Evaluation of Determinants MCQ Quiz - Objective Question with Answer for Evaluation of Determinants - Download Free PDF

Calculation:

Statement I

$|M^2|=|M\times M|=|M|\cdot |M|=|M{|}^2$

⇒ Statement I is correct.

Statement II

For a non-singular matrix, $M\times M^{-1}=I$, where $I$ is the identity matrix.

$|M|\cdot |M^{-1}|=|I|=1$

⇒ Statement II is incorrect unless $|M|=\pm 1$.

Statement III

The determinant of a matrix is equal to the determinant of its transpose:

$|M|=|M^T|$

⇒ Statement III is correct.

Out of the three statements, two are correct: I and III.

Calculation:

Given,

Let ω be a non-real cube root of unity, so ${\omega }^3=1$ and $1+\omega +{\omega }^2=0$.

Consider the determinant

$\mathrm{\Delta }(x)=\left|\begin{array}{ccc}x+1& \omega & {\omega }^2\\ \omega & x+{\omega }^2& 1\\ {\omega }^2& 1& x+\omega \end{array}\right|=0.$

Step 1 — Column operation:  Replace the first column by $C_1-C_2$:

$\mathrm{\Delta }(x)=\left|\begin{array}{ccc}{x}^2-1& 2k+1& 1\\ k-1& k+2& 1\\ 0& 3& 1\end{array}\right|.$

Step 2 — Expansion along the third row:

$\mathrm{\Delta }(x)=-3\left|\begin{array}{cc}{k}^2-1& 1\\ k-1& 1\end{array}\right|+\left|\begin{array}{cc}{k}^2-1& 2k+1\\ k-1& k+2\end{array}\right|,$

which simplifies to

$\mathrm{\Delta }(x)=x(x^2-1)-x(\omega +{\omega }^2)=x(x^2-1)+x=x^3.$

Step 3 — Equate to zero:

$\mathrm{\Delta }(x)=0⟹x^3=0⟹x=0.$

∴ The root of the equation is  $x=0$.

Hence, the correct answer is Option 1.

Concept:

When A² + B² + C² = 0, it implies A = B = C = 0 (since the squares of real numbers are non-negative).

Substitute the values of A, B, and C for determinant calculation into the matrix

Calculation:

$\left|\begin{array}{ccc}1& cos0& cos0\\ cos0& 1& cos0\\ cos0& cos0& 1\end{array}\right|$

Since, Cos0 =1

Thus Matrix becomes 

$\left|\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right|$

Now determinant = 1[(1×1 - 1×1)] - 1[(1×1 - 1×1)] + 1[(1×1 - 1×1)]

= = 1(0) - 1(0) + 1(0) = 0

∴ The value of the determinant is 0.

Hence, the correct answer is Option 2.

Calculation:

Determinant Δ = $a(ei-fh)-b(di-fg)+c(dh-eg)$

Now, For our matrix, 

$a=2,b=3+i,c=-1,d=3-i,e=0,f=i,g=-1,h=-i,i=1$

calculate the subdeterminants

⇒ $ei-fh=(0)(1)-(i)(-i)=0-(-1)=1$

⇒ $di-fg=(3-i)(1)-(i)(-1)=3-i+i=3$

⇒ $dh-eg=(3-i)(-i)-(0)(-1)=-3i+i2=-3i-1=-1-3i$

⇒ Δ = $2(1)-(3+i)(3)+(-1)(-1-3i)$

⇒ Δ = $2-9-3i+1+3i$

⇒ $\Delta =-6+0i$

Since we are given that $\Delta =A+iB$ comparing the real and imaginary parts, we find:

A  = -6 and B = 0

Thus A + B = -6 + 0 = - 6

Hence, the Correct answer is Option 2.

Calculation:

Given,

Δ  =  $\left|\begin{array}{ccc}k(k+2)& 2k+1& 1\\ 2k+1& k+2& 1\\ 3& 3& 1\end{array}\right|$

Simplify the determinant by the column operation  $C_1\to C_1-C_2$:

$\mathrm{\Delta }=\left|\begin{array}{ccc}{k}^2-1& 2k+1& 1\\ k-1& k+2& 1\\ 0& 3& 1\end{array}\right|$

Expanding along the third row,

$\mathrm{\Delta }=-3\left|\begin{array}{cc}{k}^2-1& 1\\ k-1& 1\end{array}\right|+\left|\begin{array}{cc}{k}^2-1& 2k+1\\ k-1& k+2\end{array}\right|=(k-1{)}^3.$

Thus  $\mathrm{\Delta }=(k-1{)}^3$.

