Practice Start Test

Matrices and Determinants

JEE Mathematics · 271 questions · Page 23 of 28 · Click an option or "Show Solution" to reveal answer

Q221
The system of equations α x+y+z=α−1x+αy+z=α−1x+y+α z=α−1\begin{array}{ll}{\alpha \,x + y + z = \alpha - 1} \\ {x + \alpha y + z = \alpha - 1} \\ {x + y + \alpha \,z = \alpha - 1} \end{array} has no solutions, if α\alpha is :
A −2-2
B either −2-2 or 11
C not −2-2
D 11
Correct Answer
Option A
Solution
ax+y+z=α−1ax + y + z = \alpha - 1
x+α y+z=α−1;x + \alpha \,y + z = \alpha - 1;
x+y+zα=α−1x + y + z\alpha = \alpha - 1
Δ=∣α111α111α∣\Delta = \left| \begin{array}{lll}\alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{array} \right|
=α(α2−1)−1(α−1)+1(1−α)= \alpha \left( {{\alpha ^2} - 1} \right) - 1\left( {\alpha - 1} \right) + 1\left( {1 - \alpha } \right)
=α(α−1)(α+1)−1(α−1)−1(α−1)= \alpha \left( {\alpha - 1} \right)\left( {\alpha + 1} \right) - 1\left( {\alpha - 1} \right) - 1\left( {\alpha - 1} \right)

For infinite solutions,

Δ=0\Delta = 0
⇒(α−1)[α2+α−1−1]=0\Rightarrow \left( {\alpha - 1} \right)\left[ {{\alpha ^2} + \alpha - 1 - 1} \right] = 0
⇒(α−1)[α2+α−2]=0\Rightarrow \left( {\alpha - 1} \right)\left[ {{\alpha ^2} + \alpha - 2} \right] = 0
⇒α=−2,1;\Rightarrow \alpha = - 2,1;

But

α≠1.   \alpha \ne 1.\,\,\,

∴\therefore

  α=−2\,\,\alpha = - 2
Q222
If a>0a>0 and discriminant of  ax2+2bx+c\,a{x^2} + 2bx + c is −ve-ve, then ∣abax+bbcbx+cax+bbx+c0∣\left| \begin{array}{lll}a & b & {ax + b} \\ b & c & {bx + c} \\ {ax + b} & {bx + c} & 0 \end{array} \right| is equal to
A +ve+ve
B (ac−b2)(ax2+2bx+c)\left( {ac - {b^2}} \right)\left( {a{x^2} + 2bx + c} \right)
C −ve-ve
D 00
Correct Answer
Option C
Solution

We have

∣abax+bbcbx+cax+bbx+c0∣\left| \begin{array}{lll}a & b & {ax + b} \\ b & c & {bx + c} \\ {ax + b} & {bx + c} & 0 \end{array} \right|

By

   R3→R3−(xR1+R2);\,\,\,{R_3} \to {R_3} - \left( {x{R_1} + {R_2}} \right);
=∣abax+bbcbx+c00−(ax2+2bx+C)∣= \left| \begin{array}{lll}a & b & {ax + b} \\ b & c & {bx + c} \\ 0 & 0 & { - \left( {a{x^2} + 2bx + C} \right)} \end{array} \right|
=(ax2+2bx+c)(b2−ac)= \left( {a{x^2} + 2bx + c} \right)\left( {{b^2} - ac} \right)
=(+)(−)=−ve.= \left( + \right)\left( - \right) = - ve.
Q223
Let a,b,ca, b, c be such that b(a+c)≠0b\left( {a + c} \right) \ne 0 if ∣aa+1a−1−bb+1b−1cc−1c+1∣+∣a+1b+1c−1a−1b−1c+1(−1)n+2a(−1)n+1b(−1)nc∣=0\left| \begin{array}{lll}a & {a + 1} & {a - 1} \\ { - b} & {b + 1} & {b - 1} \\ c & {c - 1} & {c + 1} \end{array} \right| + \left| \begin{array}{lll}{a + 1} & {b + 1} & {c - 1} \\ {a - 1} & {b - 1} & {c + 1} \\ {{{\left( { - 1} \right)}^{n + 2}}a} & {{{\left( { - 1} \right)}^{n + 1}}b} & {{{\left( { - 1} \right)}^n}c} \end{array} \right| = 0 then the value of nn :
A any even integer
B any odd integer
C any integer
D zero
Correct Answer
Option B
Solution
∣aa+1a−1−bb+1b−1cc−1c+1∣+∣a+1b+1c−1a−1b−1c+1(−1)n+2a(−1)n+1b(−1)nc∣=0\left| \begin{array}{lll}a & {a + 1} & {a - 1} \\ { - b} & {b + 1} & {b - 1} \\ c & {c - 1} & {c + 1} \end{array} \right| + \left| \begin{array}{lll}{a + 1} & {b + 1} & {c - 1} \\ {a - 1} & {b - 1} & {c + 1} \\ {{{\left( { - 1} \right)}^{n + 2}}a} & {{{\left( { - 1} \right)}^{n + 1}}b} & {{{\left( { - 1} \right)}^n}c} \end{array} \right| = 0
⇒∣aa+1a−1−bb+1b−1cc−1c+1∣+∣a+1a−1(−1)n+2ab+1b−1(−1)n+1bc−1c+1(−1)nc∣=0\Rightarrow \left| \begin{array}{lll}a & {a + 1} & {a - 1} \\ { - b} & {b + 1} & {b - 1} \\ c & {c - 1} & {c + 1} \end{array} \right| + \left| \begin{array}{lll}{a + 1} & {a - 1} & {{{\left( { - 1} \right)}^{n + 2}}a} \\ {b + 1} & {b - 1} & {{{\left( { - 1} \right)}^{n + 1}}b} \\ {c - 1} & {c + 1} & {{{\left( { - 1} \right)}^n}c} \end{array} \right| = 0

