(0-0)=-1\ne 0.$$ So $A_2$ is invertible. ### (iii) $e=f=1, d=0$ $$A_3=\begin{bmatrix}1&0&1\\0&0&1\\1&1&0\end{bmatrix}.$$ Determinant: $$\det A_3= 1\begin{vmatrix}0&1\\1&0\end{vmatrix} -0\begin{vmatrix}0&1\\1&0\end{vmatrix} +1\begin{vmatrix}0&0\\1&1\end{vmatrix} =-1+0+0=-1\ne 0.$$ So $A_3$ is invertible. Thus for diagonal pattern $(1,0,0)$, exactly **2** matrices are invertible. --- By symmetry, when the single diagonal 1 is at $(2,2)$ or $(3,3)$, the count will be the same. Hence total invertible matrices in Case II: $$3\times 2=6.$$ --- ## 5. Total count - Case I contributed $0$. - Case II contributed $6$. Therefore, the required number of matrices is $$6.$$ --- ## 6. Match with options The number is $6$, which lies in the range: - at least 4 but less than 7. So the correct option is $$\boxed{\text{B}}.$$ --- ## 7. Comparison with stored answer Stored correct answer: $\text{B}$. Our derived answer is also $\text{B}$, so they agree.

Matrices and Determinants question

Matrices and Determinants question

1. General form of a symmetric $3\times 3$ binary matrix

More from Matrices and Determinants

Matrices and Determinants question

Matrices and Determinants question

1. General form of a symmetric $3\times 3$ binary matrix

More from Matrices and Determinants

1. General form of a symmetric 3×33\times 33×3 binary matrix

More from Matrices and Determinants

1. General form of a symmetric 3×33\times 33×3 binary matrix

More from Matrices and Determinants

1. General form of a symmetric $3\times 3$ binary matrix

1. General form of a symmetric $3\times 3$ binary matrix