Sign analysis

• $k>0$: if \(0, Δ < 0; if $k>1$, Δ > 0  ⇒ Statement I is false.
• $k<0$:  $\mathrm{\Delta }<0$  ⇒ Statement II is true.
• $k=0$:  $\mathrm{\Delta }=(-1{)}^3=-1\ne 0$  ⇒ Statement III is false.

∴ Only Statement II is correct  ⇒  exactly one statement is true.

Hence, the correct answer is Option 2.

Concept:

Properties of Determinant of a Matrix:

• If each entry in any row or column of a determinant is 0, then the value of the determinant is zero.
• For any square matrix say A, |A| = |AT|.
• If we interchange any two rows (columns) of a matrix, the determinant is multiplied by -1.
• If any two rows (columns) of a matrix are same then the value of the determinant is zero.

Calculation:

$\left|\begin{array}{ccc}\mathrm{x}-\mathrm{a}& \mathrm{y}-\mathrm{b}& \mathrm{z}-\mathrm{c}\\ \mathrm{a}& \mathrm{b}& \mathrm{c}\\ \mathrm{x}& \mathrm{y}& \mathrm{z}\end{array}\right|$

Apply R3 → R3 - R2

= $\left|\begin{array}{ccc}\mathrm{x}-\mathrm{a}& \mathrm{y}-\mathrm{b}& \mathrm{z}-\mathrm{c}\\ \mathrm{a}& \mathrm{b}& \mathrm{c}\\ \mathrm{x}-\mathrm{a}& \mathrm{y}-\mathrm{b}& \mathrm{z}-\mathrm{c}\end{array}\right|$

As we can see that the first and the third row of the given matrix are equal. 

We know that, if any two rows (columns) of a matrix are same then the value of the determinant is zero.

$\left|\begin{array}{ccc}\mathrm{x}-\mathrm{a}& \mathrm{y}-\mathrm{b}& \mathrm{z}-\mathrm{c}\\ \mathrm{a}& \mathrm{b}& \mathrm{c}\\ \mathrm{x}& \mathrm{y}& \mathrm{z}\end{array}\right|$ = 0

Concept:

$\mathrm{i}=\sqrt{-1}$

i² = -1 , i³ = - i, i⁴ = 1, i⁶ = - 1 , i⁸ = 1 , i⁹ = i, i ¹² = 1, and i¹⁵ = - i

Calculations: 

Given determinant is $\left|\begin{array}{ccc}\mathrm{i}& {\mathrm{i}}^2& {\mathrm{i}}^3\\ {\mathrm{i}}^4& {\mathrm{i}}^6& {\mathrm{i}}^8\\ {\mathrm{i}}^9& {\mathrm{i}}^{12}& {\mathrm{i}}^{15}\end{array}\right|$

Since, we have, 

$\mathrm{i}=\sqrt{-1}$

i2 = -1 , i3 = - i, i4 = 1, i6 = - 1 , i8 = 1 , i9 = i, i 12 = 1, and i15 = - i

=$\left|\begin{array}{ccc}\mathrm{i}& -1& -\mathrm{i}\\ 1& -1& 1\\ \mathrm{i}& 1& -\mathrm{i}\end{array}\right|$

=i(i - 1) + 1(-i - i) - i (1 + i)

= i² - i - 2i - i - i²

= - 4i

Concept:

If $\mathrm{A}=\left[\begin{array}{cc}{\mathrm{a}}_{11}& {\mathrm{a}}_{12}\\ {\mathrm{a}}_{21}& {\mathrm{a}}_{22}\end{array}\right]$ then determinant of A is given by:

|A| = a11 × a22 – a21 × a12

|An| = |A|n 

Calculation:

Given that,

$\mathrm{A}=\left[\begin{array}{cc}\mathrm{x}& 2\\ 4& \mathrm{x}\end{array}\right]$ and |A2| = 64

⇒ |A| = x2 - 8           .... (1)

Given |A2| = 64

⇒ |A|2 = 64          [∵ |An| = |A|n]

⇒ |A| = (64)1/2 = 8       ....(2)

From equation 1 and 2

⇒ x2 - 8 = 8

⇒ x2 = 16

Concept:

Property of determinants:

If A and B are two square matrices then |AB| = |A||B|

Calculation:

Given: A = $\left[\begin{array}{cc}2& 5\\ 2& 1\end{array}\right]$ and B = $\left[\begin{array}{cc}4& -3\\ 1& 5\end{array}\right]$

Now,

|A| = 2 × 1 - 5 × 2 = 2 - 10 = -8

|B| = 4 × 5 - (-3 × 1) = 20 + 3 = 23

As we know that, |AB| = |A||B|

= -8 × 23 = -184

Concept:

Elementary row or column transformations do not change the value of the determinant of a matrix.