(Taking transpose of second determinant)

                         C1⇔C3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_1} \Leftrightarrow {C_3}
⇒∣aa+1a−1−bb+1b−1cc−1c+1∣−∣(−1)n+2aa−1a+1(−1)n+2(−b)b−1b+1(−1)n+2cc+1c−1∣=0\Rightarrow \left| \begin{array}{lll}a & {a + 1} & {a - 1} \\ { - b} & {b + 1} & {b - 1} \\ c & {c - 1} & {c + 1} \end{array} \right| - \left| \begin{array}{lll}{{{\left( { - 1} \right)}^{n + 2}}a} & {a - 1} & {a + 1} \\ {{{\left( { - 1} \right)}^{n + 2}}\left( { - b} \right)} & {b - 1} & {b + 1} \\ {{{\left( { - 1} \right)}^{n + 2}}c} & {c + 1} & {c - 1} \end{array} \right| = 0
                         C2⇔C3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_2} \Leftrightarrow {C_3}
⇒∣aa+1a−1−bb+1b−1cc−1c+1∣+(1)n+2∣aa+1a−1−bb+1b−1cc−1c+1∣=0\Rightarrow \left| \begin{array}{lll}a & {a + 1} & {a - 1} \\ { - b} & {b + 1} & {b - 1} \\ c & {c - 1} & {c + 1} \end{array} \right| + {\left( 1 \right)^{n + 2}}\left| \begin{array}{lll}a & {a + 1} & {a - 1} \\ { - b} & {b + 1} & {b - 1} \\ c & {c - 1} & {c + 1} \end{array} \right| = 0
⇒[1+(−1)n+2]∣aa+1a−1−bb+1b−1cc−1c+1∣=0\Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left| \begin{array}{lll}a & {a + 1} & {a - 1} \\ { - b} & {b + 1} & {b - 1} \\ c & {c - 1} & {c + 1} \end{array} \right| = 0
                      C2−C1,C3−C1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_2} - {C_1},{C_3} - {C_1}
⇒[1+(−1)n+2]∣a1−1−b2b+12b−1c−11∣=0\Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left| \begin{array}{lll}a & 1 & { - 1} \\ { - b} & {2b + 1} & {2b - 1} \\ c & { - 1} & 1 \end{array} \right| = 0
R1+R3{R_1} + {R_3}
⇒[1+(−1)n+2]∣a+c00−b2b+12b−1c−11∣=0\Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left| \begin{array}{lll}{a + c} & 0 & 0 \\ { - b} & {2b + 1} & {2b - 1} \\ c & { - 1} & 1 \end{array} \right| = 0
⇒[1+(−1)n+2](a+c)(2b+1+2b−1)=0\Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left( {a + c} \right)\left( {2b + 1 + 2b - 1} \right) = 0
⇒4b(a+c)[1+(−1)n+2]=0\Rightarrow 4b\left( {a + c} \right)\left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right] = 0
⇒1+(−1)n+2=0\Rightarrow 1 + {\left( { - 1} \right)^{n + 2}} = 0
     \,\,\,\,\,

as

     b(a+c)≠0\,\,\,\,\,b\left( {a + c} \right) \ne 0
⇒n\Rightarrow n

should be an odd integer.

