Question

The largest value of a third-order determinant whose elements are $0$ or $1$ is

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is C $2$

Let $\Delta =\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|$ be a determinant of order $3.$ Then,

$\Delta =a_1{b}_2{c}_3+a_3{b}_1{c}_2+a_2{b}_3{c}_1-a_1{b}_3{c}_2-a_2{b}_1{c}_3-a_3{b}_2{c}_1$

$=(a_1{b}_2{c}_3+a_3{b}_1{c}_2+a_2{b}_3{c}_1)-(a_1{b}_3{c}_2+a_2{b}_1{c}_3+a_3{b}_2{c}_1)$

Since each element of $\Delta$ is either $1$ or $0,$ therefore the value of the determinant cannot exceed $3.$

Clearly, the value of $\Delta$ is maximum when the value of each term in first bracket is $1$ and the value of each term in the second bracket is zero. But $a_1{b}_2{c}_3=a_3{b}_1{c}_2=1$ implies that every element of the determinant $\Delta$ is $1$ and in that case $\Delta =0.$ So, we can choose $a_1{b}_2{c}_3=0$.Thus one of the determinant with largest value whose elememts are $0$ or $1$ is

$\Delta =\left|\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right|=2$