Calculation:

$\left|\begin{array}{ccc}1& 1& 1\\ 1& 1+\mathrm{a}& 1\\ 1+\mathrm{b}& 1& 1\end{array}\right|$

Applying R₂ → R₂ – R₁, R₃ → R₃ – R₁, we get

= $\left|\begin{array}{ccc}1& 1& 1\\ 0& \mathrm{a}& 0\\ \mathrm{b}& 0& 0\end{array}\right|$

Now, Expanding along C₃

= 1 (0 – ab) – 0 + 0 = -ab

Concept:

Elementary row or column transformations do not change the value of the determinant of a matrix.

Calculation:

$\left|\begin{array}{ccc}1& 1& 1\\ 1& 1+\mathrm{x}& 1\\ 1& 1& 1+\mathrm{y}\end{array}\right|$

Applying R₂ → R₂ – R₁, R₃ → R₃ – R₁, we get

$=\left|\begin{array}{ccc}1& 1& 1\\ 0& \mathrm{x}& 0\\ 0& 0& \mathrm{y}\end{array}\right|$

Now, Expanding along C₁

= 1 (xy – 0) – 0 + 0 = xy

CONCEPT:

• If $A=\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]$ is a square matrix of order 3, then determinant of A is given by |A| = a₁₁ × {(a₂₂ × a₃₃) – (a₂₃ × a₃₂)} - a₁₂ × {(a₂₁ × a₃₃) – (a₂₃ × a₃₁)} + a₁₃ × {(a₂₁ × a₃₂) – (a₂₂ × a₃₁)}
• If A is a matrix of order n, then |k ⋅ A| = kn ⋅ |A|, where k ∈ R.

CALCULATION:

Given: $A=\left[\begin{array}{ccc}3& 4& 9\\ 11& 6& 7\\ 8& 9& 5\end{array}\right]$ and |2A| = k

⇒ |A| = 3 × (30 - 63) - 4 × (55 - 56) + 9 × (99 - 48)

⇒ |A| = - 99 + 4 + 459 = 364

As we know that, if A is a matrix of order n, then |k ⋅ A| = kn ⋅ |A|, where k ∈ R.

⇒ |2A| = 2³ ⋅ 364 = 2912

Hence, the correct option is 3.

Concept:

Properties of Determinant of a Matrix:

• If each entry in any row or column of a determinant is 0, then the value of the determinant is zero.
• For any square matrix say A, |A| = |AT|.
• If we interchange any two rows (columns) of a matrix, the determinant is multiplied by -1.
• If any two rows (columns) of a matrix are same then the value of the determinant is zero.

Calculation:

$\left|\begin{array}{ccc}2& 7& 37\\ 3& 6& 33\\ 4& 5& 29\end{array}\right|$

Apply C2 → 5C2 + C₁, we get

= $\left|\begin{array}{ccc}2& 37& 37\\ 3& 33& 33\\ 4& 29& 29\end{array}\right|$

As we can see that the second and the third column of the given matrix are equal. 

We know that, if any two rows (columns) of a matrix are same then the value of the determinant is zero.

∴$\left|\begin{array}{ccc}2& 7& 37\\ 3& 6& 33\\ 4& 5& 29\end{array}\right|$ = 0

Calculation:

Given $\left|\begin{array}{ccc}\mathrm{x}& 2& 3\\ 1& \mathrm{x}& 1\\ 3& 2& \mathrm{x}\end{array}\right|$ = 0

⇒ x(x² - 2) - 2(x - 3) + 3(2 - 3x) = 0

Now, x³ - 2x - 2x + 6 + 6 - 9x = 0

⇒ x³ - 13x + 12 = 0

∵ x = 3 is a root of the equation, ∴ (x - 3) = 0

If we divide (x3 - 13x + 12) from (x - 3) we will get (x2 + 3x - 4)

⇒ (x - 3)(x² + 3x - 4) = 0

⇒ x2 + 3x - 4 = 0

⇒ (x + 4)(x - 1) = 0

⇒ x = 1, -4

Concept:

If $A=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]$ then determinant of A is given by: |A| = (a­₁₁ × a₂₂) – (a₁₂ – a₂₁).

Calculation:

Let, $\mathrm{A}=\left|\begin{array}{cc}2+i& 2-i\\ 1+i& i-1\end{array}\right|$

⇒ |A| = (2 + i) (i – 1) – (2 – i) (1 + i)

= 2i + i² – 2 – i – (2 – i + 2i – i²)

= i – 1 – 2 – (2 + i + 1)                   (∵ i² = -1)

= i – 3 – 2 – i – 1

= -6

∴ |A| is real number.