Q224
Let AA be a square matrix all of whose entries are integers. Then which one of the following is true?
A If det A=±1,A = \pm 1, then A−1{A^{ - 1}} exists but all its entries are not necessarily integers
B If det A≠±1,A \ne \pm 1, then A−1{A^{ - 1}} exists and all its entries are non integers
C If det A=±1,A = \pm 1, then A−1{A^{ - 1}} exists but all its entries are integers
D If det A=±1,A = \pm 1, then A−1{A^{ - 1}} need not exists
Correct Answer
Option C
Solution

As all entries of square matrix

AA

are integers, therefore all co-factors should also be integers. If det

A=±1  A = \pm 1\,\,

then

A−1  {A^{ - 1}}\,\,

exists. Also all entries of

A−1{A^{ - 1}}

are integers.

Q225
The number of values of kk, for which the system of equations : (k+1)x+8y=4kkx+(k+3)y=3k−1\begin{array}{ll}{\left( {k + 1} \right)x + 8y = 4k} \\ {kx + \left( {k + 3} \right)y = 3k - 1} \end{array} $ has no solution, is
A infinite
B 1
C 2
D 3
Correct Answer
Option B
Solution

From the given system, we have

k+1k=8k+3≠4k3k−1{{k + 1} \over k} = {8 \over {k + 3}} \ne {{4k} \over {3k - 1}}

( as System has no solution)

⇒k2+4k+3=8k\Rightarrow {k^2} + 4k + 3 = 8k
⇒k=1,3\Rightarrow k = 1,3

If

k=1k = 1

then

81+3≠4.12{8 \over {1 + 3}} \ne {{4.1} \over 2}

which is false And if

k=3k = 3

Then

86≠4.39−1{8 \over 6} \ne {{4.3} \over {9 - 1}}

which is true, therefore

k=3k=3

Hence for only one value of

k.k.

System has no solution.

Q226
If a1,a2,a3,........,an,.....{a_1},{a_2},{a_3},........,{a_n},..... are in G.P., then the determinant Δ=∣log⁡anlog⁡an+1log⁡an+2log⁡an+3log⁡an+4log⁡an+5log⁡an+6log⁡an+7log⁡an+8∣\Delta = \left| \begin{array}{lll}{\log {a_n}} & {\log {a_{n + 1}}} & {\log {a_{n + 2}}} \\ {\log {a_{n + 3}}} & {\log {a_{n + 4}}} & {\log {a_{n + 5}}} \\ {\log {a_{n + 6}}} & {\log {a_{n + 7}}} & {\log {a_{n + 8}}} \end{array} \right|$ is equal to :
A 11
B 00
C 44
D 22
Correct Answer
Option B
Solution

As

    a1,a2,a3,.........\,\,\,\,{a_1},{a_2},{a_3},.........

are in

G.P.G.P.

∴\therefore Using

an=arn−1,   {a_n} = a{r^{n - 1}},\,\,\,

we get the given determinant, as

       ∣log⁡arn−1log⁡arnlog⁡arn+1log⁡arn+2log⁡arn+3log⁡arn+4log⁡arn+5log⁡arn+6log⁡arn+7∣\,\,\,\,\,\,\,\left| \begin{array}{lll}{\log a{r^{n - 1}}} & {\log a{r^n}} & {\log a{r^{n + 1}}} \\ {\log a{r^{n + 2}}} & {\log a{r^{n + 3}}} & {\log a{r^{n + 4}}} \\ {\log a{r^{n + 5}}} & {\log a{r^{n + 6}}} & {\log a{r^{n + 7}}} \end{array} \right|

Operating

C3−C2{C_3} - {C_2}

and

C2−C1{C_2} - {C_1}

and using

log⁡m−log⁡n=log⁡mn    \log m - \log n = \log {m \over n}\,\,\,\,

we get

=∣log⁡arn−1log⁡rlog⁡rlog⁡arn+2log⁡rlog⁡rlog⁡arn+5log⁡rlog⁡r∣= \left| \begin{array}{lll}{\log a{r^{n - 1}}} & {\log r} & {\log r} \\ {\log a{r^{n + 2}}} & {\log r} & {\log r} \\ {\log a{r^{n + 5}}} & {\log r} & {\log r} \end{array} \right|
=0=0

(two columns being identical)

Q227
The number of values of kk for which the linear equations 4x+ky+2z=0,kx+4y+z=04x + ky + 2z = 0,kx + 4y + z = 0 and 2x+2y+z=02x+2y+z=0 possess a non-zero solution is :
A 22
B 11
C zero
D 33
Correct Answer
Option A
Solution
Δ=0⇒∣4k2k41221∣=0\Delta = 0 \Rightarrow \left| \begin{array}{lll}4 & k & 2 \\ k & 4 & 1 \\ 2 & 2 & 1 \end{array} \right| = 0
⇒4(4−2)−k(k−2)+\Rightarrow 4\left( {4 - 2} \right) - k\left( {k - 2} \right) +
                    2(2k−8)=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\left( {2k - 8} \right) = 0
⇒8−k2+2k+4k−16=0\Rightarrow 8 - {k^2} + 2k + 4k - 16 = 0
⇒k2−6k+8=0\Rightarrow {k^2} - 6k + 8 = 0
⇒(k−4)(k−2)=0,k=4,2\Rightarrow \left( {k - 4} \right)\left( {k - 2} \right) = 0,k = 4,2
Q228
Let a,b,ca, b, c be any real numbers. Suppose that there are real numbers x,y,zx, y, z not all zero such that x=cy+bz,x=cy+bz, y=az+cx,y=az+cx, and z=bx+ay.z=bx+ay. Then a2+b2+c2+2abc{a^2} + {b^2} + {c^2} + 2abc is equal to :
A 22
B −1-1
C 00
D 11
Correct Answer
Option D
Solution

The given equations are

−x+cy+bz=0cx−y+az=0bx+ay−z=0\begin{array}{ll}{ - x + cy + bz = 0} \\ {cx - y + az = 0} \\ {bx + ay - z = 0} \end{array}

As

x,y,zx,y,z

are not all zero ∴\therefore The above system should not have unique (zero) solution

⇒Δ=0⇒∣−1cbc−1aba−1∣=0\Rightarrow \Delta = 0 \Rightarrow \left| \begin{array}{lll}{ - 1} & c & b \\ c & { - 1} & a \\ b & a & { - 1} \end{array} \right| = 0
⇒−1(1−a2)−c(−c−ab)+b(ac+b)=0\Rightarrow - 1\left( {1 - {a^2}} \right) - c\left( { - c - ab} \right) + b\left( {ac + b} \right) = 0
⇒−1+a2+b2+c2+2abc=0\Rightarrow - 1 + {a^2} + {b^2} + {c^2} + 2abc = 0
⇒a2+b2+c2+2abc=1\Rightarrow {a^2} + {b^2} + {c^2} + 2abc = 1
Q229
Let AA be a  2×2\,2 \times 2 matrix Statement - 1 : adj(adj A)=Aadj\left( {adj\,A} \right) = A Statement - 2 :∣adj A∣=∣A∣\left| {adj\,A} \right| = \left| A \right|
A statement - 1 is true, statement - 2 is true; statement - 2 is not a correct explanation for statement - 1.
B statement - 1 is true, statement - 2 is false.
C statement - 1 is false, statement -2 is true
D statement -1 is true, statement - 2 is true; statement - 2 is a correct explanation for statement - 1.
Correct Answer
Option D
Solution

We know that

∣adj(adj  A)∣=∣Adj  A∣2−1\left| {adj\left( {adj\,\,A} \right)} \right| = {\left| {Adj\,\,A} \right|^{2 - 1}}

b

               \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
               \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
            \,\,\,\,\,\,\,\,\,\,\,\,
                    =∣A∣2−1=∣A∣\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left| A \right|^{2 - 1}} = \left| A \right|

∴\therefore

      \,\,\,\,\,\,

Both the statements are true and statement

−2-2

is a correct explanation for statement -

1.1.
Q230
Let PP and QQ be 3×33 \times 3 matrices P≠Q.P \ne Q. If P3=Q3{P^3} = {Q^3} and P2Q=Q2P{P^2}Q = {Q^2}P then determinant of (P2+Q2)\left( {{P^2} + {Q^2}} \right) is equal to :
A −2-2
B 11
C 00
D −1-1
Correct Answer
Option C
Solution

Given

P3=q3                              ...(1){P^3} = {q^3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)
P2Q=Q2p                      ...(2){P^2}Q = {Q^2}p\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)

Subtracting

(1)(1)

and

(2)(2)

, we get

P3−P2Q=Q3−Q2P{P^3} - {P^2}Q = {Q^3} - {Q^2}P
⇒P2(P−Q)+Q2(P−Q)=0\Rightarrow {P^2}\left( {P - Q} \right) + {Q^2}\left( {P - Q} \right) = 0
⇒(P2+Q2)(P−Q)=0\Rightarrow \left( {{P^2} + {Q^2}} \right)\left( {P - Q} \right) = 0
⇒∣p2+Q2∣=0\Rightarrow \left| {{p^2} + {Q^2}} \right| = 0

as

P≠QP \ne Q
Ready for a full JEE mock test? Timed · full syllabus · instant results
Take a Mock Test →